Functions: Chapter Test

20 multiple-choice questions and 10 core practice problems, drawn from across the chapter and mixed together.

Multiple choice

20 questions, 100 points in total, 5 points each. Answer in any order and change your mind as often as you like. When you submit, your answers lock and every question shows its worked solution.

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Multiple choice 0 / 20 answered

Question 1 of 20

1

If $g(x) = 5 - 2x$ , what is $g(-3)$ ?

Answer choices for question 1 A $11$ B $-1$ C $-9$ D $6$
2

Advanced. This question goes beyond core Algebra I. It is not required by the course.

An operation is defined by $a \otimes b = 4a - b$ . What is $6 \otimes 7$ ?

Answer choices for question 2 A $-4$ B $17$ C $22$ D $24$
3

What is the domain of $h(x) = \sqrt{x + 3}$ ?

Answer choices for question 3 A all numbers $x \ge 3$ B all numbers except $-3$ C all numbers $x \ge -3$ D all numbers $x \le -3$
4

Advanced. This question goes beyond core Algebra I. It is not required by the course.

A courier's fee in dollars for a parcel of mass $m$ kilograms is $c(m) = 4m + 6$ , and the loyalty points awarded for a fee of $c$ dollars are $p(c) = 2c - 5$ . How many loyalty points does a $3$ kilogram parcel earn?

Answer choices for question 4 A $31$ B $10$ C $18$ D $1$
5

Advanced. This question goes beyond core Algebra I. It is not required by the course.

What is the inverse of $f(x) = 3x + 8$ ?

Answer choices for question 5 A $f^{-1}(x) = \dfrac{1}{3x + 8}$ B $f^{-1}(x) = \dfrac{x + 8}{3}$ C $f^{-1}(x) = \dfrac{x - 8}{3}$ D $f^{-1}(x) = 3x - 8$
6

Advanced. This question goes beyond core Algebra I. It is not required by the course.

Let $f(x) = x + 7$ and $g(x) = x^2 - 9$ . Which inputs are excluded from the domain of $\dfrac{f}{g}$ ?

Answer choices for question 6 A $3$ only B $-7$ C no inputs are excluded D $3$ and $-3$
7

Advanced. This question goes beyond core Algebra I. It is not required by the course.

A rule's graph is drawn below, crossed by one dashed vertical line and one dashed horizontal line. Taken together, what do the two line tests say about this graph?

Answer choices for question 7 A It is not a function, because a horizontal line meets it twice. B It is a function, but it is not one-to-one, so it has no inverse function. C It is a function and it is one-to-one, so it has an inverse function. D It is neither a function nor one-to-one, since both dashed lines meet the curve.
8

Advanced. This question goes beyond core Algebra I. It is not required by the course.

Let $f(x) = \dfrac{4}{x - 2}$ and $g(x) = x + 5$ . What is the domain of $f \circ g$ ?

Answer choices for question 8 A all numbers except $2$ B all numbers except $-5$ C all numbers D all numbers except $-3$
9

Advanced. This question goes beyond core Algebra I. It is not required by the course.

An operation is defined by $a \triangle b = a^2 + 2b$ . Which expression is $(k - 3) \triangle 5$ in simplest form?

Answer choices for question 9 A $k^2 - 6k + 19$ B $k^2 + 19$ C $k^2 - 6k + 14$ D $k^2 - 6k - 1$
10

Advanced. This question goes beyond core Algebra I. It is not required by the course.

Let $f(x) = x^2 + 3$ and $g(x) = 2x$ . What is $(f \circ g)(x)$ ?

Answer choices for question 10 A $2x^2 + 6$ B $2x^2 + 3$ C $4x^2 + 3$ D $x^2 + 2x + 3$
11

Advanced. This question goes beyond core Algebra I. It is not required by the course.

A function $f$ has domain all numbers except $4$ , and its range is all numbers except $0$ . What are the domain and the range of $f^{-1}$ ?

Answer choices for question 11 A Domain all numbers except $4$ ; range all numbers except $0$ . B Domain all numbers except $0$ ; range all numbers except $4$ . C Domain all numbers except $0$ ; range all numbers except $0$ . D Domain all numbers except $-4$ ; range all numbers except $0$ .
12

Advanced. This question goes beyond core Algebra I. It is not required by the course.

For $f(x) = x^2$ on all numbers and $g(x) = \sqrt{x}$ , the composition $f(g(x))$ collapses to $x$ . What does $g(f(x))$ give at the input $-5$ , and what does that settle?

Answer choices for question 12 A $-5$ , so both directions hold and the two rules are inverses. B $25$ , so the two rules are not inverses. C It has no value, because a square root cannot accept $-5$ . D $5$ , so the second direction fails and the two rules are not inverses.
13

Advanced. This question goes beyond core Algebra I. It is not required by the course.

Let $f(x) = x^2 - 4$ and $g(x) = x + 2$ . Which statement describes $\left(\dfrac{f}{g}\right)(x)$ completely?

Answer choices for question 13 A It equals $x - 2$ , for every input except $-2$ . B It equals $x - 2$ , for every input. C It equals $x - 2$ , for every input except $2$ . D It equals $x + 2$ , for every input except $-2$ .
14

Advanced. This question goes beyond core Algebra I. It is not required by the course.

Let $f(x) = x^2$ and $g(x) = \sqrt{x}$ . The composition $(f \circ g)(x)$ simplifies to $x$ . Which inputs does it accept?

Answer choices for question 14 A every number, since the composition simplifies to $x$ B every number except $0$ C every number $x \ge 0$ D every number $x > 0$
15

Advanced. This question goes beyond core Algebra I. It is not required by the course.

Let $a \star b = 3a - b$ . What is $(4 \star 2) \star 5$ ?

Answer choices for question 15 A $11$ B $5$ C $30$ D $25$
16

Advanced. This question goes beyond core Algebra I. It is not required by the course.

Let $f(x) = \dfrac{1}{x}$ and $g(x) = \dfrac{1}{x}$ . The composition $f(g(x))$ simplifies to $x$ . Which inputs does it accept?

Answer choices for question 16 A every input, since the composition simplifies to $x$ B every input except $0$ C every input except $0$ and $1$ D only the inputs $x > 0$
17

Advanced. This question goes beyond core Algebra I. It is not required by the course.

The rule $f(x) = (x - 2)^2$ is restricted to $x \ge 2$ , which makes it one-to-one. What is its inverse, and which inputs does that inverse accept?

Answer choices for question 17 A $f^{-1}(x) = \sqrt{x} + 2$ , accepting $x \ge 0$ B $f^{-1}(x) = \sqrt{x} + 2$ , accepting $x \ge 2$ C $f^{-1}(x) = \sqrt{x - 2}$ , accepting $x \ge 2$ D $f^{-1}(x) = \dfrac{1}{(x - 2)^2}$ , accepting every input except $2$
18

Advanced. This question goes beyond core Algebra I. It is not required by the course.

For which of these pairs is $f \circ g$ the same function as $g \circ f$ ?

Answer choices for question 18 A $f(x) = 2x$ and $g(x) = x + 1$ B $f(x) = x + 4$ and $g(x) = x + 9$ C $f(x) = x^2$ and $g(x) = x + 3$ D $f(x) = 3x$ and $g(x) = x - 5$
19

Advanced. This question goes beyond core Algebra I. It is not required by the course.

Let $f(x) = \sqrt{x - 1}$ and $g(x) = x^2$ . Which inputs does $(f \circ g)(x)$ accept?

Answer choices for question 19 A $x \ge 1$ B every number C $x \le -1$ or $x \ge 1$ D $-1 \le x \le 1$
20

Advanced. This question goes beyond core Algebra I. It is not required by the course.

Which of these operations has an identity element, a single fixed number $e$ with $a \ast e = a$ and $e \ast a = a$ for every $a$ ?

Answer choices for question 20 A $a \diamond b = 3a + b$ B $a \triangle b = ab + 1$ C $a \otimes b = a - b$ D $a \star b = a + b - 6$

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Where you stand

Read and evaluate function notation, substituting a number or a whole expression for the input, and apply the one-output-per-input definition, including its graphical form, the vertical line test. Review: What Is a Function?
Find a rule's domain by excluding the inputs that make a denominator zero or put a negative under a square root, and identify an output a rule never produces. Review: What Is a Function?
Build $f + g$ , $f - g$ and $\tfrac{f}{g}$ from two rules, and exclude every input where the denominator $g(x)$ is zero, including one that canceling would hide. Review: Combining and Composing Functions
Compose two rules in the order given, working from the inside out, both at a number and as a formula, and recognize that reversing the order usually changes the function. Review: Combining and Composing Functions
Decide a composite's domain from the two-part rule, an input must be legal for the inner rule and its output legal for the outer one, rather than reading the domain off the simplified formula. Review: Combining and Composing Functions
Find an inverse by writing $y = f(x)$ , swapping $x$ and $y$ and solving for $y$ ; read $f^{-1}$ as the undo rather than the reciprocal; and trade the domain and range between a function and its inverse. Review: Inverse Functions
Decide whether a rule is one-to-one, by the horizontal line test or by producing two inputs that share an output, restrict a domain to repair a rule that is not, and verify a proposed inverse by composing in BOTH directions. Review: Inverse Functions
Evaluate a defined binary operation by substituting its inputs into the rule, including a whole expression and a nested pair worked from the inside out, and test commutativity or an identity element rather than assuming one. Review: Defining New Operations

Core practice

10 problems from across the chapter. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 The complete plotted record

The graph shows the complete function $f$ . Find $f(-2)+2f(1)$ .

The complete graph of $f$ .

Text description of this figure

A square coordinate grid with equal unit lengths on both axes. The horizontal x-axis runs from negative four to four and the vertical y-axis from negative five to five, with tick marks, number labels and gridlines at every whole number, and the origin labeled 0. Four filled dots are plotted and they are the whole graph: one three units left of the y-axis and one unit above the x-axis, at negative three, one; one two units left and three units above, at negative two, three; one one unit right and four units below, at one, negative four; and one three units right and two units above, at three, two. No line or curve joins the dots, no dot is labeled with its coordinates, and no other point is shown.
1. Hint 1
  Read the height at each requested input before doing the arithmetic.
2. Hint 2
  Keep the factor outside the function attached to the output at input one.
Answer

$-5$ .

Full solution

The plotted heights give $f(-2)=3$ and $f(1)=-4$ .
Therefore
$f(-2)+2f(1)=3+2(-4),$
which is $-5$ .
The other plotted inputs do not enter this expression.

Answer

$-5$ .

Key idea

Function notation selects outputs by input before ordinary arithmetic combines them.
Problem 1: Mark complete
Problem 2 The allowed real inputs

Let $h(x)=\dfrac{\sqrt{x+3}\,\sqrt{3-x}}{x-3}$ wherever this expression is real and defined. State its domain, and decide whether it ever produces a positive output.
1. Hint 1
  Each of the two roots, and the denominator, restricts the input.
2. Hint 2
  Within the surviving interval, compare the signs of the numerator and denominator.
Answer

Domain: $-3\le x<3$ . No positive output is produced.

Full solution

The first root requires $x+3\ge0$ , so $x\ge-3$ .
The second root requires $3-x\ge0$ , so $x\le3$ .
The denominator also requires $x\ne3$ , leaving
$-3\le x<3.$
On that domain the numerator is nonnegative and the denominator is negative, so the quotient is at most zero.

At $x=-3$ , the output is zero.
At every other allowed input the numerator is positive, so the output is negative.
Thus no positive output occurs.

Answer

Domain: $-3\le x<3$ . No positive output is produced.

Key idea

Separate domain conditions must all hold, and signs can rule out entire classes of outputs.
Problem 2: Mark complete
Problem 3 Three combined records

Advanced. This question goes beyond core Algebra I. It is not required by the course.

For all real inputs, let $f(x)=x(x-2)$ and $g(x)=x^2-2x$ . Give expanded polynomial formulas for $f+g$ and $f-g$ , and give the simplest formula and complete domain for $f/g$ .
1. Hint 1
  Check whether the two written rules produce the same outputs before combining them.
2. Hint 2
  Any input that makes the divisor output zero remains excluded after simplification.
Answer

$(f+g)(x)=2x^2-4x$ ; $(f-g)(x)=0$ , both on all real inputs. $(f/g)(x)=1$ for $x\ne0,2$ .

Full solution

Expanding the first rule gives $f(x)=x^2-2x$ , identical to $g(x)$ .
Addition and subtraction therefore give
$\begin{aligned}(f+g)(x)&=2x^2-4x\\(f-g)(x)&=0.\end{aligned}$
Their domains remain all real numbers.

The divisor is $x(x-2)$ , so it is zero at $x=0$ and $x=2$ .
For every other input the same nonzero quantity appears above and below, giving
$(f/g)(x)=1.$
The two excluded inputs do not become legal when the fraction is canceled.

Answer

$(f+g)(x)=2x^2-4x$ ; $(f-g)(x)=0$ , both on all real inputs. $(f/g)(x)=1$ for $x\ne0,2$ .

Key idea

Identical function rules can simplify a quotient to a constant without removing the original divisor restrictions.
Problem 3: Mark complete
Problem 4 The two-stage run

Advanced. This question goes beyond core Algebra I. It is not required by the course.

A cutter produces $n(t)=7t+3$ tiles in a run lasting $t$ minutes, counting three test tiles at start-up and seven more each minute. Every tile has mass $25$ grams, and each batch is packed on a tray of mass $40$ grams, so a batch of $n$ tiles has packed mass $m(n)=25n+40$ grams. Give an expanded formula for the packed mass as a function of the run length, and use it for a run of $6$ minutes. A coworker computes $n(m(6))$ instead: what number does that give, and why is it not the packed mass?
1. Hint 1
  Decide which quantity each rule takes in and which it hands back before chaining the two.
2. Hint 2
  The mass rule accepts a number of tiles, so the tile rule runs first; for the coworker's expression, work out the inner result before the outer one.
Answer

$m(n(t))=175t+115$ grams, giving $1165$ grams for a $6$ minute run. The coworker gets $1333$ , a number of tiles, not a mass.

Full solution

The mass rule needs a number of tiles, so the tile rule runs first and its whole output drops into $m$ .
That gives
$m(n(t))=25(7t+3)+40,$
which expands to $175t+115$ grams.

A run of $6$ minutes produces $n(6)=45$ tiles, and the formula gives
$175(6)+115=1165.$
The packed batch has mass $1165$ grams, which is $25$ grams for each of the $45$ tiles plus the $40$ gram tray.

The coworker's order sends the run length into the mass rule first, so $m(6)=190$ , and then
$n(190)=1333.$
That output is a number of tiles, produced by treating $6$ minutes as $6$ tiles and $190$ grams as a run length, so it is not the packed mass.
Because the two stages measure different quantities here, only one order hands each rule an input it can accept.

Answer

$m(n(t))=175t+115$ grams, giving $1165$ grams for a $6$ minute run. The coworker gets $1333$ , a number of tiles, not a mass.

Key idea

Here the two stages measure different quantities, so only one order hands each rule a quantity it is built to accept.
Problem 4: Mark complete
Problem 5 Two linked entries

Advanced. This question goes beyond core Algebra I. It is not required by the course.

The one-to-one function $f(x)=3/(x+1)$ has domain $x\ne-1$ . Two records say $f(a)=6$ and $f^{-1}(b)=2$ . Find $a$ and $b$ .
1. Hint 1
  The forward entry and the inverse entry describe pairings in opposite directions.
2. Hint 2
  Solve the first equation for its input, and reverse the second entry before substituting into the rule.
Answer

$a=-\frac12$ , $b=1$ .

Full solution

The first entry gives $3/(a+1)=6$ , where $a+1\ne0$ .
Clearing the denominator and solving gives
$\begin{aligned}3&=6a+6\\a&=-\frac12.\end{aligned}$
This allowed input produces $3/(1/2)=6$ .

The inverse record $f^{-1}(b)=2$ means $f(2)=b$ .
Hence
$b=\frac3{2+1},$
so $b=1$ .
Reversing the checked pair $(2,1)$ confirms the inverse record.

Answer

$a=-\frac12$ , $b=1$ .

Key idea

Forward and inverse notation can be combined by keeping track of which member of each pair is the original input.
Problem 5: Mark complete
Problem 6 A nested symbol

Advanced. This question goes beyond core Algebra I. It is not required by the course.

For all real inputs, define $a\diamond b=(a-b)^2$ . Write $(t\diamond1)\diamond(t\diamond(-1))$ as an expanded polynomial, and state the smallest value that expression can take and the input $t$ that produces it.
1. Hint 1
  Compute each inner rule before applying the square of a difference to the results.
2. Hint 2
  Simplify the difference of the two inner squares before squaring it, then recall which values a square can take.
Answer

$16t^2$ ; its smallest value is $0$ , produced only by $t=0$ .

Full solution

The two inner outputs are $(t-1)^2$ and $(t+1)^2$ .
Their difference is
$\begin{aligned}(t-1)^2-(t+1)^2&=-4t.\end{aligned}$
Squaring this difference gives
$(-4t)^2=16t^2.$

A square is never negative, so $16t^2\ge0$ for every real $t$ .
At $t=0$ the expression gives
$16(0)^2=0.$
Every other input has $t^2>0$ , so its output is positive.
The smallest value is therefore $0$ , produced only by $t=0$ .

Answer

$16t^2$ ; its smallest value is $0$ , produced only by $t=0$ .

Key idea

Nested custom operations take whole inner outputs, and when the result is a square that can equal zero, its least value follows from a square never being negative.
Problem 6: Mark complete
Problem 7 The simplified outputs

Advanced. This question goes beyond core Algebra I. It is not required by the course.

Let $f(u)=\sqrt u$ for $u\ge0$ , and let $g(x)=(x-2)/(x-2)$ for $x\ne2$ . Give the simplest formulas and complete domains of both $f\circ g$ and $g\circ f$ .
1. Hint 1
  A simplified output does not erase restrictions from either stage of a composition.
2. Hint 2
  In the second order, determine when the square root equals the input rejected by the outer rule.
Answer

$(f\circ g)(x)=1$ for $x\ne2$ . $(g\circ f)(x)=1$ for $x\ge0$ and $x\ne4$ .

Full solution

For $f\circ g$ , the inner rule requires $x\ne2$ and produces $1$ .
The outer rule accepts that output, so
$f(g(x))=\sqrt1,$
giving constant value $1$ for exactly $x\ne2$ .

For $g\circ f$ , the inner root requires $x\ge0$ .
Its output must not equal $2$ , so
$\sqrt x\ne2$
excludes $x=4$ .
At every surviving input, the numerator and nonzero denominator in $g$ are equal, so the output is $1$ .
The formulas agree, but their domains differ.

Answer

$(f\circ g)(x)=1$ for $x\ne2$ . $(g\circ f)(x)=1$ for $x\ge0$ and $x\ne4$ .

Key idea

Compositions with the same simplified formula can still be different functions because their legal inputs differ.
Problem 7: Mark complete
Problem 8 A restricted machine

Advanced. This question goes beyond core Algebra I. It is not required by the course.

The function $R(u)=2u+5$ accepts exactly $-2\le u\le3$ . The function $g(x)=x^2-1$ accepts all real inputs. Give an expanded formula and the complete domain for $R\circ g$ , and decide whether $(R\circ g)(3)$ exists.
1. Hint 1
  The inner output must fall inside the outer machine’s allowed interval.
2. Hint 2
  Apply both endpoints of that interval to the expression produced by the inner rule.
Answer

$(R\circ g)(x)=2x^2+3$ for $-2\le x\le2$ ; $(R\circ g)(3)$ is undefined.

Full solution

Substitution gives
$R(g(x))=2(x^2-1)+5,$
which simplifies to $2x^2+3$ .
The inner output must satisfy
$-2\le x^2-1\le3.$
The lower condition is automatic since
$x^2-1\ge-1.$
The upper condition gives
$x^2\le4,$
or
$-2\le x\le2.$

At $x=3$ , the inner output is $8$ , which exceeds the outer limit $3$ .
Thus the composition is undefined there, even though the polynomial expression alone could be evaluated.

Answer

$(R\circ g)(x)=2x^2+3$ for $-2\le x\le2$ ; $(R\circ g)(3)$ is undefined.

Key idea

A stated domain remains part of a function even when its algebraic formula accepts additional inputs.
Problem 8: Mark complete
Problem 9 The proposed reversal

Advanced. This question goes beyond core Algebra I. It is not required by the course.

Let $f(x)=\sqrt{x+1}-2$ on $x\ge-1$ . Dara proposes $g(x)=(x+2)^2-1$ on all real inputs as its inverse. Is Dara’s proposal, rule and domain together, correct? State the inverse function with its domain and range, and verify both compositions.
1. Hint 1
  Find the outputs actually produced by the original square root rule.
2. Hint 2
  The inverse input set must equal that output set, and the restriction controls the square root of a square in the check.
Answer

No. $f^{-1}(x)=(x+2)^2-1$ for $x\ge-2$ , with range $y\ge-1$ ; both compositions return $x$ on the respective domains.

Full solution

The root is nonnegative, so the range of $f$ is $y\ge-2$ .
Every such value is reached by input $(y+2)^2-1$ .
Swapping $x$ and $y$ in the original rule gives
$x=\sqrt{y+1}-2.$
Thus $y=(x+2)^2-1$ , with inverse domain $x\ge-2$ and range $y\ge-1$ .

With $g$ restricted correctly, for $x\ge-1$ the expression $g(f(x))$ is $(\sqrt{x+1})^2-1$ , which equals $x$ .
For $x\ge-2$ ,
$f(g(x))=\sqrt{(x+2)^2}-2.$
Here $x+2\ge0$ , so this equals $x+2-2$ , hence $x$ .
Both checks hold.
Dara’s unrestricted rule fails, for example, at $x=-3$ , where
$f(g(-3))=-1.$

Answer

No. $f^{-1}(x)=(x+2)^2-1$ for $x\ge-2$ , with range $y\ge-1$ ; both compositions return $x$ on the respective domains.

Key idea

An inverse formula needs the original range as its domain for both compositions to undo correctly.
Problem 9: Mark complete
Problem 10 A reduced graph

Advanced. This question goes beyond core Algebra I. It is not required by the course.

The graph shows the complete function $f$ . Delete as few plotted points as possible so the remaining function has an inverse, while retaining the point with input $-2$ . Give every inverse pair and its domain and range, and verify that the two functions undo each other.

The complete graph of $f$ .

Text description of this figure

A coordinate grid with equal unit lengths on both axes. The horizontal x-axis runs from negative three to six and the vertical y-axis from negative two to five, with tick marks, number labels and gridlines at every whole number, and the origin labeled 0. Four filled dots are plotted and they are all the points of the function: one two units left of the y-axis and one unit above the x-axis, at negative two, one; one on the y-axis four units above the x-axis, at zero, four; one three units right and one unit above, at three, one; and one five units right and one unit below, at five, negative one. No line or curve joins the dots, no dot is labeled with its coordinates, and no dot is marked out from the others.
1. Hint 1
  Each retained output must have a single original input if reversal is to be a function.
2. Hint 2
  Locate the repeated height, respect the point that must stay, and reverse the remaining pairs.
3. Hint 3
  Follow one retained pair forward and then back to check a direction of the undoing.
Answer

Delete $(3,1)$ . Inverse pairs: $(1,-2)$ , $(4,0)$ , $(-1,5)$ . Domain $\{-1,1,4\}$ ; range $\{-2,0,5\}$ . Both compositions return their inputs.

Full solution

Inputs $-2$ and $3$ both produce output $1$ .
Because the point at input $-2$ must stay, the point $(3,1)$ must be deleted.
The remaining three heights are distinct, so one deletion is sufficient and is the minimum.

The retained pairs are $(-2,1)$ , $(0,4)$ , and $(5,-1)$ .
Reversing them gives $(1,-2)$ , $(4,0)$ , and $(-1,5)$ .
The inverse domain is $\{-1,1,4\}$ and its range is $\{-2,0,5\}$ .

Each retained forward arrow is exactly reversed: the routes are $-2$ to $1$ to $-2$ , $0$ to $4$ to $0$ , and $5$ to $-1$ to $5$ .
Starting instead at $1$ , $4$ , or $-1$ follows each route in reverse and also returns to its start.
Thus both compositions are verified on all of their inputs.

Answer

Delete $(3,1)$ . Inverse pairs: $(1,-2)$ , $(4,0)$ , $(-1,5)$ . Domain $\{-1,1,4\}$ ; range $\{-2,0,5\}$ . Both compositions return their inputs.

Key idea

Removing repeated outputs can make a finite function invertible, after which reversing every pair verifies both directions.
Problem 10: Mark complete

Star problems Ten optional challenges to stretch your reasoning Advanced. This problem set goes beyond core Algebra I. You can skip it. → Continue to Chapter 12: Graphing Functions →

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