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Functions: 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, 118 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 rule, two numbers and an expression . 10 points. Question 1 of 10.

A single rule, $f(x) = 2x^2 - 5$ , is put to work below at two numbers and once at an expression.
1. Part A.
  Find $f(4)$ and $f(-4)$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Find $f(t + 1)$ , expanded and simplified.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
3. Part C.
  Compare the two outputs you computed in part A. Explain what that comparison does, or does not, show about whether $f$ is a function, and state what a rule would have to do to fail the definition.
  
  Carry your own answer forward Argue from the two outputs you computed in part A, whatever they came out to be.
  
  Explain why it works A sentence or two. Reasons, not steps. 3 points
The answer

Part A

$f(4) = 27$ and $f(-4) = 27$ .

Part B

$f(t + 1) = 2t^2 + 4t - 3$ .

Part C

The two outputs match, and that shows nothing against $f$ : the definition restricts each input to one output and says nothing about outputs being shared. A rule fails only when a single input is assigned two different outputs.

Worked solution

Part A

Replace $x$ by each input in turn, keeping the parentheses so the sign is squared along with the digit.

$f(4) = 2(4)^2 - 5 = 32 - 5 = 27, \qquad f(-4) = 2(-4)^2 - 5 = 32 - 5 = 27.$

Squaring removes the sign before the doubling and the subtraction happen, so the two inputs land on the same output.

Part B

The input is the whole expression $t + 1$ , so all of it goes into the square.

$f(t + 1) = 2(t + 1)^2 - 5 = 2(t^2 + 2t + 1) - 5 = 2t^2 + 4t + 2 - 5 = 2t^2 + 4t - 3.$

The middle term $4t$ comes from squaring the whole input; squaring only the $t$ would lose it.

Part C

A function assigns to each input exactly one output. Two different inputs arriving at the same output are two separate pairs, and each still has a single result.

$4 \mapsto 27 \quad \text{and} \quad -4 \mapsto 27$

Neither input is ambiguous, so the definition is satisfied. What the definition forbids is the reverse: one input with two outputs, as in the rule that sends $9$ to both $3$ and $-3$ because both square to $9$ . That is the only way to fail.

In one line

$f(4) = 27$ and $f(-4) = 27$ , and $f(t + 1) = 2t^2 + 4t - 3$ . The repeated output is harmless: a function may send two different inputs to one output, and it fails the definition only when one input is given two different outputs.

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

Substitutes each input in full, so that $(-4)^2$ is $16$ rather than $-16$ . . Worth 2 points.

Reports both outputs, each labelled with the input it came from. . Worth 1 point.

Part B 4 points

Substitutes the entire expression $t + 1$ into the square, not the letter alone. . Worth 2 points.

Expands the square, distributes the $2$ across all three terms, and combines the constants. . Worth 2 points.

Part C 3 points

Locates the definition's restriction on INPUTS rather than outputs, and applies it to the comparison made in part A. . Worth 2 points. needs an explanation, not just an answer

States the condition that would actually break the definition, one input with two different outputs. . Worth 1 point.
2. Undoing a rule, and a symbol that looks like a division . 12 points. Question 2 of 10.

Let $f(x) = \dfrac{x}{4} - 3$ .
1. Part A.
  Find $f^{-1}(x)$ .
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
2. Part B.
  Compute $f^{-1}(5)$ , then check that result by running the original rule $f$ on it.
  
  Carry your own answer forward Use the inverse rule you found in part A, whatever it was, and test it against the original rule $f$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  Compute $\dfrac{1}{f(5)}$ , and explain what question each of $f^{-1}(5)$ and $\dfrac{1}{f(5)}$ answers.
  
  Carry your own answer forward Set your value of $f^{-1}(5)$ from part B beside the reciprocal you compute here.
  
  Explain why it works A sentence or two. Reasons, not steps. 4 points
The answer

Part A

$f^{-1}(x) = 4x + 12$ .

Part B

$f^{-1}(5) = 32$ , and $f(32) = 8 - 3 = 5$ , the number the inverse was handed.

Part C

$\dfrac{1}{f(5)} = -\dfrac{4}{7}$ . The inverse answers which input $f$ turns into $5$ ; the reciprocal answers what one divided by the output at $5$ comes to. The two questions are unrelated.

Worked solution

Part A

Write $y = \dfrac{x}{4} - 3$ , swap the two letters so the output becomes the input, and then solve for $y$ .

$x = \frac{y}{4} - 3 \;\Longrightarrow\; x + 3 = \frac{y}{4} \;\Longrightarrow\; y = 4(x + 3) = 4x + 12.$

The rule divides by $4$ and then subtracts $3$ ; the undo adds $3$ and then multiplies by $4$ , with the steps reversed and each one turned around.

Part B

Evaluate the inverse at $5$ .

$f^{-1}(5) = 4(5) + 12 = 32.$

A correct inverse must send $5$ back to the input that $f$ turns into $5$ , so run $f$ forwards on the result.

$f(32) = \frac{32}{4} - 3 = 8 - 3 = 5.$

The round trip closes, which is what confirms the value.

Part C

Evaluate $f$ at $5$ first, then take the reciprocal of that output.

$f(5) = \frac{5}{4} - 3 = -\frac{7}{4}, \qquad \frac{1}{f(5)} = -\frac{4}{7}.$

Set that beside the value of $f^{-1}(5)$ from part B. The two are nowhere near each other, and they are not supposed to be. The inverse runs the rule backwards and hands back the input that produced a given output. The reciprocal runs the rule forwards and then flips the output into a fraction. The raised $-1$ marks the inverse under composition, not a division.

In one line

$f^{-1}(x) = 4x + 12$ , so $f^{-1}(5) = 32$ , confirmed by $f(32) = 5$ . The reciprocal at the same input is $\dfrac{1}{f(5)} = -\dfrac{4}{7}$ , a different number answering a different question: the inverse recovers an input, while the reciprocal flips an output.

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

Swaps the two letters and then SOLVES the new equation for $y$ , rather than stopping at the swap. . Worth 2 points.

Multiplies the whole of $x + 3$ by $4$ , not the $x$ alone. . Worth 2 points.

Part B 4 points

Evaluates the inverse rule at the given input correctly. . Worth 2 points.

Checks the value by running the original rule on it and comparing with the number the inverse was handed. . Worth 2 points.

Part C 4 points

Evaluates $f$ at the input first and then takes the reciprocal of that output, rather than reciprocating the rule's parts. . Worth 2 points.

States what question each of the two expressions answers, and uses that to say how the two are related. . Worth 2 points. needs an explanation, not just an answer
3. Three rules from two, and the one that loses inputs . 11 points. Question 3 of 10.

Let $f(x) = 3x + 1$ and $g(x) = x^2 - 25$ . New functions can be built from these two by ordinary arithmetic on their outputs.
1. Part A.
  Write $(f + g)(x)$ and $(f - g)(x)$ in simplest form.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
2. Part B.
  Write $\left(\dfrac{f}{g}\right)(x)$ and state every input it excludes.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
3. Part C.
  The input $-\tfrac{1}{3}$ makes $f$ zero. Decide whether it belongs to the domain of $\dfrac{f}{g}$ , and justify your verdict by contrasting it with an input that makes $g$ zero.
  
  Justify your claim State the claim, then give the reason it has to be true. 4 points
The answer

Part A

$(f + g)(x) = x^2 + 3x - 24$ and $(f - g)(x) = -x^2 + 3x + 26$ .

Part B

$\left(\dfrac{f}{g}\right)(x) = \dfrac{3x + 1}{x^2 - 25}$ , and it excludes $x = 5$ and $x = -5$ .

Part C

It does belong: it makes only the top zero, so the quotient has the value $0$ there. An input that makes $g$ zero, such as $5$ , is excluded instead, because a division by zero has no value at all.

Worked solution

Part A

Evaluate both rules at the same input and combine the outputs.

$(f + g)(x) = (3x + 1) + (x^2 - 25) = x^2 + 3x - 24.$

$(f - g)(x) = (3x + 1) - (x^2 - 25) = 3x + 1 - x^2 + 25 = -x^2 + 3x + 26.$

The subtraction reaches every term of $g$ , which is why the $-25$ arrives as $+25$ .

Part B

The quotient puts $f$ 's output over $g$ 's output.

$\left(\frac{f}{g}\right)(x) = \frac{3x + 1}{x^2 - 25}.$

It has no value at any input where the bottom rule is zero, so solve that.

$x^2 - 25 = 0 \;\Longrightarrow\; (x - 5)(x + 5) = 0 \;\Longrightarrow\; x = 5 \ \text{ or } \ x = -5.$

Both inputs are thrown out, even though $f$ is perfectly happy at each of them.

Part C

A fraction is undefined only when its denominator is zero; a zero numerator is an ordinary value. At $x = -\tfrac{1}{3}$ the denominator is $\tfrac{1}{9} - 25$ , which is not zero.

$\left(\frac{f}{g}\right)\left(-\tfrac{1}{3}\right) = \frac{0}{\tfrac{1}{9} - 25} = 0.$

So that input is admitted, and its output happens to be $0$ . At $x = 5$ the denominator is $0$ , and no such division has a value, so that input is thrown out. A zero on top produces an output; a zero underneath produces none.

In one line

$(f + g)(x) = x^2 + 3x - 24$ , $(f - g)(x) = -x^2 + 3x + 26$ , and $\left(\dfrac{f}{g}\right)(x) = \dfrac{3x + 1}{x^2 - 25}$ excluding $x = 5$ and $x = -5$ . The input $-\tfrac{1}{3}$ stays in the domain, with output $0$ , because only a zero denominator, never a zero numerator, destroys a 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

Combines the two rules at the same input and collects like terms. . Worth 2 points.

Distributes the subtraction across BOTH terms of $g$ , so the constant changes sign. . Worth 2 points.

Part B 3 points

Writes the quotient with $f$ above and $g$ below. . Worth 1 point.

Sets the denominator rule equal to zero and reports BOTH inputs that solve it. . Worth 2 points.

Part C 4 points

Reaches a verdict on whether an input making only the numerator zero belongs to the domain, and states the value the quotient takes there. . Worth 2 points.

Justifies the difference by naming division by zero as the only thing a quotient forbids. . Worth 2 points. needs an explanation, not just an answer
4. A service fee before the rate, or after it . 12 points. Question 4 of 10.

A currency desk changes dollars into euros in two steps. It first deducts a flat service fee of $3$ dollars, then converts every remaining dollar at $0.9$ euros per dollar.
1. Part A.
  Write the fee step and the conversion step as two separate rules, then give a single formula for the euros received from $d$ dollars.
  
  Model the situation Name your unknown first, then write every other quantity in terms of that one letter. 4 points
2. Part B.
  How many euros does a traveller handing over $200$ dollars receive?
  
  Carry your own answer forward Substitute into the composite formula you built in part A, whatever form it took.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  A rival desk converts the whole amount first and then deducts a flat $3$ euro fee. Write that composition, and compare what the two desks hand a traveller who brings $200$ dollars, saying which is better and by how much.
  
  Carry your own answer forward Compare against the amount you computed in part B.
  
  Compare the two methods Say what each one costs you, and when you would reach for it. 5 points
The answer

Part A

$s(d) = d - 3$ and $c(x) = 0.9x$ , so the euros received are $c(s(d)) = 0.9(d - 3) = 0.9d - 2.7$ .

Part B

$177.30$ euros.

Part C

The rival's rule is $0.9d - 3$ , giving $177$ euros on $200$ dollars, against the first desk's $177.30$ euros. The first desk is better by $0.30$ euros, because deducting before converting shrinks the fee by the same factor the money is converted at.

Worked solution

Part A

Name the two steps separately.

$s(d) = d - 3, \qquad c(x) = 0.9x.$

The fee comes off first, so $s$ is the inner rule and its output is what the conversion acts on.

$c(s(d)) = c(d - 3) = 0.9(d - 3) = 0.9d - 2.7.$

Part B

Substitute the amount into the composite, taking the fee off before the rate is applied.

$c(s(200)) = 0.9(200 - 3) = 0.9(197) = 177.3.$

The traveller receives $177.30$ euros.

Part C

The rival converts first, so the conversion is now the inner rule, and its fee is charged in euros. Writing that euro fee as $u(x) = x - 3$ ,

$u(c(d)) = 0.9d - 3.$

At $d = 200$ that gives

$0.9(200) - 3 = 180 - 3 = 177.$

The first desk handed over $177.30$ euros, so it beats the rival by $0.30$ euros. The gap is the same at every amount, since $0.9d - 2.7$ and $0.9d - 3$ sit a constant $0.30$ apart. The reason is the order: when the fee is deducted first, the fee itself is converted at $0.9$ , so it costs $2.70$ euros rather than a full $3$ .

In one line

The desk's rule is $c(s(d)) = 0.9(d - 3) = 0.9d - 2.7$ , which gives $177.30$ euros on $200$ dollars. The rival's rule is $s(c(d)) = 0.9d - 3$ , which gives $177$ euros. The first desk is better by $0.30$ euros at every amount, because its fee is deducted before the conversion and so is converted along with the rest.

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

Names two separate rules, one subtracting the fee and one multiplying by the rate. . Worth 2 points.

Places the fee step INSIDE the conversion step, matching the order the desk uses. . Worth 2 points.

Part B 3 points

Substitutes the amount into the composite, deducting the fee before applying the rate. . Worth 2 points.

Reports the result as an amount in euros rather than a bare number. . Worth 1 point.

Part C 5 points

Builds the rival's composite with the conversion inside and the fee subtracted afterwards. . Worth 2 points.

Compares the two amounts at the stated exchange, naming which is larger and the size of the gap. . Worth 2 points.

Ties the gap to the order of the two steps, not to a difference in the rate or the fee. . Worth 1 point. needs an explanation, not just an answer
5. Two steps in, one formula out . 12 points. Question 5 of 10.

Let $f(x) = \dfrac{1}{x - 3}$ and $g(x) = \dfrac{6}{x}$ .
1. Part A.
  Find a single simplified formula for $(f \circ g)(x)$ .
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
2. Part B.
  State every input that $f \circ g$ rejects, and say which of the two steps rejects each one.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
3. Part C.
  Reading a domain straight off the simplified formula from part A is a common shortcut. State the set that shortcut produces, compare it with the set you reported in part B, and identify exactly what the shortcut leaves untested.
  
  Carry your own answer forward Compare the shortcut's set with the exclusions you listed in part B.
  
  Find and correct the error Say which line first goes wrong, why it is wrong, and then do it correctly. 4 points
The answer

Part A

$(f \circ g)(x) = \dfrac{x}{6 - 3x}$ .

Part B

$x = 0$ , rejected by the inner rule, and $x = 2$ , rejected by the outer rule because $g(2) = 3$ .

Part C

The shortcut gives every input except $2$ , the one zero of the simplified denominator. It misses $x = 0$ , and what it leaves untested is whether an input is a legal input for the inner rule in the first place.

Worked solution

Part A

Drop the whole rule for $g$ into $f$ in place of its input.

$f(g(x)) = \frac{1}{\dfrac{6}{x} - 3}.$

Combine the two terms in the lower fraction over the common denominator $x$ , then divide by that fraction by multiplying by its reciprocal.

$\frac{6}{x} - 3 = \frac{6 - 3x}{x}, \qquad f(g(x)) = \frac{x}{6 - 3x}.$

Part B

An input has to be legal for $g$ first, and then $g$ 's output has to be legal for $f$ . The inner rule divides by $x$ , so it refuses $0$ outright.

$g \text{ needs } x \neq 0.$

The outer rule refuses an input of $3$ , so the second test asks where $g$ produces $3$ .

$\frac{6}{x} = 3 \;\Longrightarrow\; x = 2.$

At $x = 2$ the inner rule is perfectly content, returning $g(2) = 3$ , and the outer rule then divides by zero. So the composite rejects $0$ and $2$ .

Part C

The simplified formula breaks down where its own denominator vanishes.

$6 - 3x = 0 \;\Longrightarrow\; x = 2,$

so the shortcut produces "every input except $2$ ". That is exactly the second half of the two-part rule and nothing else: it is the condition $g(x) \neq 3$ in disguise. The first half, that $x$ must be a legal input for $g$ before anything else can happen, is never tested, and that is the half which bars $0$ .

The simplification is what destroyed the evidence. Clearing the compound fraction moved the inner denominator's $x$ up into the numerator, so the tidy formula has no way left to object at $0$ . A composite's domain is fixed by the two steps that built it, never by the formula they collapsed into.

In one line

$(f \circ g)(x) = \dfrac{x}{6 - 3x}$ , and it rejects $x = 0$ , which the inner rule refuses, and $x = 2$ , whose output $g(2) = 3$ the outer rule refuses. The shortcut's set, every input except $2$ , tests only the second of those requirements, and the simplified formula cannot reveal the first, because clearing the compound fraction moved the inner denominator up into the numerator.

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 entire rule for $g$ into $f$ 's input slot. . Worth 2 points.

Clears the compound fraction correctly, arriving at a single quotient. . Worth 2 points.

Part B 4 points

Applies both halves of the rule, testing the inner rule's own domain and then where its output is illegal for the outer rule. . Worth 2 points.

Reports both excluded inputs and attributes each to the step that rejects it. . Worth 2 points.

Part C 4 points

Names which of the two halves of the domain rule the shortcut leaves untested, rather than blaming an arithmetic slip in the simplification. . Worth 2 points. needs an explanation, not just an answer

Explains what the simplification did to the evidence, and why the final formula cannot be used to recover the domain. . Worth 2 points. needs an explanation, not just an answer
6. Cutting a rule down until it can be reversed . 14 points. Question 6 of 10.

Let $f(x) = x^2 - 6x$ , defined at first on every real number.
1. Part A.
  Give two different inputs that share an output, and say what that shows about reversing $f$ as it stands.
  
  Construct a counterexample Give one specific case, and show it breaks the claim. 3 points
2. Part B.
  Restricted to $x \ge 3$ , the rule is one-to-one. Find $f^{-1}$ for the restricted rule, and state which inputs $f^{-1}$ accepts.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 5 points
3. Part C.
  Verify the pair by composing in both directions on the restricted domain, and explain why the condition $x \ge 3$ cannot simply be carried onto $f^{-1}$ unchanged.
  
  Carry your own answer forward Compose the inverse you found in part B with the restricted rule, in both orders.
  
  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

$f(0) = 0$ and $f(6) = 0$ . Two different inputs share an output, so no rule could send that output back to a single input, and $f$ on all real numbers has no inverse function.

Part B

$f^{-1}(x) = 3 + \sqrt{x + 9}$ , accepting every input $x \ge -9$ .

Part C

Both directions collapse to the input: $f^{-1}(f(x)) = 3 + \sqrt{(x - 3)^2} = x$ for $x \ge 3$ , and $f(f^{-1}(x)) = x$ for $x \ge -9$ . The condition $x \ge 3$ described $f$ 's inputs, so the swap turns it into the inverse's outputs, while the inverse's own inputs are $f$ 's outputs.

Worked solution

Part A

Test two inputs that the completed square suggests are symmetric about the turning point.

$f(0) = 0^2 - 6(0) = 0, \qquad f(6) = 6^2 - 6(6) = 36 - 36 = 0.$

The rule is therefore not one-to-one. An inverse would have to send the output $0$ back to the input it came from, but $0$ came from both $0$ and $6$ , and a function may never give one input two outputs. So no inverse function exists on this domain.

Part B

Rewrite the rule so the input appears only once.

$f(x) = x^2 - 6x = (x - 3)^2 - 9.$

Write $y = (x - 3)^2 - 9$ with $x \ge 3$ , then swap the letters. The restriction travelled with the input, so after the swap it is a condition on $y$ .

$x = (y - 3)^2 - 9, \quad y \ge 3 \;\Longrightarrow\; (y - 3)^2 = x + 9 \;\Longrightarrow\; y = 3 + \sqrt{x + 9}.$

Only the nonnegative root survives, because $y \ge 3$ forces $y - 3 \ge 0$ . The inverse accepts exactly the outputs the restricted rule produced, and the smallest of those is $-9$ , at $x = 3$ .

Part C

Compose one way, using the completed-square form.

$f^{-1}(f(x)) = 3 + \sqrt{\bigl((x - 3)^2 - 9\bigr) + 9} = 3 + \sqrt{(x - 3)^2} = 3 + (x - 3) = x.$

The last step is legitimate precisely because the restricted domain guarantees $x - 3 \ge 0$ ; on all real numbers it would fail for every input below $3$ . Now compose the other way.

$f\bigl(f^{-1}(x)\bigr) = \left(\sqrt{x + 9}\right)^2 - 9 = (x + 9) - 9 = x.$

Both directions return the input, so the two rules genuinely undo each other. The restriction cannot be copied across because it was a statement about what $f$ is allowed to take IN. Under the swap, $f$ 's inputs become $f^{-1}$ 's outputs, so $x \ge 3$ describes the inverse's range, while the inverse's domain is the set of values $f$ actually produced, $x \ge -9$ . Domain and range trade places; they are never carried over unchanged.

In one line

$f(0) = f(6) = 0$ , so $f$ on all real numbers is not one-to-one and cannot be reversed. Restricted to $x \ge 3$ it can be: $f^{-1}(x) = 3 + \sqrt{x + 9}$ , accepting $x \ge -9$ , and both compositions collapse to the input. The condition $x \ge 3$ becomes the inverse's range rather than its domain, because the swap trades inputs for outputs.

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

Produces two specific different inputs with equal outputs and shows the arithmetic for both. . Worth 2 points.

States what the shared output means: the rule is not one-to-one, so it cannot be reversed as it stands. . Worth 1 point.

Part B 5 points

Rewrites the rule so the input appears once, then swaps the letters and solves for the new output. . Worth 2 points.

Keeps only the nonnegative root, using the restriction that moved onto the new variable. . Worth 2 points.

States the inputs the inverse accepts, matching them to the outputs the restricted rule produced. . Worth 1 point.

Part C 6 points

Carries out BOTH compositions and shows each one collapsing to the input. . Worth 2 points.

Uses the restriction to justify replacing $\sqrt{(x - 3)^2}$ by $x - 3$ , rather than asserting it. . Worth 2 points. needs an explanation, not just an answer

Explains which variable the restriction described before the swap and which it describes after, covering both the inverse's inputs and its outputs. . Worth 2 points. needs an explanation, not just an answer
7. Two orders, then a hunt for an identity . 11 points. Question 7 of 10.

An operation is defined by $a \star b = a + 2b - ab$ . Its two slots are treated differently, so how it behaves in each slot is worth checking.
1. Part A.
  Compute $3 \star 5$ and $5 \star 3$ , and say what the two results settle.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Find the only number $e$ for which $a \star e = a$ holds for every $a$ .
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
3. Part C.
  Test that candidate on the other side, and state whether $\star$ has an identity element.
  
  Carry your own answer forward Take the candidate you found in part B and place it in the FIRST slot instead.
  
  Construct a counterexample Give one specific case, and show it breaks the claim. 4 points
The answer

Part A

$3 \star 5 = -2$ and $5 \star 3 = -4$ . The two orders disagree, so $\star$ is not commutative.

Part B

$e = 0$ .

Part C

$0 \star 5 = 10$ , not $5$ , so the candidate fails from the left. In general $0 \star a = 2a$ , which returns the input only at $a = 0$ , so $\star$ has no identity element.

Worked solution

Part A

Put each pair into its slots in the order written, doing the multiplications before the additions.

$3 \star 5 = 3 + 2(5) - (3)(5) = 3 + 10 - 15 = -2,$

$5 \star 3 = 5 + 2(3) - (5)(3) = 5 + 6 - 15 = -4.$

One disagreeing pair is enough: swapping the inputs can change the answer, so the operation is not commutative.

Part B

Substitute $e$ into the second slot and gather the terms that contain it.

$a \star e = a + 2e - ae = a \;\Longrightarrow\; 2e - ae = 0 \;\Longrightarrow\; e(2 - a) = 0.$

A product is zero when one of its factors is. The factor $2 - a$ vanishes only at the single value $a = 2$ , so it cannot deliver the condition for every $a$ ; the other factor must be the one that vanishes, forcing $e = 0$ . Checking on that side: $a \star 0 = a + 0 - 0 = a$ .

Part C

An identity has to leave every input unchanged from both sides. Put the candidate in the first slot.

$0 \star a = 0 + 2a - (0)(a) = 2a.$

That returns $a$ only when $a = 0$ , and one counterexample finishes it.

$0 \star 5 = 10 \neq 5.$

The candidate survives one side and fails the other, which is exactly the risk part A flagged: because the operation is not commutative, passing on the right says nothing about the left. Part B showed $0$ was the only candidate even on the right, so no number works on both sides and $\star$ has no identity element.

In one line

$3 \star 5 = -2$ while $5 \star 3 = -4$ , so $\star$ is not commutative. The condition $a \star e = a$ forces $e = 0$ , but $0 \star a = 2a$ , which already fails at $a = 5$ . So $\star$ has no identity element, and because the operation is not commutative, passing on one side could never have settled the other.

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

Substitutes each pair in the order written, doing the multiplication before the additions. . Worth 2 points.

Compares the two results and states what a single pair of inputs can settle about the operation. . Worth 1 point.

Part B 4 points

Substitutes the unknown into the rule and gathers the terms containing it. . Worth 2 points.

Argues that the condition must hold for EVERY input, so the factor depending on $a$ cannot be the one that vanishes. . Worth 2 points. needs an explanation, not just an answer

Part C 4 points

Tests the candidate in the other slot and reports what a specific input shows. . Worth 2 points.

Reaches a verdict on whether an identity element exists and grounds it in what the two-sided test showed, rather than in one side alone. . Worth 2 points. needs an explanation, not just an answer
8. A reading built by two machines, and run backwards . 12 points. Question 8 of 10.

A conveyor scale turns a mass $m$ in kilograms into a raw count with $r(m) = 8m + 20$ , and the display turns a raw count $r$ into a percentage with $P(r) = \dfrac{r}{4} - 1$ .
1. Part A.
  Write the percentage shown on the display as a single rule in terms of the mass.
  
  Model the situation Name your unknown first, then write every other quantity in terms of that one letter. 4 points
2. Part B.
  The display reads $30$ . What mass is on the scale?
  
  Carry your own answer forward Solve using the composite rule you built in part A, whatever form it came out in.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  Write the inverse of the composite rule, and say what its input and its output each represent. Explain why those two roles are forced by the composite rather than chosen.
  
  Carry your own answer forward Invert the composite rule you built in part A, whatever form it came out in.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 5 points
The answer

Part A

$P(r(m)) = \dfrac{8m + 20}{4} - 1 = 2m + 4$ .

Part B

$13$ kilograms.

Part C

The inverse is $\dfrac{x - 4}{2}$ . Its input is a display reading and its output is the mass in kilograms that produced it. The composite took masses in and gave readings out, so reversing it swaps those roles.

Worked solution

Part A

The scale runs first, so its whole rule drops into the display's input slot.

$P(r(m)) = \frac{8m + 20}{4} - 1 = (2m + 5) - 1 = 2m + 4.$

Every term of the raw count is divided by $4$ , not just the first one.

Part B

Set the composite equal to the reading and solve.

$2m + 4 = 30 \;\Longrightarrow\; 2m = 26 \;\Longrightarrow\; m = 13.$

Running the chain forwards confirms it: $r(13) = 8(13) + 20 = 124$ and $P(124) = 31 - 1 = 30$ .

Part C

Write $y = 2m + 4$ , swap the letters, and solve for the new output.

$x = 2y + 4 \;\Longrightarrow\; y = \frac{x - 4}{2}.$

Its input is a percentage shown on the display, and its output is the mass in kilograms behind that reading. Those roles are forced rather than chosen: the composite's inputs were masses and its outputs were readings, and an inverse turns each pair (mass, reading) into the pair (reading, mass). So the inverse's domain is the set of readings and its range is the set of masses, exactly the trade the swap performs. Part B was this inverse used once, taking the reading $30$ and returning the mass $13$ .

In one line

The display shows $2m + 4$ percent for a mass of $m$ kilograms, so a reading of $30$ comes from a mass of $13$ kilograms. Reversing the chain gives $\dfrac{x - 4}{2}$ , which takes a display reading in and returns the mass behind it, because an inverse swaps a rule's inputs with its outputs.

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 entire scale rule into the display rule's input slot, in that order. . Worth 2 points.

Divides both terms of the raw count by $4$ before subtracting $1$ . . Worth 2 points.

Part B 3 points

Sets the composite equal to the displayed reading and solves for the mass. . Worth 2 points.

Reports the answer as a mass in kilograms rather than a bare number. . Worth 1 point.

Part C 5 points

Produces the inverse rule by swapping and then solving. . Worth 2 points.

Says which quantity the inverse takes in and which it returns, attaching the right unit to each. . Worth 2 points.

Explains that the two roles are forced by domain and range trading places under the swap, not chosen for convenience. . Worth 1 point. needs an explanation, not just an answer
9. One pair of rules, composed both ways . 12 points. Question 9 of 10.

Let $f(x) = x^2 - 1$ , restricted to $x \ge 0$ , and let $g(x) = \sqrt{x + 1}$ .
1. Part A.
  Find simplified formulas for $f(g(x))$ and $g(f(x))$ .
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
2. Part B.
  State the inputs each of the two composites accepts.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
3. Part C.
  Explain what decides each composite's admissible inputs, and use your results from parts A and B to state whether $f$ and $g$ reverse each other, and on which sets.
  
  Carry your own answer forward Argue from the two formulas in part A and the two sets you reported in part B.
  
  Justify your claim State the claim, then give the reason it has to be true. 4 points
The answer

Part A

$f(g(x)) = x$ and $g(f(x)) = x$ .

Part B

$f \circ g$ accepts every $x \ge -1$ , and $g \circ f$ accepts every $x \ge 0$ .

Part C

The two steps fix them, not the simplified formula: an input must be legal for the inner rule and its output legal for the outer one. Both directions return the input, so the pair reverse each other, with $f$ carrying $x \ge 0$ onto $x \ge -1$ and $g$ carrying $x \ge -1$ back onto $x \ge 0$ .

Worked solution

Part A

Drop each whole rule into the other and simplify.

$f(g(x)) = \left(\sqrt{x + 1}\right)^2 - 1 = (x + 1) - 1 = x.$

$g(f(x)) = \sqrt{(x^2 - 1) + 1} = \sqrt{x^2} = x.$

The last step is valid because $f$ is restricted to $x \ge 0$ : the root returns the nonnegative number whose square is $x^2$ , and on this domain that number is $x$ itself. Without the restriction it would fail for every negative input.

Part B

Apply the two-part rule separately to each composite.

For $f \circ g$ , the inner rule needs $x + 1 \ge 0$ , so $x \ge -1$ ; its output $\sqrt{x + 1}$ is never negative, so it always lands inside $f$ 's restricted domain $x \ge 0$ , and nothing further is lost.

$f \circ g: \quad x \ge -1.$

For $g \circ f$ , the inner rule is the restricted $f$ , so $x \ge 0$ from the start; its output then needs $f(x) + 1 \ge 0$ , which is $x^2 \ge 0$ , true for every number.

$g \circ f: \quad x \ge 0.$

Part C

A composite's admissible inputs are decided by the two steps that built it, never by the formula they collapse into. Simplifying to $x$ erases every trace of the square root and of the restriction, but both were applied before the simplification and no amount of tidying undoes them.

$f \circ g: \ x \ge -1, \qquad g \circ f: \ x \ge 0.$

The two sets differ because the composites run their steps in opposite orders, so a different rule goes first and imposes its own demand first. Both compositions returning the input is precisely the two-direction test for a reversing pair, so $f$ on $x \ge 0$ and $g$ on $x \ge -1$ undo each other. The sets confirm the trade: $f$ takes inputs $x \ge 0$ and produces outputs $x \ge -1$ , and $g$ takes those very outputs as its inputs and hands back the original inputs.

In one line

Both composites simplify to $x$ , but $f \circ g$ accepts only $x \ge -1$ and $g \circ f$ only $x \ge 0$ , because each is limited by the two steps that built it rather than by its final formula. Since both directions return the input, $f$ on $x \ge 0$ and $g$ on $x \ge -1$ reverse each other, and the two sets trade places between them.

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 each whole rule into the other and simplifies both composites to the input. . Worth 2 points.

Justifies replacing $\sqrt{x^2}$ by $x$ from the restriction rather than assuming it. . Worth 2 points. needs an explanation, not just an answer

Part B 4 points

Checks the inner rule's own domain and then whether its output is legal for the outer rule, for each composite separately. . Worth 2 points.

Reports two different sets, each attached to the composite it belongs to. . Worth 2 points.

Part C 4 points

States that the two-step requirement, not the simplified formula, fixes each composite's admissible inputs. . Worth 2 points. needs an explanation, not just an answer

Uses what BOTH compositions produced to reach a verdict on the pair, naming the set each rule works on. . Worth 2 points.
10. What cancelling cannot give back . 12 points. Question 10 of 10.

Let $f(x) = \dfrac{x^2 - 25}{x + 5}$ .
1. Part A.
  Find $f(3)$ and $f(-1)$ , working from the rule exactly as it is written.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Simplify the rule, state the one input $f$ does not accept, and state the one output $f$ never produces.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 6 points
3. Part C.
  Taken on its own, the simplified rule from part B is defined at every real number. Explain why it is still not the same function as $f$ , and name the single input-output pair that one of the two owns and the other does not.
  
  Carry your own answer forward Use the excluded input and the unreachable output you identified in part B.
  
  Explain why it works A sentence or two. Reasons, not steps. 3 points
The answer

Part A

$f(3) = -2$ and $f(-1) = -6$ .

Part B

$f(x) = x - 5$ at every input except $x = -5$ , and the output $-10$ is never produced.

Part C

They agree everywhere $f$ is defined, but the simplified rule accepts one input more. It owns the pair $(-5, -10)$ , which $f$ does not, and a function is the pairing of inputs with outputs it carries, so the two are different functions however alike the formulas look.

Worked solution

Part A

Substitute each input into the top and the bottom, then divide.

$f(3) = \frac{3^2 - 25}{3 + 5} = \frac{-16}{8} = -2, \qquad f(-1) = \frac{(-1)^2 - 25}{-1 + 5} = \frac{-24}{4} = -6.$

Part B

Factor the top and divide out the common factor, remembering that the division was only ever available where the bottom is not zero.

$\frac{x^2 - 25}{x + 5} = \frac{(x - 5)(x + 5)}{x + 5} = x - 5, \qquad x \neq -5.$

The refused input is $-5$ , where the original denominator vanishes. For the outputs: at every input it does accept, $f$ behaves exactly like $x - 5$ , and $x - 5$ reaches $-10$ only at $x = -5$ . That input is unavailable, so $-10$ is the one number $f$ never returns.

Part C

A function is settled by which inputs it accepts and what it returns for each, not by how its formula is written. Compare the two at the one input where they can differ.

$\text{the simplified rule sends } -5 \mapsto -10, \qquad f \text{ gives } -5 \text{ no output at all.}$

Everywhere else the two agree exactly. But the simplified rule owns the pair $(-5, -10)$ , which $f$ does not, so the two collections of pairs are different and the rules are different functions. That single missing pair is also why $-10$ never appears among $f$ 's outputs. Cancelling a common factor rewrites how a rule looks; it cannot hand back an input the original rule never accepted.

In one line

$f(3) = -2$ and $f(-1) = -6$ . The rule simplifies to $x - 5$ at every input except $-5$ , and the single output it never produces is $-10$ . It is not the same function as $x - 5$ taken on its own, which owns the pair $(-5, -10)$ , because a function is decided by the pairs it holds rather than by the appearance of its formula.

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

Substitutes each input into both the numerator and the denominator, keeping the signs straight. . Worth 2 points.

Reports both outputs, each paired with the input that produced it. . Worth 1 point.

Part B 6 points

Factors and divides out the common factor, carrying the restriction along with the simplified rule. . Worth 2 points.

Takes the excluded input from the ORIGINAL denominator rather than from the simplified rule. . Worth 2 points.

Identifies the unreachable output by asking which input would have been needed to produce it. . Worth 2 points.

Part C 3 points

Explains that a function is the pairing of inputs with outputs it carries, so a rule accepting one extra input is a different function. . Worth 2 points. needs an explanation, not just an answer

Names the one pair that separates the two rules, the input where one has a value and the other has none. . Worth 1 point.

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 rule, two numbers and an expression 10 points

Part A 3

Substitutes each input in full, so that $(-4)^2$ is $16$ rather than $-16$ . 2
Reports both outputs, each labelled with the input it came from. 1

Part B 4

Substitutes the entire expression $t + 1$ into the square, not the letter alone. 2
Expands the square, distributes the $2$ across all three terms, and combines the constants. 2

Part C 3

Locates the definition's restriction on INPUTS rather than outputs, and applies it to the comparison made in part A. 2
States the condition that would actually break the definition, one input with two different outputs. 1

2. Undoing a rule, and a symbol that looks like a division 12 points

Part A 4

Swaps the two letters and then SOLVES the new equation for $y$ , rather than stopping at the swap. 2
Multiplies the whole of $x + 3$ by $4$ , not the $x$ alone. 2

Part B 4

Evaluates the inverse rule at the given input correctly. 2
Checks the value by running the original rule on it and comparing with the number the inverse was handed. 2

Part C 4

Evaluates $f$ at the input first and then takes the reciprocal of that output, rather than reciprocating the rule's parts. 2
States what question each of the two expressions answers, and uses that to say how the two are related. 2

3. Three rules from two, and the one that loses inputs 11 points

Part A 4

Combines the two rules at the same input and collects like terms. 2
Distributes the subtraction across BOTH terms of $g$ , so the constant changes sign. 2

Part B 3

Writes the quotient with $f$ above and $g$ below. 1
Sets the denominator rule equal to zero and reports BOTH inputs that solve it. 2

Part C 4

Reaches a verdict on whether an input making only the numerator zero belongs to the domain, and states the value the quotient takes there. 2
Justifies the difference by naming division by zero as the only thing a quotient forbids. 2

4. A service fee before the rate, or after it 12 points

Part A 4

Names two separate rules, one subtracting the fee and one multiplying by the rate. 2
Places the fee step INSIDE the conversion step, matching the order the desk uses. 2

Part B 3

Substitutes the amount into the composite, deducting the fee before applying the rate. 2
Reports the result as an amount in euros rather than a bare number. 1

Part C 5

Builds the rival's composite with the conversion inside and the fee subtracted afterwards. 2
Compares the two amounts at the stated exchange, naming which is larger and the size of the gap. 2
Ties the gap to the order of the two steps, not to a difference in the rate or the fee. 1

5. Two steps in, one formula out 12 points

Part A 4

Substitutes the entire rule for $g$ into $f$ 's input slot. 2
Clears the compound fraction correctly, arriving at a single quotient. 2

Part B 4

Applies both halves of the rule, testing the inner rule's own domain and then where its output is illegal for the outer rule. 2
Reports both excluded inputs and attributes each to the step that rejects it. 2

Part C 4

Names which of the two halves of the domain rule the shortcut leaves untested, rather than blaming an arithmetic slip in the simplification. 2
Explains what the simplification did to the evidence, and why the final formula cannot be used to recover the domain. 2

6. Cutting a rule down until it can be reversed 14 points

Part A 3

Produces two specific different inputs with equal outputs and shows the arithmetic for both. 2
States what the shared output means: the rule is not one-to-one, so it cannot be reversed as it stands. 1

Part B 5

Rewrites the rule so the input appears once, then swaps the letters and solves for the new output. 2
Keeps only the nonnegative root, using the restriction that moved onto the new variable. 2
States the inputs the inverse accepts, matching them to the outputs the restricted rule produced. 1

Part C 6

Carries out BOTH compositions and shows each one collapsing to the input. 2
Uses the restriction to justify replacing $\sqrt{(x - 3)^2}$ by $x - 3$ , rather than asserting it. 2
Explains which variable the restriction described before the swap and which it describes after, covering both the inverse's inputs and its outputs. 2

7. Two orders, then a hunt for an identity 11 points

Part A 3

Substitutes each pair in the order written, doing the multiplication before the additions. 2
Compares the two results and states what a single pair of inputs can settle about the operation. 1

Part B 4

Substitutes the unknown into the rule and gathers the terms containing it. 2
Argues that the condition must hold for EVERY input, so the factor depending on $a$ cannot be the one that vanishes. 2

Part C 4

Tests the candidate in the other slot and reports what a specific input shows. 2
Reaches a verdict on whether an identity element exists and grounds it in what the two-sided test showed, rather than in one side alone. 2

8. A reading built by two machines, and run backwards 12 points

Part A 4

Substitutes the entire scale rule into the display rule's input slot, in that order. 2
Divides both terms of the raw count by $4$ before subtracting $1$ . 2

Part B 3

Sets the composite equal to the displayed reading and solves for the mass. 2
Reports the answer as a mass in kilograms rather than a bare number. 1

Part C 5

Produces the inverse rule by swapping and then solving. 2
Says which quantity the inverse takes in and which it returns, attaching the right unit to each. 2
Explains that the two roles are forced by domain and range trading places under the swap, not chosen for convenience. 1

9. One pair of rules, composed both ways 12 points

Part A 4

Substitutes each whole rule into the other and simplifies both composites to the input. 2
Justifies replacing $\sqrt{x^2}$ by $x$ from the restriction rather than assuming it. 2

Part B 4

Checks the inner rule's own domain and then whether its output is legal for the outer rule, for each composite separately. 2
Reports two different sets, each attached to the composite it belongs to. 2

Part C 4

States that the two-step requirement, not the simplified formula, fixes each composite's admissible inputs. 2
Uses what BOTH compositions produced to reach a verdict on the pair, naming the set each rule works on. 2

10. What cancelling cannot give back 12 points

Part A 3

Substitutes each input into both the numerator and the denominator, keeping the signs straight. 2
Reports both outputs, each paired with the input that produced it. 1

Part B 6

Factors and divides out the common factor, carrying the restriction along with the simplified rule. 2
Takes the excluded input from the ORIGINAL denominator rather than from the simplified rule. 2
Identifies the unreachable output by asking which input would have been needed to produce it. 2

Part C 3

Explains that a function is the pairing of inputs with outputs it carries, so a rule accepting one extra input is a different function. 2
Names the one pair that separates the two rules, the input where one has a value and the other has none. 1

Multiple choice

Free response