Algebra 3 Quiz: Composite Functions

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Composite FunctionsQuestion 1 of 20

Let $f(x)=\frac{1}{x-2}$ and $g(x)=\sqrt{x}$ . Which set is the domain of $(f\circ g)(x)$ ?

x\ge0

and

x\ne2

x\ge0

x\ge0

and

x\ne4

x>0

and

x\ne4

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Algebra 3 Quiz

Algebra 3 Quiz: Composite Functions

Practice Composite Functions in Algebra 3 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Composite Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 3.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Let $f(x)=\frac{1}{x-2}$ and $g(x)=\sqrt{x}$ . Which set is the domain of $(f\circ g)(x)$ ?

$x\ge0$ and $x\ne2$
$x\ge0$
$x\ge0$ and $x\ne4$ (correct answer)
$x>0$ and $x\ne4$

Explanation: Whenever you see a composition like

(f\circ g)(x)

, remember that the domain must satisfy both functions: the input has to work in

g

, and the output of

g

has to work in

f

. Start with

g(x)=\sqrt{x}

, so you already need

x\ge 0

. Now compute the composition:

(f\circ g)(x)=f(\sqrt{x})=\frac{1}{\sqrt{x}-2}.

The denominator forces

\sqrt{x}\ne 2

,which means

x\ne 4

. Thus the domain is

x\ge 0

and

x\ne 4

. The choice that says "

x\ge0

and

x\ne2

" is a classic trap: it nowhere checks

f

's restriction on

\sqrt{x}

instead of on

x

. Actually

x=2

is fine, since

\sqrt2-2\ne0

. The choice that simply says "

x\ge0

" misses the denominator issue entirely. And the choice that says "

x>0

and

x\ne4

" incorrectly excludes

x=0

, but

\sqrt0=0

, so

f(0)=1/(0-2)

is defined. So for composition problems, run the input through the inner function first,then ask whether that result is allowed by the outer function. Check both layers and you will avoid the most common domain mistakes.

Question 2

A store applies a 20-dollar-off coupon before adding 7% sales tax. If an item's original price is $p$ dollars, which expression gives the final price $F(p)$ ?

$F(p)=1.07p-21.40$ (correct answer)
$F(p)=1.07p-20$
$F(p)=0.07p-20$
$F(p)=p-20$

Explanation: When you see a coupon applied before tax, think of it as a two-step process: first adjust the price, then apply the multiplier for sales tax. The item starts at

p

dollars, and the 20-dollar coupon is applied first, so the taxable price is

p-20

. Adding 7% tax means multiplying by

1.07

, giving

F(p)=1.07(p-20)

. Distributing, this becomes

F(p)=1.07p-21.40

, which is the right expression. The choice

F(p)=1.07p-20

is the most common trap: it applies the coupon after tax, effectively charging tax on the original price and then subtracting 20 from the total. That would be correct only if the store taxed first and then applied the coupon. The choice

F(p)=0.07p-20

uses only the tax rate instead of the full multiplier

1.07

, so it calculates the tax as a separate amount and then subtracts the coupon, which misses the original price entirely. Finally,

F(p)=p-20

ignores the sales tax completely. A good strategy for these problems is to identify the order of operations in the wording. If a discount happens before tax, write

(p-\text{discount})

first, then multiply by

1+\text{tax rate}

. If tax is added first, multiply first and subtract after. Expanding the parentheses shows why the 21.40 appears.

Question 3

Let $f(x)=2x+1$ for $x<0$ and $f(x)=x^2-1$ for $x\ge0$ . If $g(x)=2-x$ , find $(f\circ g)(2)$ .

$-3$
$1$
$5$
$-1$ (correct answer)

Explanation: Whenever you see a composite function with a piecewise definition, work from the inside out. First evaluate the inner function, then decide which piece of

f

applies using the value you got, not the original

x

. Here,

g(2)=2-2=0

, so

(f\circ g)(2)=f(0)

. Since

0

satisfies

x\ge 0

, you must use the second branch:

f(0)=0^2-1=-1

. So the value is

-1

. Now look at the wrong answers to see what they represent. Choosing

1

means you used the first branch

2x+1

for

f(0)

, but that branch requires

x<0

, and

0

is not less than

0

. Choosing

5

means you evaluated

f(2)

instead of

f(g(2))

2(2)+1=5

, which also uses the wrong branch since

2\ge 0

. Choosing

-3

comes from reversing the composition or stacking errors: if you incorrectly compute

f(2)=5

, then

g(5)=2-5=-3

, but the composition asks for

f(g(2))

, not

g(f(2))

. A useful habit for piecewise composition: evaluate the inner function first, write down its exact output, and then check which interval that output falls into before choosing a branch. This prevents the most common traps of using the wrong piece or the wrong order.

Question 4

Let $f(x)=3x-1$ and $g(x)=x^2+2$ . If $h(x)=(f\circ g)(x)-(g\circ f)(x)$ , which expression is equivalent to $h(x)$ ?

$-6x^2-6x+2$
$6x^2-6x-2$
$-6x^2+6x+2$ (correct answer)
$6x^2+6x+2$

Explanation: Whenever you see function composition like

(f\circ g)(x)

, read it as

f(g(x))

: apply the inner function first, then plug that entire result into the outer function. On questions like this, the safest process is to compute

f(g(x))

and

g(f(x))

separately before subtracting. Here

f(g(x))=3(x^2+2)-1=3x^2+5

. And

g(f(x))=(3x-1)^2+2=9x^2-6x+3

. Subtracting:

(3x^2+5)-(9x^2-6x+3)=-6x^2+6x+2

. So the expression

-6x^2+6x+2

is correct. The distractor

-6x^2-6x+2

usually comes from squaring

(3x-1)

9x^2+6x+1

; the cross term inside

g(f(x))

has the wrong sign, which flips the middle term in the final answer. The choice

6x^2-6x-2

is exactly what you get if you reverse the order and compute

(g\circ f)-(f\circ g)

instead. The remaining choice

6x^2+6x+2

combines that reversal with a sign error in the squared binomial and a constant slip; testing your result at

x=0

x=1

would reveal it doesn't match. Study tip: When subtracting one composite expression from another, put parentheses around the entire second expression and distribute the minus sign term-by-term. Also remember

(a-b)^2=a^2-2ab+b^2

— the middle sign is easy to drop.

Question 5

Let $f(x)=\frac{x-1}{x+2}$ and $g(x)=2x+1$ . Which value is excluded from the domain of $(f\circ g)(x)$ ?

$x=0$
$x=-2$
$x=-\frac{3}{2}$ (correct answer)
$x=\frac{1}{2}$

Explanation: When you see a composition like

(f\circ g)(x)

, the key is to remember that the input to

f

is not

x

; it is

g(x)

. So you need

g(x)

to be defined, and then you need

g(x)

to be in the domain of

f

. Here,

g(x)=2x+1

is defined for all real

x

, and

f

is undefined only when its denominator is zero:

x+2=0 \quad\Rightarrow\quad x=-2.

So in the composition,

f(g(x))=\frac{(2x+1)-1}{(2x+1)+2}=\frac{2x}{2x+3},

you must exclude any

x

where

g(x)=-2

2x+1=-2 \quad\Rightarrow\quad 2x=-3 \quad\Rightarrow\quad x=-\frac{3}{2}.

Thus

x=-\frac{3}{2}

is the excluded value. The choice

x=-2

is a trap: it is excluded from

f

's own domain, but that does not matter unless

g(x)

actually equals

-2

, which happens at

-\frac{3}{2}

, not at

-2

. The choice

x=0

makes the numerator zero, giving

f(g(0))=f(1)=0

, which is perfectly allowed. The choice

x=\frac{1}{2}

gives

g\left(\frac12\right)=2

, and

f(2)=\frac14

, so it is also valid. Study tip: for a composition, always substitute

g(x)

into

f

's denominator and solve

g(x)=\text{excluded value}

, rather than using

x

alone.

Question 6

Let $f(x)=\sqrt{x}$ and $g(x)=\frac{x-1}{x+2}$ . What is the domain of $(f\circ g)(x)$ ?

$x\le-2$ or $x\ge1$
$x<-2$ or $x\ge1$ (correct answer)
$-2<x\le1$
$x>-2$

Explanation: Whenever you see a composite function like

(f\circ g)(x)

, think in two layers: first,

g(x)

must be defined; second, its output must be in the domain of

f

. Since

f(x)=\sqrt{x}

, you need

g(x)\ge 0

. So here you need

\frac{x-1}{x+2}\ge 0

and

x\ne -2

. Solve that rational inequality by testing intervals around the critical values

x=-2

and

x=1

. For

x<-2

, numerator and denominator are both negative, so the fraction is positive — good. For

-2<x<1

, the fraction is negative — not allowed. For

x>1

, it is positive — good. At

x=1

, the fraction equals

0

, and

\sqrt{0}

is defined, so include

1

. At

x=-2

g(x)

is undefined, so exclude it. Thus the domain is

x<-2 \quad \text{or} \quad x\ge 1.

The choice saying

x\le -2

x\ge1

is close but wrongly includes

x=-2

. The choice

-2<x\le1

includes exactly the interval where the fraction is negative, so it fails. And

x>-2

also covers that negative interval while omitting the valid region

x<-2

. The correct domain keeps the denominator-zero value excluded but includes the numerator-zero value. A useful habit for composite domain problems: make a sign chart for the inside function's output, exclude values where it is undefined, and include values where the outside function can accept the result. That two-step check avoids most domain traps.

Question 7

Let $f(x)=\frac{1}{x+1}$ and $g(x)=\frac{1}{x}$ . Which set is the domain of $(f\circ g)(x)$ ?

$x\ne0$ and $x\ne1$
$x\ne-1$
$x\ne0$
$x\ne0$ and $x\ne-1$ (correct answer)

Explanation: When you see composition

(f\circ g)(x)

, you're plugging one function into another. The domain is not automatically the domain of the outer function; it's all inputs that are allowed in the inner function and whose outputs are then allowed in the outer function. Start with

g(x)=\frac1x

, which requires

x\ne0

. Now compute

f(g(x))=f\left(\frac1x\right)=\frac{1}{\frac1x+1}

. The outer function

f

is undefined when its input equals

-1

. So you also need

\frac1x\ne -1

. Solving

\frac1x=-1

gives

x=-1

. Therefore

x=0

is excluded by

g

, and

x=-1

is excluded because it makes

g(x)=-1

,which is outside the domain of

f

. The correct domain is

x\ne0

and

x\ne-1

. Why the others miss the full picture: the choice

x\ne0

and

x\ne1

includes an unnecessary restriction — at

x=1

g(1)=1

, then

f(1)=\frac12

, so it's perfectly defined. The choice

x\ne-1

alone forgets that

g

itself is undefined at

0

. The choice

x\ne0

alone misses the composition break at

x=-1

, where

g(-1)=-1

, causing division by zero in

f

. The correct answer must combine both restrictions: the inner function's domain and the outer function's domain applied to the inner output. Strategy: for composition questions, find the inner domain first, then set the inner output equal to whatever makes the outer function undefined, and exclude those

x

-values too. This two-step check prevents missing restrictions.

Question 8

Let $f(x)=x^2-3x$ and $g(x)=2x+1$ . Find all $x$ for which $(f\circ g)(x)=(g\circ f)(x)$ .

$x=\frac{-2\pm\sqrt{10}}{2}$ (correct answer)
$x=-2\pm\sqrt{10}$
$x=\frac{-2\pm\sqrt{10}}{4}$
$x=\frac{2\pm\sqrt{10}}{2}$

Explanation: Whenever you see a composition question like this, your first step should be to build each composite separately. Remember

(f\circ g)(x)

means plug

g(x)

into

f

, while

(g\circ f)(x)

means plug

f(x)

into

g

. Here,

(f\circ g)(x)=(2x+1)^2-3(2x+1)=4x^2-2x-2

, and

(g\circ f)(x)=2(x^2-3x)+1=2x^2-6x+1

. Set them equal:

4x^2-2x-2=2x^2-6x+1 \Rightarrow 2x^2+4x-3=0.

By the quadratic formula,

x=\frac{-4\pm\sqrt{4^2-4(2)(-3)}}{2(2)} =\frac{-4\pm\sqrt{40}}{4} =\frac{-4\pm 2\sqrt{10}}{4} =\frac{-2\pm\sqrt{10}}{2}.

So the correct expression is

\frac{-2\pm\sqrt{10}}{2}

. The choice

-2\pm\sqrt{10}

is double the correct answer; it treats the fraction

\frac{-4\pm2\sqrt{10}}{4}

as if only the numerator were divided by

2

. The choice

\frac{-2\pm\sqrt{10}}{4}

comes from factoring

2

out of the numerator but then failing to cancel it with the denominator

4

. The choice

\frac{2\pm\sqrt{10}}{2}

uses

+4

instead of

-4

for the

-b

term in the quadratic formula — a sign error. For your study tip: whenever you get a result involving radicals, always check whether you can simplify the fraction by canceling common factors, and double-check the sign on

b

in the quadratic formula. Compositions just ask you to substitute carefully — once the two expressions are set equal, it's just a quadratic.

Question 9

Let $f(x)=3x-2$ . If $(f\circ g)(x)=x$ , which could be $g(x)$ ?

$g(x)=\frac{x-2}{3}$
$g(x)=\frac{x+2}{3}$ (correct answer)
$g(x)=3x+2$
$g(x)=x+2$

Explanation: When you see

(f\circ g)(x)=x

, you're being asked to undo

f

g

must be the inverse of

f

, because plugging

g(x)

into

f

returns the original

x

. Start with

f(g(x))=x

, so

3g(x)-2=x

. Solve for

g(x)

: add

2

to both sides to get

3g(x)=x+2

, then divide by

3

. This gives

g(x)=\frac{x+2}{3}

. You can verify:

f\left(\frac{x+2}{3}\right)=3\left(\frac{x+2}{3}\right)-2=x+2-2=x

. The distractor

g(x)=\frac{x-2}{3}

comes from using

x-2

instead of

x+2

, a sign error; plugging it in gives

x-4

, not

x

. The expression

g(x)=3x+2

misunderstands "undoing" as "doing same operation with different numbers"; plugging it in gives

9x+4

, definitely not

x

. Similarly,

g(x)=x+2

only reverses the subtraction before multiplication; plugging it in gives

3x+4

, not

x

. All three fail the defining test that composition equals identity. So on this exam, when a composition equals

x

, solve directly by isolating

g(x)

, then verify by substituting back into

f

. Watch the order and sign of inverse operations: for

3x-2

, add

2

first, evinthen divide by

3

Question 10

If $(f\circ g)(x)=x^2+4x+7$ and $g(x)=x+2$ , which of the following is $f(x)$ ?

$f(x)=x^2+4x+7$
$f(x)=x^2+3$ (correct answer)
$f(x)=x^2+7$
$f(x)=x^2+4x+3$

Explanation: When you see a composite function like

(f\circ g)(x)

, remember it means

f(g(x))

. Here

g(x)=x+2

, so you need a function

f

such that

f(x+2)=x^2+4x+7

. Let

y=x+2

, so

x=y-2

. Substitute into the composite:

f(y)=(y-2)^2+4(y-2)+7

. Expanding,

y^2-4y+4+4y-8+7=y^2+3

. Since

y

is just a placeholder,

f(x)=x^2+3

. The choice

x^2+4x+7

is the original composite itself—it confuses the input

x

with the input

x+2

. The choice

x^2+7

misses the constant adjustment from substituting

x=y-2

, likely because it ignores the linear terms. The choice

x^2+4x+3

also fails: if you try

f(x)=x^2+4x+3

, then

f(x+2)=x^2+8x+15

, not the given composite. Your takeaway: when solving for an outer function, substitute

u=g(x)

, rewrite the composite entirely in terms of

u

, then replace

u

with

x

. This prevents the common trap of treating the composite expression as already being

f(x)

Question 11

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

$x\ge 0$
$x\ge 5$
$x\le -5$ or $x\ge 5$
$\text{all real numbers}$ (correct answer)

Explanation: Whenever you see a composition like

(f\circ g)(x)

, your first move is to write it out:

f(g(x))

. Don't just look at

f

's domain and apply it to

x

directly. Here,

f(x)=\sqrt{x-5}

, so

f(g(x))=\sqrt{g(x)-5}

. Since

g(x)=x^2+5

, this becomes

\sqrt{(x^2+5)-5}=\sqrt{x^2}.

Now ask: when is

\sqrt{x^2}

defined? The radicand is

x^2

, and

x^2\ge 0

for every real number

x

. So the domain is all real numbers. The choice $x\ge 0$ is too restrictive: negative values of

x

still produce a nonnegative

x^2

, so they are perfectly valid.
The choice $x\ge 5$ confuses the range of

g

—which is always at least

5

—with a restriction on

x

itself;

x

can be anything.
The choice $x\le -5$ or $x\ge 5$ would be the domain for something like

\sqrt{x^2-25}

, but here the radicand simplifies to

x^2

, not

x^2-25

. That choice over-restricts by solving an inequality that doesn't apply. For your study strategy: always substitute and simplify the composition first. A restricted domain like

x\ge 5

for

f

can disappear after composing, because

g

's outputs may already satisfy

f

's requirements. On exam day, simplify the composed expression before deciding the domain.

Question 12

Let $f(x)=\sqrt{x+3}$ and $g(x)=x^2-1$ . Which statement about the domains of $f\circ g$ and $g\circ f$ is true?

Domain of $f\circ g$ is all real numbers; domain of $g\circ f$ is $x\ge-3$ . (correct answer)
Domain of $f\circ g$ is all real numbers; domain of $g\circ f$ is all real numbers.
Domain of $f\circ g$ is $x\ge-3$ ; domain of $g\circ f$ is all real numbers.
Domain of $f\circ g$ is $x\ge-1$ ; domain of $g\circ f$ is $x\ge-3$ .

Explanation: Whenever you see composite functions, first ask: what input

x

keeps every evaluation legal? For

f\circ g

x

goes into

g

first.

g(x)=x^2-1

is a polynomial, so its domain is all real numbers. Then its output must be in the domain of

f

, which requires the input to

f

to be

\ge -3

. So you need

g(x)+3=(x^2-1)+3=x^2+2\ge 0

. Since

x^2+2

is always positive,

f\circ g

has domain all real numbers. For

g\circ f

x

goes into

f

first, so you need

x+3\ge 0

, meaning

x\ge -3

. The outer function

g

accepts any real output, so the domain of

g\circ f

is exactly

x\ge -3

. The statement saying both domains are all real numbers comes from simplifying

g(f(x))=x+2

and forgetting that the original

f

is undefined for

x<-3

. The statement claiming

f\circ g

x\ge -3

and

g\circ f

is all real numbers reverses which composition carries the restriction:

f\circ g

has no restriction, while

g\circ f

requires

x\ge -3

. The statement claiming

f\circ g

x\ge -1

has no basis:

x^2+2\ge 0

holds for every real

x

, not just

x\ge -1

. Study tip: To find the domain of a composition, first restrict inputs to the inner function's domain, then require the inner output to satisfy the outer function's domain. Never rely only on the simplified final formula.

Question 13

Let $f(x)=x^2-4$ and $g(x)=\sqrt{x+2}$ . How many real solutions does the equation $(f \circ g)(x)=(g \circ f)(x)$ have?

0 (correct answer)
1
2
3

Explanation: Compute

f(g(x))=x-2

for

x\ge -2

, and

g(f(x))=\sqrt{x^2-2}

for

x^2\ge 2

. The equation

x-2=\sqrt{x^2-2}

forces

x\ge 2

. Squaring gives

x^2-4x+4=x^2-2

, hence

x=3/2

, which contradicts

x\ge 2

. So there are no real solutions.

Question 14

Suppose $f(x)=2x-5$ and $(f \circ g)(x)=6x+1$ . What is $(g \circ f)(x)$ ?

$6x-11$
$6x-12$ (correct answer)
$6x+3$
$3x-2$

Explanation: From

2g(x)-5=6x+1

, solve

g(x)=3x+3

. Then

g(f(x))=3(2x-5)+3=6x-12

. The other expressions come from sign errors or composing in the wrong order.

Question 15

Let $f(x)=\sqrt{x+1}$ and $g(x)=x^2-1$ . Which statement is true?

$(f \circ g)(x)=x$ for all real $x$ .
$(f \circ g)(x)=|x|$ for all real $x$ . (correct answer)
$(g \circ f)(x)=x$ for all real $x$ .
$(g \circ f)(x)=|x|$ for all real $x$ .

Explanation:

(f\circ g)(x)=\sqrt{x^2}=|x|

for all real

x

(g\circ f)(x)=x

only for

x\ge -1

, not for all real

x

. Thus the true statement is the one saying

f\circ g

equals

|x|

Question 16

Let $f(x)=x^2-1$ and $g(x)=2\sin x$ . How many values of $x$ in $[0,2\pi)$ satisfy $(f \circ g)(x)=0$ ?

2
3
4 (correct answer)
5

Explanation:

(f\circ g)(x)=4\sin^2 x-1

. Setting this to zero gives

\sin x=\pm\frac12

. In

[0,2\pi)

\sin x=\frac12

\pi/6

and

5\pi/6

, and

\sin x=-\frac12

7\pi/6

and

11\pi/6

, for 4 solutions.

Question 17

Let $f(x)=\frac{1}{x^2}$ and $g(x)=\sqrt{x+2}$ . What is the domain of $(f \circ g)(x)$ ?

$[-2,\infty)$
$(-2,0)\cup(0,\infty)$
$(-\infty,-2]\cup[2,\infty)$
$(-2,\infty)$ (correct answer)

Explanation:

(f\circ g)(x)=\frac{1}{(\sqrt{x+2})^2}=\frac{1}{x+2}

. The square root requires

x\ge -2

, and the denominator requires

x+2\neq 0

, so

x>-2

. The interval including -2 makes the denominator zero, and the other intervals are not related to the actual restrictions.

Question 18

Let $f(x)=\sqrt{x-a}$ and $g(x)=x^2-2$ . If the domain of $(f \circ g)(x)$ is $(-\infty,-3]\cup[3,\infty)$ , what is the value of $a$ ?

5
6
7 (correct answer)
9

Explanation:

(f\circ g)(x)=\sqrt{x^2-2-a}

. Its domain is

x^2\ge a+2

. For the boundary to be

\pm 3

a+2=9

, so

a=7

. The value 9 is the value of

a+2

, not

a

Question 19

The radius of a circular oil slick $t$ hours after noon is $r(t)=\sqrt{t^2-4}$ meters. The area of the slick is $A(r)=\pi r^2$ square meters. If the model is valid for all real $t$ for which the radius is defined, which expression gives $A$ as a function of $t$ , and what is its domain?

$A(t)=\pi(t^2-4)$ , domain $(-\infty,-2]\cup[2,\infty)$ (correct answer)
$A(t)=\pi(t^2-4)$ , domain $[2,\infty)$
$A(t)=\pi\sqrt{t^2-4}$ , domain $(-\infty,-2]\cup[2,\infty)$
$A(t)=\pi|t^2-4|$ , domain $(-\infty,\infty)$

Explanation:

r

is defined when

t^2-4\ge 0

, i.e.

|t|\ge 2

. Substitution gives

A=\pi(\sqrt{t^2-4})^2=\pi(t^2-4)

. The domain is all real

t

with

|t|\ge 2

, not only

t\ge 2

, and the area uses

r^2

, not

r

Question 20

Let $f(x)=\frac{x-1}{x+2}$ and $g(x)=\frac{2}{x-3}$ . Which expression is equal to $(f \circ g)(x)$ , and what is its domain?

$\frac{5-x}{2(x-2)}$ , domain $x\neq 2$
$\frac{5-x}{2(x-2)}$ , domain $x\neq 2,3$ (correct answer)
$\frac{x-5}{2(x-2)}$ , domain $x\neq 2,3$
$\frac{5-x}{2(x-3)}$ , domain $x\neq 2,3$

Explanation: Compute

f(g(x))=\frac{2/(x-3)-1}{2/(x-3)+2}=\frac{5-x}{2(x-2)}

. The domain excludes

x=3

because

g

is undefined, and

x=2

because

g(2)=-2

, where

f

is undefined. Omitting the

x=3

restriction, using the opposite numerator sign, or keeping

x-3

in the denominator are the common errors.