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Functions and Their Graphs: Chapter Test

20 multiple-choice questions and 10 free-response questions, drawn from across the chapter and mixed together.

Free response

10 questions in parts, 119 points in total. Work them out on paper. There are no hints here: reveal each question's answer, worked solution, and rubric when you are ready to mark that one.

Free response · work it on paper

Question 1 of 10

1. One curve, and a question that waits on a declaration . 9 points. Question 1 of 10.

The relation $x = y^2 - 4y$ is a set of points in the plane. Each part below declares which coordinate is fed in.
1. Part A.
  Declare $y$ to be the input and state the rule that produces the output. Then declare $x$ to be the input, solve for $y$ , and report every output the input $x = 5$ receives.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
2. Part B.
  Declare $y$ to be the input again and decide whether that function is one-to-one. Name two inputs that settle it, and say what the verdict predicts about the reading with $x$ as the input.
  
  Justify your claim State the claim, then give the reason it has to be true. 3 points
3. Part C.
  A classmate insists that 'is this curve a function' can be settled by looking at the curve alone, with nothing further supplied. Identify the part of that position which is sound, and the precise step at which it fails. Then say how the vertical line test and the horizontal line test are related to the two readings.
  
  Carry your own answer forward Argue from the two readings your own earlier work produced, expected or not. The credit here is for locating the verdict in the pairing of a curve with a declared input, not for reproducing one particular answer.
  
  Explain why it works A sentence or two. Reasons, not steps. 3 points
The answer

Part A

With $y$ as the input the rule is $x = y^2 - 4y$ , one output for each input. With $x$ as the input, $y = 2 \pm \sqrt{x + 4}$ , so $x = 5$ receives the two outputs $y = 5$ and $y = -1$ .

Part B

It is not one-to-one: the inputs $y = 0$ and $y = 4$ both produce $x = 0$ . Two inputs sharing an output is exactly what makes the reversed reading fail, so with $x$ as the input the relation cannot be a function.

Part C

Right that the set of points is fixed. What is not fixed is which coordinate is fed in, so the verdict belongs to the curve together with a declared input: this curve is a function of $y$ and not of $x$ . The two line tests are one test asked about the two readings, the vertical one about $x$ and the horizontal one about $y$ .

Worked solution

Part A

With $y$ declared the input, the relation already reads as a rule: square the input, subtract four times it, and hand back the single number that results. One output every time.

With $x$ declared the input, the output $y$ has to be dug out, so complete the square.

$y^2 - 4y - x = 0 \ \Longrightarrow\ (y - 2)^2 = x + 4 \ \Longrightarrow\ y = 2 \pm \sqrt{x + 4}$

At $x = 5$ this gives $y = 2 \pm 3$ , the two outputs $5$ and $-1$ . Both check against the original relation, since $5^2 - 4(5) = 5$ and $(-1)^2 - 4(-1) = 5$ . One input with two outputs is what a function forbids.

Part B

Write the $y$ reading as $g(y) = y^2 - 4y = (y - 2)^2 - 4$ , which takes the same value at inputs equally far either side of $2$ .

$g(0) = 0 = g(4)$

Two different inputs share the output $0$ , so $g$ is not one-to-one. Now swap the coordinates of every pair, which is what changing the declaration does: those two pairs become $(0, 0)$ and $(0, 4)$ , which share a first coordinate. So the swapped reading is not a function, and the failure part A found at $x = 5$ is the same failure seen at $x = 0$ .

Part C

What is right. The relation is one fixed set of points. Nothing about the picture changes when the question changes, and the classmate is right to resist the idea that a curve is somehow two objects.

Where it breaks. Being a function is not a property of the set of points; it is a property of that set together with a choice of which coordinate is the input. The choice comes from outside the set, and the two choices give opposite answers here:

$y \text{ as input: one output each} \qquad x \text{ as input: } x = 5 \text{ gives } y = 5 \text{ and } y = -1$

So the honest form of the question is always 'a function of which variable', and the vertical line test is the answer for one reading while the horizontal line test is the answer for the other.

In one line

With $y$ as the input the relation is a function, $x = y^2 - 4y$ , but not a one-to-one one, since $g(0) = g(4) = 0$ . With $x$ as the input it is not a function at all: $y = 2 \pm \sqrt{x + 4}$ , so $x = 5$ carries the two outputs $5$ and $-1$ . 'Is this a function' therefore has no answer until the input variable is declared.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Solves the relation for $y$ and keeps both signs the square root carries. . Worth 2 points.

Reports how many outputs each declaration produces at the stated input, naming which declaration each count belongs to. . Worth 1 point.

Part B 3 points

Exhibits two inputs with a shared output and argues from them to the verdict on one-to-oneness. . Worth 2 points. needs an explanation, not just an answer

Connects the shared output to the failure of the reading with $x$ as the input, through the swap of coordinates. . Worth 1 point. needs an explanation, not just an answer

Part C 3 points

Grants that the set of points is fixed, then locates the verdict in the pairing of that set with a declared input. . Worth 2 points. needs an explanation, not just an answer

Names the two line tests as the same test asked about the two different readings. . Worth 1 point.
2. The slot, the quotient, and the one input the cancelling costs . 10 points. Question 2 of 10.

Let $f(x) = \dfrac{3}{x + 1}$ on its natural domain. Every part below puts an expression rather than a number into the slot of $f$ .
1. Part A.
  Form $\dfrac{f(x + h) - f(x)}{h}$ and simplify it to a single fraction. State every restriction the simplified form has to carry.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
2. Part B.
  Evaluate your simplified expression at $x = 2$ and $h = 1$ , then compute $\dfrac{f(3) - f(2)}{1}$ directly and compare the two. Say what the number measures for this rule.
  
  Carry your own answer forward Use whichever simplified expression you produced in part A, even if it was not the expected one, and compare it honestly against the direct computation. The credit here is for carrying out both routes and reading the result, not for landing on one particular number.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  A classmate says the simplified fraction IS the difference quotient, on the grounds that the two agree everywhere the fraction is defined. Decide whether the two are the same expression, say what the classmate's position gets right, and justify your verdict from the equality test; if they are not the same, name an input that separates them.
  
  Carry your own answer forward Run the argument on whichever simplified form you produced in part A, and say honestly what your own cancelling step did to the inputs the expression accepts. The credit is for the account of what equality of expressions demands, not for a particular fraction.
  
  Explain why it works A sentence or two. Reasons, not steps. 3 points
The answer
Part A

Simplifying gives the bare formula $\dfrac{-3}{(x + 1)(x + h + 1)}$ ; what represents the difference quotient is that formula restricted by $h \ne 0$ , carrying $x \ne -1$ and $x + h \ne -1$ as well.
- $-\dfrac{3}{(x + 1)(x + h + 1)}$ is the same fraction with the sign written outside; what is not the same is an answer with no restrictions attached, since the restrictions are part of what the expression means
Part B

Both routes give $-\tfrac{1}{4}$ . It is the slope of the line through the two points of the graph at inputs $2$ and $3$ , so the output falls by a quarter of a unit per unit of input across that step.

Part C

Right that the two agree wherever both are defined. But the bare formula, on its own natural domain, accepts $h = 0$ , where the quotient reads $\tfrac{0}{0}$ : equality needs the same domain, not matching values alone. What the quotient equals is that formula restricted by $h \ne 0$ .
Worked solution

Part A

Put the whole expression $x + h$ into the slot, then combine the two fractions over a common denominator rather than expanding anything.

$\frac{3}{x + h + 1} - \frac{3}{x + 1} = \frac{3(x + 1) - 3(x + h + 1)}{(x + 1)(x + h + 1)} = \frac{-3h}{(x + 1)(x + h + 1)}$

Every surviving term carries a factor of $h$ , which is the check that the subtraction was done correctly. Dividing by $h$ cancels it:

$\frac{f(x + h) - f(x)}{h} = \frac{-3}{(x + 1)(x + h + 1)}, \qquad h \ne 0.$

The restrictions $x \ne -1$ and $x + h \ne -1$ come from $f$ having to be defined at both inputs; $h \ne 0$ comes from the division that produced the bare formula, and it is what separates that formula from the quotient it is standing in for.

Part B

From the simplified form, with $x = 2$ and $h = 1$ :

$\frac{-3}{(2 + 1)(2 + 1 + 1)} = \frac{-3}{12} = -\frac{1}{4}$

Directly, $f(3) = \tfrac{3}{4}$ and $f(2) = 1$ , so the quotient is $\tfrac{3}{4} - 1 = -\tfrac{1}{4}$ . The two agree, as they must for $h \ne 0$ . The quantity is the secant slope from input $2$ to input $3$ , the average rate of change of $f$ across that step, and it is negative because this rule falls as the input grows.

Part C

What is right. For every $h \ne 0$ the two expressions produce the same number, and part B checked one such case.

Where it breaks. The equality test demands the same domain, the same codomain and the same values, not values alone. At $h = 0$ the original quotient has a zero denominator and no value, while the bare formula, read on its own natural domain, is defined:

$\frac{f(x + h) - f(x)}{h}\bigg|_{h = 0} = \frac{0}{0}, \qquad \frac{-3}{(x + 1)(x + h + 1)}\bigg|_{h = 0} = \frac{-3}{(x + 1)^2}$

So $h = 0$ is the input that separates them: it lies in one domain and not the other. The cancelling step is a division by $h$ , and dividing is exactly what removed that input from the record. Attach the restriction $h \ne 0$ to the formula and the two objects do coincide; leave it off and they do not.

In one line

$\dfrac{f(x + h) - f(x)}{h} = \dfrac{-3}{(x + 1)(x + h + 1)}$ for $h \ne 0$ , with $x \ne -1$ and $x + h \ne -1$ . At $x = 2$ , $h = 1$ both routes give $-\tfrac{1}{4}$ , the average rate of change from input $2$ to input $3$ . Taken on its own natural domain, the bare formula is a different expression from the quotient, because it accepts $h = 0$ and the quotient does not; restricted by $h \ne 0$ , it is the quotient.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Substitutes the whole expression $x + h$ into the slot and combines the two fractions over a common denominator. . Worth 2 points.

Cancels the factor of $h$ and records all three restrictions the simplified fraction carries. . Worth 2 points.

Part B 3 points

Evaluates the simplified expression and the direct difference quotient, and sets the two results side by side. . Worth 2 points.

States what the number measures for this rule, as a change in output per unit of input across the step. . Worth 1 point.

Part C 3 points

Grants the agreement on the overlap and then appeals to the domain, not the values, as what settles equality. . Worth 2 points. needs an explanation, not just an answer

Names $h = 0$ as the disagreeing input and traces it to the division that produced the simplified form. . Worth 1 point.
3. Everything the factored form already knows . 11 points. Question 3 of 10.

Here $f(x) = (2 - x)(x + 6)^2$ , written as a product from the start. Nothing below calls for a plotted point: each answer is to be read out of the factors, or out of algebra performed on them.
1. Part A.
  List every zero of $f$ , then give the sign of $f$ on each interval the zeros cut.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
2. Part B.
  Expand $f$ , factor out the highest power of $x$ , and use that form to settle both ends. State a threshold beyond which your argument holds.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
3. Part C.
  Decide, at each of the two zeros, whether the curve passes through the axis there or stays on the side it arrived from, and make the power on each factor carry the argument.
  
  Carry your own answer forward Argue from the factored form and from whichever sign chart you produced in part A, even if it was not the expected one. The credit here is for tying each verdict to the power on its factor rather than to a drawing.
  
  Justify your claim State the claim, then give the reason it has to be true. 4 points
The answer

Part A

Zeros at $x = -6$ and $x = 2$ . $f > 0$ on $x < -6$ and on $-6 < x < 2$ , and $f < 0$ on $x > 2$ .

Part B

$f(x) = x^3\left(-1 - \tfrac{10}{x} - \tfrac{12}{x^2} + \tfrac{72}{x^3}\right)$ . For $\lvert x \rvert \ge 24$ the bracket is negative, so $f$ rises on the far left and falls on the far right.

Part C

At $x = -6$ the factor is squared, so it is never negative and the sign of $f$ is the same on both sides: the graph meets the axis and turns back. At $x = 2$ the factor $(2 - x)$ carries the power $1$ , so the sign flips there and the graph crosses.

Worked solution

Part A

A product is zero exactly when one of its factors is, so the zeros are $x = 2$ from $(2 - x)$ and $x = -6$ from $(x + 6)^2$ . Those two cut the line into three intervals, and inside each one no factor changes sign.

$x < -6:\ (+)(+) = +, \qquad -6 < x < 2:\ (+)(+) = +, \qquad x > 2:\ (-)(+) = -$

The squared factor is positive away from $-6$ and contributes a $+$ on both sides of it, so the only sign change comes from $(2 - x)$ , which turns negative once $x$ passes $2$ .

Part B

Expand first, then pull out the highest power.

$f(x) = (2 - x)(x^2 + 12x + 36) = -x^3 - 10x^2 - 12x + 72 = x^3\left(-1 - \frac{10}{x} - \frac{12}{x^2} + \frac{72}{x^3}\right)$

Now bound the correction terms. Once $\lvert x \rvert \ge 24$ we have $\left\lvert \tfrac{10}{x} \right\rvert \le \tfrac{5}{12}$ , $\left\lvert \tfrac{12}{x^2} \right\rvert \le \tfrac{1}{48}$ and $\left\lvert \tfrac{72}{x^3} \right\rvert \le \tfrac{1}{192}$ , which together stay below $\tfrac{1}{2}$ . So the bracket sits between $-\tfrac{3}{2}$ and $-\tfrac{1}{2}$ and is certainly negative, and $f$ has the opposite sign to $x^3$ out there: large and positive on the far left, large and negative on the far right. No limits are needed, only the stated threshold.

Part C

The verdict is decided by the multiplicity of each root, not by the fact that it is a root.

At $x = -6$ . Group the factor as $(x + 6)^2$ , a square, so it is positive for every $x \ne -6$ and contributes the same $+$ on both sides. Part A found $f > 0$ on both $x < -6$ and $-6 < x < 2$ , which is that fact in the sign chart: no sign change, so the graph touches the axis at $(-6, 0)$ and turns back.

At $x = 2$ . The factor $(2 - x)$ has power $1$ , an odd power, so it carries its own sign:

$2 - x > 0 \text{ for } x < 2, \qquad 2 - x < 0 \text{ for } x > 2$

The sign of $f$ therefore flips across $2$ , and the graph crosses the axis there. Neither verdict looked at a curve; both came from the powers.

In one line

The zeros are $x = -6$ and $x = 2$ , with $f > 0$ on $x < -6$ and on $-6 < x < 2$ and $f < 0$ on $x > 2$ . Expanded, $f(x) = -x^3 - 10x^2 - 12x + 72 = x^3\left(-1 - \tfrac{10}{x} - \tfrac{12}{x^2} + \tfrac{72}{x^3}\right)$ , whose bracket is negative once $\lvert x \rvert \ge 24$ , so the graph rises on the far left and falls on the far right. The squared factor makes the graph touch and turn back at $-6$ ; the power $1$ makes it cross at $2$ .

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Reads both zeros off the factors and uses them to cut the line into the intervals to be tested. . Worth 1 point.

Gives each factor a fixed sign on each interval and multiplies those signs to reach the sign of $f$ there. . Worth 2 points.

Part B 4 points

Expands the product and factors out the highest power of $x$ to isolate a bracket of correction terms. . Worth 2 points.

Bounds the correction terms beyond a stated threshold and reads the behaviour of both ends from the sign of the leading power. . Worth 2 points.

Part C 4 points

Argues that an even power holds one sign on both sides of its root, so the graph meets the axis there and turns back. . Worth 2 points. needs an explanation, not just an answer

Argues that an odd power carries the sign of its single factor, so the sign flips and the graph crosses. . Worth 2 points. needs an explanation, not just an answer
4. Two excluded inputs, only one of which leaves a gap . 11 points. Question 4 of 10.

Let $f(x) = \dfrac{x^3 + 2x^2}{x^2 + 5x + 6}$ on its natural domain, declared into the real numbers.
1. Part A.
  Read the rejected inputs off the denominator before simplifying anything, then give the exact coordinates of the one point the graph is missing.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
2. Part B.
  Give the simplified rule that agrees with $f$ wherever $f$ is defined, and say what the graph does at each excluded input your factorisation rejected.
  
  Carry your own answer forward Continue from the factorisation you produced in part A, whatever it was. The credit here is for cancelling only what the two parts genuinely share and for separating a cancelled exclusion from one that survives.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
3. Part C.
  Set $f$ beside $g(x) = \dfrac{x^2}{x + 3}$ , with $g$ carrying the domain its own formula implies and declared, like $f$ , into the real numbers. Say whether one rule or two rules are in front of you, name which of the three equality conditions is responsible and exhibit an input that witnesses it together with the value one rule gives there, and give the domain a declaration would have to hand $g$ to close the difference.
  
  Carry your own answer forward Use the simplified rule and the excluded inputs you produced in parts A and B, even if they were not the expected ones, and compare honestly from them. The credit is for running all three conditions of function equality, not for a particular pair of excluded values.
  
  Compare the two methods Say what each one costs you, and when you would reach for it. 4 points
The answer

Part A

The excluded inputs are $x = -2$ and $x = -3$ . The missing point is $(-2, 4)$ .

Part B

$\dfrac{x^2}{x + 3}$ , on the inputs $f$ allows. At $x = -3$ the offending factor does not cancel, so the graph runs off along the vertical line $x = -3$ instead of losing a single point.

Part C

They are not the same function. Their values agree wherever both are defined, but $-2$ lies in $g$ 's natural domain and not in $f$ 's, so the equal-domains condition fails. Declaring $g$ on the inputs $x \ne -3$ and $x \ne -2$ repairs it.

Worked solution

Part A

Factor both parts. The denominator decides the domain.

$f(x) = \frac{x^2(x + 2)}{(x + 2)(x + 3)}$

The denominator vanishes at $x = -2$ and at $x = -3$ , so both are excluded. For every other input the shared factor cancels, leaving $\tfrac{x^2}{x + 3}$ . The excluded input whose factor cancelled is $-2$ , and the height the graph would have had there is found by evaluating the simplified rule at it:

$\frac{(-2)^2}{-2 + 3} = \frac{4}{1} = 4$

So the graph is missing exactly the point $(-2, 4)$ .

Part B

Cancelling the shared factor gives the agreeing rule, valid at every input $f$ accepts.

$f(x) = \frac{x^2(x + 2)}{(x + 2)(x + 3)} = \frac{x^2}{x + 3}, \qquad x \ne -2$

At $x = -3$ the factor $(x + 3)$ is still in the denominator after the cancelling, so the rule is not merely missing a value there: the outputs grow without bound as inputs approach $-3$ , and the graph shoots off along a vertical line. Two rules can exclude an input and look completely different near it, and which one happens depends on whether the offending factor cancels.

Part C

Two functions are equal exactly when they share a domain, a codomain and every value, so run all three.

Values. For every input both accept, the cancelling of part B shows the outputs match, so this condition holds.

Codomain. Both are declared into the real numbers, so this holds too.

Domain. The natural domain of $g$ excludes only $-3$ , while the natural domain of $f$ excludes $-3$ and $-2$ :

$g(-2) = \frac{4}{1} = 4, \qquad f(-2) \ \text{is undefined}$

One condition fails, so they are different functions. Since a domain is data rather than something read off a formula, the repair is a declaration, not an edit to the rule: declare $g$ on the inputs $x \ne -3$ and $x \ne -2$ , and the three conditions all hold. The hole at $(-2, 4)$ is where the difference is visible on the page.

In one line

The excluded inputs are $x = -2$ and $x = -3$ , and the graph is missing the single point $(-2, 4)$ . The agreeing rule is $\tfrac{x^2}{x + 3}$ ; at $x = -3$ the factor survives the cancelling, so the graph runs off along a vertical line there instead. Taken on its own natural domain, $g(x) = \tfrac{x^2}{x + 3}$ is a different function from $f$ , because $-2$ lies in its domain and not in $f$ 's; declaring $g$ on $x \ne -3$ and $x \ne -2$ makes the two equal.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Factors the denominator and names both inputs it rejects, before any cancelling is done. . Worth 2 points.

Identifies which excluded input leaves a single missing point and reads its height off the simplified rule. . Worth 2 points.

Part B 3 points

Cancels only the factor the numerator and denominator share, and keeps the excluded input attached to the simplified rule. . Worth 2 points.

Says what happens at the excluded input whose factor survives, distinguishing it from a single missing point. . Worth 1 point.

Part C 4 points

Tests all three conditions of function equality and identifies the one that fails. . Worth 2 points. needs an explanation, not just an answer

Exhibits an input that lies in one domain and not the other, with the value one rule gives there. . Worth 1 point.

States a declared domain for $g$ under which all three conditions hold, leaving both rules untouched. . Worth 1 point.
5. Four numbers, and the coordinate each one is allowed to touch . 12 points. Question 5 of 10.

A function $f$ has domain $-4 \le x \le 14$ and range $-2 \le y \le 6$ . Its graph is transformed by the rule $y = -\tfrac{1}{2}\,f(2x + 8) + 3$ .
1. Part A.
  Rewrite the inside in the form $b(x - h)$ , then name each of the four moves the rule calls for, with its exact amount.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
2. Part B.
  Give the domain and the range of the transformed graph, and say which endpoint of the old range became which endpoint of the new one.
  
  Carry your own answer forward Use the four values $a$ , $b$ , $h$ and $k$ you produced in part A, whatever they were, and transform the endpoints honestly from them. The credit here is for sending the domain through the inside pair and the range through the outside pair, and for reporting an interval from its smaller value up.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  Explain why the two inside numbers reached only the first coordinate of a point and the two outside numbers only the second, and why the shift could not be read off the $+8$ until the inside had been factored.
  
  Carry your own answer forward Argue from the factored inside you produced in part A and from the endpoints you moved in part B, whatever they were. The credit here is for the account of why each side reaches one coordinate only, not for a particular pair of intervals.
  
  Explain why it works A sentence or two. Reasons, not steps. 4 points
The answer

Part A

$2x + 8 = 2(x + 4)$ , so $b = 2$ , $h = -4$ , $a = -\tfrac{1}{2}$ and $k = 3$ : compress horizontally by $2$ , shift left $4$ , halve each height and reflect it, then shift up $3$ .

Part B

Domain $-6 \le x \le 3$ and range $0 \le y \le 4$ . The old top $6$ became the new bottom $0$ , and the old bottom $-2$ became the new top $4$ .

Part C

The inside acts on the number handed to $f$ , so finding where a point lands means solving the inside for $x$ , which touches the input alone; the outside acts on the height $f$ has already returned. Solving is also why the factor comes out first: it divides the shift, making the move left $4$ rather than left $8$ .

Worked solution

Part A

Pull the input multiplier out of the inside before reading anything, and read the outside straight off.

$2x + 8 = 2(x + 4) \quad\Longrightarrow\quad b = 2,\ h = -4, \qquad a = -\tfrac{1}{2},\ k = 3$

On the inside, $b = 2$ divides every input by $2$ , a horizontal compression by $2$ , and $h = -4$ shifts by $-4$ , that is left $4$ . On the outside, $a = -\tfrac{1}{2}$ halves every height and flips its sign, and $k = 3$ then raises the result by $3$ . Read the $+8$ without factoring and the shift comes out as $8$ , which is four times too far.

Part B

The two halves never mix: the inside pair moves the domain and the outside pair moves the range. A point $(u, v)$ lands at $\left(\tfrac{u}{b} + h,\ av + k\right)$ , so send each endpoint through the matching half.

$-4 \mapsto \tfrac{-4}{2} - 4 = -6, \quad 14 \mapsto \tfrac{14}{2} - 4 = 3, \qquad -2 \mapsto -\tfrac{1}{2}(-2) + 3 = 4, \quad 6 \mapsto -\tfrac{1}{2}(6) + 3 = 0$

The domain endpoints keep their order, since $b = 2$ is positive. The range endpoints swap, since $a = -\tfrac{1}{2}$ is negative, so the interval has to be reported from the smaller value up: $0 \le y \le 4$ .

Part C

Why the halves do not mix. A point $(u, v)$ on $f$ reappears at the input $x$ where the number handed to $f$ equals $u$ , and at the height obtained by putting $v$ through the outside arithmetic. The first is an equation in $x$ alone and the second a computation on $v$ alone:

$b(x - h) = u \ \Longrightarrow\ x = \frac{u}{b} + h, \qquad y = av + k$

Neither expression contains the other's variable, so an inside change can never move a height and an outside change can never move an input.

Why the factoring comes first. Solving undoes the inside operations in reverse order, dividing by $b$ before adding $h$ . That is exactly what the factored form displays. Left unfactored, the $+8$ looks like the whole shift, but the division by $b = 2$ reaches it too:

$2x + 8 = u \ \Longrightarrow\ x = \frac{u}{2} - 4$

The shift is $-4$ , a move left $4$ , not left $8$ .

In one line

The inside factors as $2(x + 4)$ , so the moves are a horizontal compression by $2$ , a shift left $4$ , a halving of every height with a reflection, and a shift up $3$ . The transformed graph has domain $-6 \le x \le 3$ and range $0 \le y \le 4$ , with the old top $6$ arriving at the new bottom $0$ . The inside reaches only the input because finding a landing input means solving the inside for $x$ , and that same solving divides the shift, making it left $4$ rather than left $8$ .

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Factors the inside into the form $b(x - h)$ before naming any shift. . Worth 2 points.

Names all four moves with their exact amounts and directions, keeping the inside pair separate from the outside pair. . Worth 2 points.

Part B 4 points

Sends the domain endpoints through the inside pair only and the range endpoints through the outside pair only. . Worth 2 points.

Reports each interval from its smaller value up and names which old endpoint became which new one. . Worth 2 points.

Part C 4 points

Argues that finding a new input means solving the inside for $x$ , so the inside numbers reach the first coordinate only. . Worth 2 points. needs an explanation, not just an answer

Argues that the outside acts on a height already produced, so it reaches the second coordinate only. . Worth 1 point. needs an explanation, not just an answer

Ties the factoring step to the fact that the input multiplier divides the shift, and states the amount the unfactored reading would give. . Worth 1 point.
6. What survives two stages, and what a superscript is counting . 12 points. Question 6 of 10.

Let $f(x) = \sqrt{2x + 1}$ and $g(x) = \dfrac{1}{x - 5}$ , each on its natural domain.
1. Part A.
  Give a simplified formula for $(g \circ f)(x)$ together with its domain, naming which requirement removes each excluded input.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
2. Part B.
  Let $p(x) = 5 - 3x$ . Say what the superscript in $p^2$ instructs you to do, compute $p^2$ and $\big(p(x)\big)^2$ separately, and decide whether the two are the same function; if they are not, give one input at which they disagree.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
3. Part C.
  Decide whether $f \circ g$ can be formed for this pair at all. If it can, give one real number in its domain and one that is not, justifying each from what the two rules accept.
  
  Justify your claim State the claim, then give the reason it has to be true. 4 points
The answer

Part A

$(g \circ f)(x) = \dfrac{1}{\sqrt{2x + 1} - 5}$ on $x \ge -\tfrac{1}{2}$ with $x \ne 12$ : the radical forces the first condition, and $\sqrt{2x + 1} \ne 5$ forces the second.

Part B

The superscript counts compositions, so $p^2 = p \circ p$ : $p^2(x) = 9x - 10$ , while $\big(p(x)\big)^2 = 9x^2 - 30x + 25$ . They are not the same function: at $x = 0$ they give $-10$ and $25$ .

Part C

It can. Its domain is every $x \le 3$ together with every $x > 5$ . The input $0$ belongs, since $g(0) = -\tfrac{1}{5}$ leaves the radicand $\tfrac{3}{5}$ ; the input $4$ does not, since $g(4) = -1$ leaves the radicand $-1$ .

Worked solution

Part A

In $g \circ f$ the right function acts first, so substitute the rule of $f$ into the input slot of $g$ .

$(g \circ f)(x) = g\big(\sqrt{2x + 1}\big) = \frac{1}{\sqrt{2x + 1} - 5}$

Two conditions decide the domain. First $f$ must accept $x$ , so $2x + 1 \ge 0$ and $x \ge -\tfrac{1}{2}$ . Second $g$ must accept the output $\sqrt{2x + 1}$ , and $g$ rejects only the input $5$ :

$\sqrt{2x + 1} = 5 \ \Longleftrightarrow\ 2x + 1 = 25 \ \Longleftrightarrow\ x = 12$

So the domain is $x \ge -\tfrac{1}{2}$ with $12$ removed. The second condition constrains an output of $f$ , so finding the excluded input needs the values $f$ produces.

Part B

The superscript on a function name counts compositions, so $p^2$ means apply $p$ and then apply $p$ again to the result. Wrap the whole first output in the rule.

$p^2(x) = p(5 - 3x) = 5 - 3(5 - 3x) = 9x - 10$

Squaring the output is a different operation entirely:

$\big(p(x)\big)^2 = (5 - 3x)^2 = 9x^2 - 30x + 25$

One is linear and the other is not, so they cannot be the same function. At $x = 0$ they give $-10$ and $25$ . Iterating a rule and squaring its value are unrelated operations that happen to share a symbol.

Part C

Substituting gives the formula, and then the two conditions decide which inputs survive.

$(f \circ g)(x) = f\!\left(\frac{1}{x - 5}\right) = \sqrt{\frac{2}{x - 5} + 1}$

First $g$ must accept $x$ , so $x \ne 5$ . Second $f$ must accept the output of $g$ , so the radicand must be nonnegative:

$\frac{2}{x - 5} + 1 = \frac{x - 3}{x - 5} \ge 0 \quad\Longleftrightarrow\quad x \le 3 \ \text{ or } \ x > 5$

The quotient is nonnegative when both parts share a sign, which happens below $3$ and above $5$ ; between them the numerator and denominator disagree. So $0$ is in the domain, with radicand $\tfrac{3}{5}$ , and $4$ is not, with radicand $-1$ , even though $g$ accepts $4$ perfectly well. The exclusion is decided by an output of $g$ , not by the domains of the two rules.

In one line

$(g \circ f)(x) = \dfrac{1}{\sqrt{2x + 1} - 5}$ on $x \ge -\tfrac{1}{2}$ with $x \ne 12$ . For $p(x) = 5 - 3x$ , $p^2(x) = 9x - 10$ while $\big(p(x)\big)^2 = 9x^2 - 30x + 25$ , and at $x = 0$ they give $-10$ and $25$ . The other composite $f \circ g$ can be formed, with domain every $x \le 3$ together with every $x > 5$ : the input $0$ belongs and the input $4$ does not.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Substitutes the rule of the inner function into the input slot of the outer one and simplifies. . Worth 1 point.

Imposes both conditions, the inner function's own requirement and the requirement on its output, and names which excluded input each produces. . Worth 3 points.

Part B 4 points

Computes the composition by substituting the whole first output back into the rule. . Worth 2 points.

Computes the square of the output separately and gives an input at which the two results differ. . Worth 2 points.

Part C 4 points

Builds the composite and states both conditions, the one on the inner input and the one on the inner output. . Worth 2 points.

Produces one input inside the domain and one outside it, with the value of the inner rule at each as the reason. . Worth 2 points. needs an explanation, not just an answer
7. Undoing a rule, and the checks that decide it . 14 points. Question 7 of 10.

Let $f(x) = 12 - (x - 5)^2$ , declared on all real numbers. Each part below fixes a set of allowed inputs and asks what can be said about undoing the rule on it.
1. Part A.
  Keep only the inputs $x \ge 5$ . Give the rule that undoes what is left, and state which numbers that rule is entitled to accept.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
2. Part B.
  Verify the pair from both sides: show that $f^{-1}\big(f(x)\big) = x$ for every $x \ge 5$ and that $f\big(f^{-1}(x)\big) = x$ for every $x \le 12$ , naming where each restriction is used.
  
  Carry your own answer forward Run both checks on whichever candidate you produced in part A, even if it was not the expected one, and report honestly what each composition returns. The credit here is for testing both orders and for naming where a restriction is used, not for reaching a particular pair of identities.
  
  Complete the derivation Each line should follow from the one above it. Say what lets you take each step. 5 points
3. Part C.
  A classmate now takes $f$ on ALL real numbers, checks that $f\big(g(x)\big) = x$ for $g(x) = 5 - \sqrt{12 - x}$ , and declares $g$ to be the inverse of $f$ . Decide whether $g$ is the inverse of $f$ on all real numbers, and justify the verdict from the compositions themselves, giving a value at which one of them fails if there is such a value.
  
  Justify your claim State the claim, then give the reason it has to be true. 5 points
The answer
Part A

$f^{-1}(x) = 5 + \sqrt{12 - x}$ , with domain $x \le 12$ .
- $5 + \sqrt{12 - x}$ may be written $\sqrt{12 - x} + 5$ ; what is not the same is $5 - \sqrt{12 - x}$ , which returns the inputs of the other restriction
Part B

$f^{-1}(f(x)) = 5 + \sqrt{(x - 5)^2} = 5 + \lvert x - 5 \rvert = x$ , where $x \ge 5$ is what removes the bars. $f(f^{-1}(x)) = 12 - \big(\sqrt{12 - x}\big)^2 = x$ , where $x \le 12$ is what lets the root exist.

Part C

The conclusion is wrong, because one composition is not enough. In the other order $g(f(x)) = 5 - \lvert x - 5 \rvert$ , which returns $x$ only for $x \le 5$ . At $x = 9$ it gives $1$ , not $9$ , so $g$ undoes $f$ from one side only.
Worked solution

Part A

On $x \ge 5$ the rule is one-to-one, because $(x - 5)^2$ is increasing there and the outside arithmetic does not repeat a value. Write $y = 12 - (x - 5)^2$ and solve for the input.

$(x - 5)^2 = 12 - y \ \Longrightarrow\ x - 5 = \sqrt{12 - y} \ \Longrightarrow\ x = 5 + \sqrt{12 - y}$

The positive root is the right one because the restriction makes $x - 5 \ge 0$ . Renaming the input gives $f^{-1}(x) = 5 + \sqrt{12 - x}$ . Its domain is the set of values $f$ actually produces, and since $(x - 5)^2 \ge 0$ those are exactly the numbers $y \le 12$ .

Part B

One direction. Feed $x$ through $f$ and then through the candidate:

$f^{-1}\big(f(x)\big) = 5 + \sqrt{12 - \big(12 - (x - 5)^2\big)} = 5 + \sqrt{(x - 5)^2} = 5 + \lvert x - 5 \rvert$

The square root of a square is the absolute value, not the bare expression. The restriction $x \ge 5$ makes $x - 5$ nonnegative, so $\lvert x - 5 \rvert = x - 5$ and the whole thing collapses to $x$ . Without the restriction this step is where the argument would fail.

The other direction. Feed $x$ through the candidate and then through $f$ :

$f\big(f^{-1}(x)\big) = 12 - \big(5 + \sqrt{12 - x} - 5\big)^2 = 12 - (12 - x) = x$

Here the restriction $x \le 12$ is what makes $\sqrt{12 - x}$ exist at all, and squaring it returns $12 - x$ exactly. Both compositions return their own input on their own domain, so the pair is genuinely inverse.

Part C

The check the classmate ran is genuine: for $x \le 12$ ,

$f\big(g(x)\big) = 12 - \big(5 - \sqrt{12 - x} - 5\big)^2 = 12 - (12 - x) = x.$

But an inverse has to undo $f$ from both directions, so run the other order on the unrestricted $f$ :

$g\big(f(x)\big) = 5 - \sqrt{(x - 5)^2} = 5 - \lvert x - 5 \rvert$

That equals $x$ when $x \le 5$ and equals $10 - x$ when $x > 5$ . The input $x = 9$ settles it: $f(9) = 12 - 16 = -4$ , and $g(-4) = 5 - \sqrt{16} = 1$ , which is not $9$ . Squaring collapsed the two inputs $1$ and $9$ onto the single output $-4$ , and no rule can send that one output back to two places. A one-sided inverse is not an inverse.

In one line

On $x \ge 5$ the inverse is $f^{-1}(x) = 5 + \sqrt{12 - x}$ , with domain $x \le 12$ , and both compositions return their own input once the restrictions remove the absolute value and supply the root. On all real numbers, $g(x) = 5 - \sqrt{12 - x}$ satisfies $f(g(x)) = x$ but gives $g(f(9)) = 1 \ne 9$ , so it undoes $f$ from one side only and is not an inverse.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Solves for the input and chooses the sign of the root that the restriction forces. . Worth 2 points.

Gives the inverse's domain as the set of values the restricted rule produces, not as the original domain. . Worth 2 points.

Part B 5 points

Carries out both compositions rather than one, each on its own domain. . Worth 2 points.

Treats the square root of a square as an absolute value and says what the restriction does to it. . Worth 2 points. needs an explanation, not just an answer

Names the restriction each direction depends on, and what would fail without it. . Worth 1 point. needs an explanation, not just an answer

Part C 5 points

Computes the composition the classmate did not run, on the unrestricted domain. . Worth 2 points.

Names one input at which that composition fails to return its own value, with the two intermediate values. . Worth 1 point. needs an explanation, not just an answer

Argues that a two-sided requirement is what the failure is about, rather than an arithmetic slip in the classmate's check. . Worth 2 points. needs an explanation, not just an answer
8. A quintic taken apart factor by factor . 12 points. Question 8 of 10.

Let $f(x) = x^5 - 10x^3 + 9x$ , on all real numbers.
1. Part A.
  Factor $f$ completely and list every zero.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
2. Part B.
  Classify $f$ as even, odd or neither, justifying the verdict from $f(-x)$ . Then find the sign of $f$ on $0 < x < 1$ by multiplying factor signs, and use your classification to name the sign on $-1 < x < 0$ without multiplying anything there.
  
  Carry your own answer forward Use the factored form you produced in part A, whatever it was, for the sign count. The credit here is for testing the parity by substituting $-x$ and for transferring a sign across the origin by oddness rather than by recomputing it.
  
  Justify your claim State the claim, then give the reason it has to be true. 5 points
3. Part C.
  A classmate claims that a polynomial built only from odd powers must cross the axis at every one of its zeros. Prove or disprove the claim; if it is false, give a counterexample.
  
  Construct a counterexample Give one specific case, and show it breaks the claim. 4 points
The answer

Part A

$f(x) = x(x - 1)(x + 1)(x - 3)(x + 3)$ , with zeros $x = -3$ , $-1$ , $0$ , $1$ and $3$ .

Part B

$f(-x) = -x^5 + 10x^3 - 9x = -f(x)$ , and the domain is all real numbers, so $f$ is odd. On $0 < x < 1$ the five factor signs multiply to a positive value, and oddness negates every output under $x \mapsto -x$ , so $f < 0$ on $-1 < x < 0$ .

Part C

The claim is false. Take $x^5 - 2x^3 + x = x(x - 1)^2(x + 1)^2$ : every power is odd, yet the factors at $1$ and $-1$ are squared, so the sign does not change across either and the graph touches and turns back there. It crosses only at $0$ .

Worked solution

Part A

Take out the common factor first, then treat what is left as a quadratic in $x^2$ .

$f(x) = x\big(x^4 - 10x^2 + 9\big) = x\big(x^2 - 1\big)\big(x^2 - 9\big) = x(x - 1)(x + 1)(x - 3)(x + 3)$

A product is zero exactly when a factor is, so the zeros are $-3$ , $-1$ , $0$ , $1$ and $3$ . Each factor appears once, so every zero has multiplicity $1$ .

Part B

Oddness. The domain is all real numbers, which holds $-x$ whenever it holds $x$ , so the question can be asked. Substitute:

$f(-x) = (-x)^5 - 10(-x)^3 + 9(-x) = -x^5 + 10x^3 - 9x = -f(x)$

The sign on $0 < x < 1$ . No factor has a root inside that interval, so each holds a fixed sign there. Reading the factored form left to right at any such input:

$(+)(-)(+)(-)(+) = +$

so $f > 0$ on $0 < x < 1$ .

The sign on $-1 < x < 0$ . Every input of that interval is the negative of an input of the first, and oddness says $f(-x) = -f(x)$ . Negating a positive value gives a negative one, so $f < 0$ on $-1 < x < 0$ , and no factor signs had to be multiplied there. That is the labour symmetry saves: one half of the graph is proved from the other.

Part C

The parity of the exponents in the expanded rule and the multiplicity of a root are two different things, and the claim confuses them. Build a polynomial that has only odd powers and still carries an even-power factor.

$x\big(x^2 - 1\big)^2 = x\big(x^4 - 2x^2 + 1\big) = x^5 - 2x^3 + x$

Every exponent on the right is odd, so the rule satisfies the classmate's description. Factored, it is $x(x - 1)^2(x + 1)^2$ , and the factors at $1$ and $-1$ carry the even power $2$ . A square is never negative, so it contributes the same sign on both sides of its root: the sign of the product is unchanged across $1$ and across $-1$ , and the graph meets the axis there and turns back. Only the factor $x$ has an odd power, so $0$ is the one zero at which the graph crosses. What decides crossing is the multiplicity of the root, not the exponents left standing after the expansion.

In one line

$f(x) = x(x - 1)(x + 1)(x - 3)(x + 3)$ , with zeros $-3$ , $-1$ , $0$ , $1$ and $3$ . It is odd, since $f(-x) = -f(x)$ on a domain symmetric about $0$ ; $f > 0$ on $0 < x < 1$ , so oddness gives $f < 0$ on $-1 < x < 0$ with no further work. The claim about odd powers is false: $x^5 - 2x^3 + x = x(x - 1)^2(x + 1)^2$ has only odd powers yet touches the axis at $1$ and at $-1$ .

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Removes the common factor and factors the remaining quartic as a quadratic in $x^2$ . . Worth 2 points.

Lists all five zeros, each read from its own linear factor. . Worth 1 point.

Part B 5 points

Notes that the domain holds $-x$ whenever it holds $x$ , then computes $f(-x)$ and compares it with $-f(x)$ . . Worth 2 points. needs an explanation, not just an answer

Assigns each factor a fixed sign on the stated interval and multiplies them. . Worth 1 point.

Transfers the sign to the reflected interval using the odd identity rather than by recomputing the factor signs there. . Worth 2 points. needs an explanation, not just an answer

Part C 4 points

Produces a polynomial whose expanded form carries only odd powers and whose factored form carries an even power. . Worth 2 points.

Identifies the root at which the graph turns back and ties the verdict to the even power on its factor. . Worth 2 points. needs an explanation, not just an answer
9. Two input maps, two orders, one measurable gap . 14 points. Question 9 of 10.

Two moves are to act on the input of a function $f$ : a horizontal compression by $4$ , which divides every input by $4$ , and a shift right $8$ . They can be applied in either order, and the parts below compare the two orders.
1. Part A.
  Write the rule for each order, giving the inside both factored as $b(x - h)$ and expanded.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
2. Part B.
  Take $(20, 9)$ to be a point of the graph of $f$ . Send it through each order, report the horizontal gap between the two places it arrives, and say whether that gap depends on the point you started from.
  
  Carry your own answer forward Use the two rules you produced in part A, whatever they were, and land the point honestly under each. The credit here is for applying the landing rule to both orders and for reporting the separation, not for arriving at a particular pair of inputs.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  Now take the compression factor to be any $b \ne 0$ and the shift to be any $h$ . Prove that the two finished graphs are always a horizontal translation of each other, give that translation in terms of $b$ and $h$ , and say exactly when the two orders agree as transformations, meaning that they deliver the same graph for every function $f$ .
  
  Carry your own answer forward Generalise from the two landing rules your own earlier work produced, expected or not, and check the formula you reach against the separation you measured. The credit here is for an argument in letters that never mentions a particular input, not for recovering one number.
  
  Complete the derivation Each line should follow from the one above it. Say what lets you take each step. 6 points
The answer

Part A

Compression first: $y = f\big(4(x - 8)\big) = f(4x - 32)$ . Shift first: $y = f\big(4(x - 2)\big) = f(4x - 8)$ .

Part B

Compression first: $(13, 9)$ . Shift first: $(7, 9)$ . The two landing points sit at the same height and are $6$ apart horizontally, a gap that does not depend on the point chosen.

Part C

Compression first lands $u$ at $\tfrac{u}{b} + h$ , shift first at $\tfrac{u}{b} + \tfrac{h}{b}$ , a constant $h - \tfrac{h}{b}$ apart: a horizontal translation. As transformations, for every $f$ at once, the orders agree exactly when $h = 0$ or $b = 1$ ; one translation-invariant graph, a constant rule say, can coincide anyway.

Worked solution

Part A

Track where an input $u$ of $f$ ends up, since that is what a rule on the input side has to reproduce.

Compression first. The compression sends $u$ to $\tfrac{u}{4}$ , then the shift sends that to $\tfrac{u}{4} + 8$ . Matching the landing rule $\tfrac{u}{b} + h$ gives $b = 4$ and $h = 8$ :

$y = f\big(4(x - 8)\big) = f(4x - 32)$

Shift first. The shift sends $u$ to $u + 8$ , and the compression that follows divides the whole thing: $\tfrac{u + 8}{4} = \tfrac{u}{4} + 2$ . So $b = 4$ and $h = 2$ :

$y = f\big(4(x - 2)\big) = f(4x - 8)$

The compression reaches the shift in the second order and not in the first, which is why the two expanded insides differ.

Part B

Apply the landing rule $\left(\tfrac{u}{b} + h,\ v\right)$ with $u = 20$ and $v = 9$ ; nothing acts on the output, so the height is untouched in both orders.

$\frac{20}{4} + 8 = 13, \qquad \frac{20}{4} + 2 = 7$

So the landings are $(13, 9)$ and $(7, 9)$ , a horizontal separation of $13 - 7 = 6$ . That gap is not an artefact of this particular point: it is the difference between the two shifts, $8 - 2$ , and every point of the graph is displaced by the same amount.

Part C

Write the two input maps as functions: the compression is $C(u) = \tfrac{u}{b}$ on landing positions, and the shift is $T(u) = u + h$ . Composing them in the two orders gives two landing maps.

$(T \circ C)(u) = \frac{u}{b} + h, \qquad (C \circ T)(u) = \frac{u + h}{b} = \frac{u}{b} + \frac{h}{b}$

Subtract them:

$(T \circ C)(u) - (C \circ T)(u) = h - \frac{h}{b} = h\left(1 - \frac{1}{b}\right)$

The difference does not contain $u$ , so it is the same number for every point of the graph. A displacement that is constant across every point is a horizontal translation, and its size is $h\left(1 - \tfrac{1}{b}\right)$ ; with $b = 4$ and $h = 8$ that is $8 - 2 = 6$ , which is the separation part B measured.

A product of two factors is zero exactly when one of them is, so the two landing maps agree precisely when $h = 0$ , meaning there is no shift to be divided, or when $b = 1$ , meaning the compression divides nothing.

That is a statement about the maps, so it is the condition for the two orders to agree as transformations: for every $f$ at once. A single graph can still coincide with itself under the two orders without the condition holding. A constant rule is the plain case, since its graph is unchanged by any horizontal translation, so both orders return the same picture for every $b$ and every $h$ while the landing maps differ. Away from those special graphs the two maps do not commute, and the order is part of the transformation rather than a detail that can be rearranged.

In one line

Compression first gives $y = f\big(4(x - 8)\big) = f(4x - 32)$ and shift first gives $y = f\big(4(x - 2)\big) = f(4x - 8)$ , under which $(20, 9)$ lands at $(13, 9)$ and $(7, 9)$ , a horizontal gap of $6$ . In general the two orders land an input at $\tfrac{u}{b} + h$ and $\tfrac{u}{b} + \tfrac{h}{b}$ , a constant $h\left(1 - \tfrac{1}{b}\right)$ apart, so the finished graphs are always a horizontal translation of each other. As transformations, applied to every $f$ at once, the two orders agree exactly when $h = 0$ or $b = 1$ ; one translation-invariant graph, a constant rule say, can coincide under both orders anyway.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Builds each order by tracking where an input lands, rather than by reading the moves off a finished formula. . Worth 2 points.

Gives both insides in factored and expanded form, with the shift in each matching its own order. . Worth 2 points.

Part B 4 points

Applies the landing rule under both orders and leaves the height untouched, since neither move acts on the output. . Worth 2 points.

Gives the horizontal separation of the two landings and says that it is the same for every point of the graph. . Worth 2 points.

Part C 6 points

Writes both orders as landing maps in the letters $b$ and $h$ , with the compression dividing the shift in exactly one of them. . Worth 2 points.

Shows the difference of the two landings carries no input variable, and concludes from that alone that the graphs are a translation apart. . Worth 2 points. needs an explanation, not just an answer

States the translation in terms of $b$ and $h$ and gives both conditions under which it is zero. . Worth 2 points. needs an explanation, not just an answer
10. A rule that never returns certain numbers . 14 points. Question 10 of 10.

A rule is declared by cases, with codomain all of the real numbers:

$f(x) = \begin{cases} 2x & \text{if } x \le 1, \\ x + 6 & \text{if } x > 1. \end{cases}$
1. Part A.
  Give the set of values $f$ actually produces, treating the two cases separately, and name every real number the rule never outputs.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
2. Part B.
  Give the rule that undoes $f$ , case by case, with the inputs each case accepts.
  
  Carry your own answer forward Attach each case of the inverse to the set of values you found that case produces in part A, whatever they were. The credit here is for undoing each case on its own values and for keeping the two sets apart, not for a particular pair of intervals.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
3. Part C.
  The rule was declared with codomain all real numbers. Decide whether, as declared, $f$ has a two-sided inverse; examine both compositions, and if one of them cannot be the identity on the declared codomain, name it and give a value at which it fails; then, if a change is needed, state the one that repairs it without touching the rule.
  
  Carry your own answer forward Argue from the set of produced values you found in part A and the candidate you built in part B, whatever they were, and work from any value of the declared codomain your own work leaves unreached. The credit is for asking whether the declaration or the rule is responsible, not for a particular value.
  
  Justify your claim State the claim, then give the reason it has to be true. 6 points
The answer

Part A

The first case produces every $y \le 2$ and the second every $y > 7$ . The numbers satisfying $2 < y \le 7$ are never output.

Part B

$g(x) = \tfrac{x}{2}$ on $x \le 2$ , and $g(x) = x - 6$ on $x > 7$ . No number is claimed by both cases, and none between $2$ and $7$ is claimed at all.

Part C

As declared it has none. Any candidate $g$ would need $f(g(5)) = 5$ , and $5$ is not a value $f$ ever produces, so $f \circ g$ cannot be the identity on the declared codomain. Declaring the codomain to be the set of values produced repairs it, leaving one-to-oneness as the only thing left to check.

Worked solution

Part A

The cases are disjoint and cover every real number, so $f$ is a genuine function; take the values one case at a time.

$x \le 1 \ \Longrightarrow\ 2x \le 2, \qquad x > 1 \ \Longrightarrow\ x + 6 > 7$

Each case is a line, so it produces every value in its own interval and nothing beyond it. Together they produce all of $y \le 2$ and all of $y > 7$ , and the boundary $2$ is attained at $x = 1$ while $7$ is only approached, since $x > 1$ is strict. So the values with $2 < y \le 7$ are never output: the range has a gap, even though the domain has none.

Part B

Each case of $f$ is one-to-one on its own piece, and the two pieces produce sets with no number in common, so no output is ever reached twice and $f$ is one-to-one overall. Undo each case on the values that case produces.

$y = 2x \ (x \le 1) \ \Longrightarrow\ x = \frac{y}{2} \ (y \le 2), \qquad y = x + 6 \ (x > 1) \ \Longrightarrow\ x = y - 6 \ (y > 7)$

Call that rule $g$ , holding the name $f^{-1}$ back until the declaration has been settled. It is itself a rule in two cases, with domain exactly the set part A found. Checking one value each way: $f(1) = 2$ and $g(2) = 1$ ; $f(4) = 10$ and $g(10) = 4$ .

Part C

One-to-oneness is not the whole condition when the codomain is declared larger than the set of values produced, and here it is.

Which composition fails. Suppose $g$ were a two-sided inverse. Then $f \circ g$ would have to be the identity on the whole declared codomain, so $f(g(y)) = y$ for every real $y$ . But $f(g(y))$ is a genuine output of $f$ , and part A showed $f$ never outputs anything strictly between $2$ and $7$ :

$f\big(g(5)\big) = 5 \ \text{is impossible, since } 5 \ \text{is not a value } f \ \text{produces.}$

So $f \circ g = \operatorname{id}$ fails, at $5$ and at every value in the gap. The other composition is fine: $g \circ f = \operatorname{id}$ holds for the rule of part B, because $f$ is one-to-one.

The repair. Nothing is wrong with the rule, so nothing about the rule needs changing. What was wrong was the declaration: shrink the codomain to the set of values $f$ actually produces, every $y \le 2$ together with every $y > 7$ . With the codomain taken as the range, reaching every declared value is automatic, and one-to-oneness is then the only condition to check, which part B verified; only at that point does the rule of part B earn the name $f^{-1}$ . This is why the clean statement 'invertible exactly when one-to-one' carries a hypothesis about the codomain; with a larger codomain the condition is one-to-one and onto.

In one line

The rule produces every $y \le 2$ and every $y > 7$ , and never a value with $2 < y \le 7$ . Its undoing rule is $g(x) = \tfrac{x}{2}$ on $x \le 2$ and $x - 6$ on $x > 7$ . With the codomain declared to be all real numbers there is no two-sided inverse, since $f \circ g$ would have to return values such as $5$ that $f$ never produces; declaring the codomain to be the set of produced values repairs it, and one-to-oneness is then all that remains to check.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Works out the values each case produces, keeping the boundary input with the case whose condition it satisfies. . Worth 2 points.

Reports the gap between the two sets, open at the attained end and closed at the approached one. . Worth 2 points.

Part B 4 points

Undoes each case separately and attaches it to the values that case produces rather than to the original inputs. . Worth 2 points.

Notes that the two cases of the inverse share no input, which is what makes the rule single-valued. . Worth 2 points.

Part C 6 points

Names which of the two compositions cannot be the identity on the declared codomain, and why the other one is unaffected. . Worth 2 points. needs an explanation, not just an answer

Exhibits one value of the declared codomain that no input reaches, and argues from it that no candidate can succeed. . Worth 2 points. needs an explanation, not just an answer

States the change to the declaration that repairs it, and says what condition is then left to check. . Worth 2 points.

Marking rubrics

Work every question first. Then mark your own paper against these, giving yourself the points for each element you actually wrote down.

1. One curve, and a question that waits on a declaration 9 points

Part A 3

Solves the relation for $y$ and keeps both signs the square root carries. 2
Reports how many outputs each declaration produces at the stated input, naming which declaration each count belongs to. 1

Part B 3

Exhibits two inputs with a shared output and argues from them to the verdict on one-to-oneness. 2
Connects the shared output to the failure of the reading with $x$ as the input, through the swap of coordinates. 1

Part C 3

Grants that the set of points is fixed, then locates the verdict in the pairing of that set with a declared input. 2
Names the two line tests as the same test asked about the two different readings. 1

2. The slot, the quotient, and the one input the cancelling costs 10 points

Part A 4

Substitutes the whole expression $x + h$ into the slot and combines the two fractions over a common denominator. 2
Cancels the factor of $h$ and records all three restrictions the simplified fraction carries. 2

Part B 3

Evaluates the simplified expression and the direct difference quotient, and sets the two results side by side. 2
States what the number measures for this rule, as a change in output per unit of input across the step. 1

Part C 3

Grants the agreement on the overlap and then appeals to the domain, not the values, as what settles equality. 2
Names $h = 0$ as the disagreeing input and traces it to the division that produced the simplified form. 1

3. Everything the factored form already knows 11 points

Part A 3

Reads both zeros off the factors and uses them to cut the line into the intervals to be tested. 1
Gives each factor a fixed sign on each interval and multiplies those signs to reach the sign of $f$ there. 2

Part B 4

Expands the product and factors out the highest power of $x$ to isolate a bracket of correction terms. 2
Bounds the correction terms beyond a stated threshold and reads the behaviour of both ends from the sign of the leading power. 2

Part C 4

Argues that an even power holds one sign on both sides of its root, so the graph meets the axis there and turns back. 2
Argues that an odd power carries the sign of its single factor, so the sign flips and the graph crosses. 2

4. Two excluded inputs, only one of which leaves a gap 11 points

Part A 4

Factors the denominator and names both inputs it rejects, before any cancelling is done. 2
Identifies which excluded input leaves a single missing point and reads its height off the simplified rule. 2

Part B 3

Cancels only the factor the numerator and denominator share, and keeps the excluded input attached to the simplified rule. 2
Says what happens at the excluded input whose factor survives, distinguishing it from a single missing point. 1

Part C 4

Tests all three conditions of function equality and identifies the one that fails. 2
Exhibits an input that lies in one domain and not the other, with the value one rule gives there. 1
States a declared domain for $g$ under which all three conditions hold, leaving both rules untouched. 1

5. Four numbers, and the coordinate each one is allowed to touch 12 points

Part A 4

Factors the inside into the form $b(x - h)$ before naming any shift. 2
Names all four moves with their exact amounts and directions, keeping the inside pair separate from the outside pair. 2

Part B 4

Sends the domain endpoints through the inside pair only and the range endpoints through the outside pair only. 2
Reports each interval from its smaller value up and names which old endpoint became which new one. 2

Part C 4

Argues that finding a new input means solving the inside for $x$ , so the inside numbers reach the first coordinate only. 2
Argues that the outside acts on a height already produced, so it reaches the second coordinate only. 1
Ties the factoring step to the fact that the input multiplier divides the shift, and states the amount the unfactored reading would give. 1

6. What survives two stages, and what a superscript is counting 12 points

Part A 4

Substitutes the rule of the inner function into the input slot of the outer one and simplifies. 1
Imposes both conditions, the inner function's own requirement and the requirement on its output, and names which excluded input each produces. 3

Part B 4

Computes the composition by substituting the whole first output back into the rule. 2
Computes the square of the output separately and gives an input at which the two results differ. 2

Part C 4

Builds the composite and states both conditions, the one on the inner input and the one on the inner output. 2
Produces one input inside the domain and one outside it, with the value of the inner rule at each as the reason. 2

7. Undoing a rule, and the checks that decide it 14 points

Part A 4

Solves for the input and chooses the sign of the root that the restriction forces. 2
Gives the inverse's domain as the set of values the restricted rule produces, not as the original domain. 2

Part B 5

Carries out both compositions rather than one, each on its own domain. 2
Treats the square root of a square as an absolute value and says what the restriction does to it. 2
Names the restriction each direction depends on, and what would fail without it. 1

Part C 5

Computes the composition the classmate did not run, on the unrestricted domain. 2
Names one input at which that composition fails to return its own value, with the two intermediate values. 1
Argues that a two-sided requirement is what the failure is about, rather than an arithmetic slip in the classmate's check. 2

8. A quintic taken apart factor by factor 12 points

Part A 3

Removes the common factor and factors the remaining quartic as a quadratic in $x^2$ . 2
Lists all five zeros, each read from its own linear factor. 1

Part B 5

Notes that the domain holds $-x$ whenever it holds $x$ , then computes $f(-x)$ and compares it with $-f(x)$ . 2
Assigns each factor a fixed sign on the stated interval and multiplies them. 1
Transfers the sign to the reflected interval using the odd identity rather than by recomputing the factor signs there. 2

Part C 4

Produces a polynomial whose expanded form carries only odd powers and whose factored form carries an even power. 2
Identifies the root at which the graph turns back and ties the verdict to the even power on its factor. 2

9. Two input maps, two orders, one measurable gap 14 points

Part A 4

Builds each order by tracking where an input lands, rather than by reading the moves off a finished formula. 2
Gives both insides in factored and expanded form, with the shift in each matching its own order. 2

Part B 4

Applies the landing rule under both orders and leaves the height untouched, since neither move acts on the output. 2
Gives the horizontal separation of the two landings and says that it is the same for every point of the graph. 2

Part C 6

Writes both orders as landing maps in the letters $b$ and $h$ , with the compression dividing the shift in exactly one of them. 2
Shows the difference of the two landings carries no input variable, and concludes from that alone that the graphs are a translation apart. 2
States the translation in terms of $b$ and $h$ and gives both conditions under which it is zero. 2

10. A rule that never returns certain numbers 14 points

Part A 4

Works out the values each case produces, keeping the boundary input with the case whose condition it satisfies. 2
Reports the gap between the two sets, open at the attained end and closed at the approached one. 2

Part B 4

Undoes each case separately and attaches it to the values that case produces rather than to the original inputs. 2
Notes that the two cases of the inverse share no input, which is what makes the rule single-valued. 2

Part C 6

Names which of the two compositions cannot be the identity on the declared codomain, and why the other one is unaffected. 2
Exhibits one value of the declared codomain that no input reaches, and argues from it that no candidate can succeed. 2
States the change to the declaration that repairs it, and says what condition is then left to check. 2

Multiple choice

Free response