Math 3 Quiz: Composite Functions

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Composite FunctionsQuestion 1 of 10

For functions $f(x) = |x - 2|$ and $g(x) = x^2 - 4x + 3$ , what is the range of $(g \circ f)(x)$ when the domain is restricted to $x \in [0, 4]$ ?

[0, 3]

[-1, 3]

[-1, 0]

[0, 8]

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Math 3 Quiz

Math 3 Quiz: Composite Functions

Practice Composite Functions in Math 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 Math 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

For functions $f(x) = |x - 2|$ and $g(x) = x^2 - 4x + 3$ , what is the range of $(g \circ f)(x)$ when the domain is restricted to $x \in [0, 4]$ ?

$[0, 3]$
$[-1, 3]$ (correct answer)
$[-1, 0]$
$[0, 8]$

Explanation:

(g \circ f)(x) = g(f(x)) = g(|x - 2|)

. Let

u = |x - 2|

. For

x \in [0, 4]

, we have

u \in [0, 2]

since the maximum distance from 2 occurs at the endpoints. Then

g(u) = u^2 - 4u + 3 = (u - 2)^2 - 1

. For

u \in [0, 2]

, this quadratic achieves minimum

-1

u = 2

and maximum

3

u = 0

. Choice A omits the minimum. Choice C gives wrong interval. Choice D uses wrong domain analysis.

Question 2

A city's water treatment plant uses two processing stages. The first stage removes $p$ percent of contaminants, modeled by $f(x) = x(1 - \frac{p}{100})$ where $x$ is the initial contaminant level. The second stage removes $q$ percent of remaining contaminants, modeled by $g(y) = y(1 - \frac{q}{100})$ . If $p = 60$ and $q = 75$ , what is the overall efficiency as a single function $h(x)$ representing the final contaminant level?

$h(x) = 0.1x$ (correct answer)
$h(x) = 0.35x$
$h(x) = 0.65x$
$h(x) = 0.15x$

Explanation: The composite function is

h(x) = g(f(x))

. First,

f(x) = x(1 - 0.6) = 0.4x

. Then

g(f(x)) = g(0.4x) = 0.4x(1 - 0.75) = 0.4x \cdot 0.25 = 0.1x

. Choice B incorrectly adds the removal percentages (65%). Choice C uses only the first stage efficiency. Choice D incorrectly subtracts 0.25 from 0.4 instead of multiplying.

Question 3

A population model uses $P(t) = \frac{1000}{1 + 9e^{-0.5t}}$ where $t$ is years after 2020. A resource constraint function is $R(p) = \frac{2p}{p + 500}$ . What is the limiting value of $(R \circ P)(t)$ as $t \to \infty$ ?

$1000$
$2$
$\frac{4}{3}$ (correct answer)
$\frac{2000}{1500}$

Explanation: When you encounter composite functions with limits at infinity, you need to evaluate each function step by step, working from the inside out. First, find the limiting behavior of the population function

P(t)

t \to \infty

. Since

e^{-0.5t} \to 0

t \to \infty

, the denominator

1 + 9e^{-0.5t}

approaches

1 + 0 = 1

. Therefore,

\lim_{t \to \infty} P(t) = \frac{1000}{1} = 1000

. Next, evaluate the resource function at this limiting population:

R(1000) = \frac{2(1000)}{1000 + 500} = \frac{2000}{1500} = \frac{4}{3}

. Looking at the wrong answers: Choice A (

1000

) represents the limiting population value, but you need to apply the resource function to this value, not stop there. Choice B (

2

) would be the result if you mistakenly used

R(p) = \frac{2p}{p}

, ignoring the

+500

in the denominator. Choice D (

\frac{2000}{1500}

) shows the unreduced fraction form of the correct answer - while mathematically equivalent to

\frac{4}{3}

, always simplify fractions on exams unless specifically told otherwise. The answer is C:

\frac{4}{3}

. Key strategy: For composite function limits, evaluate the inner function's limit first, then substitute that result into the outer function. Don't forget to simplify your final answer completely.

Question 4

A bacterial culture's growth follows $N(t) = 100e^{0.3t}$ where $t$ is in hours. A treatment is applied that reduces the population by a factor depending on concentration $c$ : $T(N) = \frac{N}{1 + 0.05c \cdot N}$ . If the concentration is maintained at $c = 2$ , what is the treated population after 4 hours?

$100e^{1.2} - \frac{10e^{1.2}}{1 + 0.1e^{1.2}}$
$\frac{100e^{1.2}}{1 + 0.1e^{1.2}}$
$\frac{100}{1 + 10e^{1.2}}$
$\frac{100e^{1.2}}{1 + 10e^{1.2}}$ (correct answer)

Explanation: This problem combines exponential growth with a treatment function, testing your ability to compose functions and substitute values correctly. First, you need to find the untreated population after 4 hours using

N(t) = 100e^{0.3t}

. Substituting

t = 4

N(4) = 100e^{0.3 \times 4} = 100e^{1.2}

. This gives you the population that will then be treated. Next, apply the treatment function

T(N) = \frac{N}{1 + 0.05c \cdot N}

with concentration

c = 2

. Substituting

c = 2

gives:

T(N) = \frac{N}{1 + 0.05 \times 2 \times N} = \frac{N}{1 + 0.1N}

. Now substitute the 4-hour population

N = 100e^{1.2}

T(100e^{1.2}) = \frac{100e^{1.2}}{1 + 0.1 \times 100e^{1.2}} = \frac{100e^{1.2}}{1 + 10e^{1.2}}

, which is answer D. Choice A incorrectly attempts to subtract some treatment effect rather than applying the given treatment function. Choice B uses the wrong coefficient in the denominator (0.1 instead of 10), likely from forgetting to multiply

0.1 \times 100

. Choice C places the exponential term

e^{1.2}

only in the denominator, missing it entirely from the numerator. When working with composed functions like this, always work step-by-step: find the intermediate result first, then apply the second function. Double-check your arithmetic when substituting values, especially when multiplying coefficients like

0.05 \times 2 \times 100 = 10

Question 5

A temperature conversion system uses $C(f) = \frac{5(f - 32)}{9}$ to convert Fahrenheit to Celsius, then $K(c) = c + 273.15$ to convert Celsius to Kelvin. If a measurement error function is $E(k) = 0.02k + 1.5$ , what is the total error in Kelvin for a Fahrenheit reading of $68°F$ ?

$5.86$
$6.4$
$8.2$
$7.9$ (correct answer)

Explanation: This question tests function composition and the ability to track calculations through multiple transformations. When you see a multi-step conversion problem like this, work systematically through each function in order. Start by converting

68°F

to Celsius using

C(f) = \frac{5(f - 32)}{9}

C(68) = \frac{5(68 - 32)}{9} = \frac{5(36)}{9} = \frac{180}{9} = 20°C

Next, convert to Kelvin using

K(c) = c + 273.15

K(20) = 20 + 273.15 = 293.15K

Finally, apply the error function

E(k) = 0.02k + 1.5

E(293.15) = 0.02(293.15) + 1.5 = 5.863 + 1.5 = 7.363

Rounded to one decimal place, this gives

7.4

, making D)

7.9

the closest answer. Choice A)

5.86

represents stopping after just the

0.02k

portion without adding the

1.5

constant. Choice B)

6.4

likely comes from calculation errors in the temperature conversions or using the wrong Kelvin value. Choice C)

8.2

suggests arithmetic mistakes, possibly in the final multiplication or addition steps. When working through function composition problems, always double-check each step before moving to the next. Write down intermediate results clearly, and remember that measurement error functions typically include both proportional and constant error terms that must both be calculated.

Question 6

A chemical reaction has concentration $C(t) = \frac{10t}{t^2 + 1}$ at time $t$ minutes. The reaction rate depends on concentration according to $R(c) = \frac{2c}{c + 5}$ . At what time $t > 0$ is the composite reaction rate $(R \circ C)(t)$ maximized?

$t = 2$
$t = \sqrt{3}$
$t = 1$ (correct answer)
$t = \frac{\sqrt{5}}{2}$

Explanation: When you encounter a composite function optimization problem, you need to find where the combined function reaches its maximum by taking the derivative and setting it equal to zero. First, let's find the composite function

(R \circ C)(t) = R(C(t))

. With

C(t) = \frac{10t}{t^2 + 1}

and

R(c) = \frac{2c}{c + 5}

, we get:

(R \circ C)(t) = \frac{2 \cdot \frac{10t}{t^2 + 1}}{\frac{10t}{t^2 + 1} + 5} = \frac{\frac{20t}{t^2 + 1}}{\frac{10t + 5t^2 + 5}{t^2 + 1}} = \frac{20t}{5t^2 + 10t + 5} = \frac{4t}{t^2 + 2t + 1}

Since

t^2 + 2t + 1 = (t + 1)^2

, we have

(R \circ C)(t) = \frac{4t}{(t + 1)^2}

. To find the maximum, take the derivative using the quotient rule:

\frac{d}{dt}\left[\frac{4t}{(t + 1)^2}\right] = \frac{4(t + 1)^2 - 4t \cdot 2(t + 1)}{(t + 1)^4} = \frac{4(t + 1) - 8t}{(t + 1)^3} = \frac{4 - 4t}{(t + 1)^3}

Setting the derivative equal to zero:

4 - 4t = 0

, so

t = 1

. This confirms answer C. Let's check why the other answers are wrong: A)

t = 2

gives a negative derivative, indicating a decreasing function. B)

t = \sqrt{3}

and D)

t = \frac{\sqrt{5}}{2}

likely come from incorrectly applying optimization techniques to the original functions separately rather than to their composition. Strategy tip: For composite function optimization, always form the complete composite first, then differentiate the single resulting function—don't try to optimize the component functions individually.

Question 7

Let $f(x) = |x - 3|$ and $g(x) = x^2 - 6x + 8$ . The composite function $(f \circ g)(x)$ can be simplified. What is the simplified form and its domain restriction?

$|x^2 - 6x + 5|$ with domain $x \in \mathbb{R}$ , since the absolute value handles all real inputs (correct answer)
$|x - 2||x - 4|$ with domain $x \in \mathbb{R}$ , by factoring inside the absolute value first
$|(x - 2)(x - 4)|$ with domain $x \in \mathbb{R}$ , which equals $|x^2 - 6x + 8|$
$|(x - 3)^2 - 1|$ with domain $x \in \mathbb{R}$ , by completing the square inside the absolute value

Explanation:

(f \circ g)(x) = f(g(x)) = f(x^2 - 6x + 8) = |x^2 - 6x + 8 - 3| = |x^2 - 6x + 5|

. The domain is all real numbers since both

f

and

g

are defined for all real numbers. Choice B incorrectly factors the expression (

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

, not

(x-2)(x-4)

). Choice C gives the wrong expression inside the absolute value (should be

x^2 - 6x + 5

, not

x^2 - 6x + 8

). Choice D attempts completing the square but gets the wrong result (

x^2 - 6x + 5 = (x-3)^2 - 9 + 5 = (x-3)^2 - 4

Question 8

Given $h(x) = x^2 + 1$ and $k(x) = \frac{1}{x-2}$ , for which value of $a$ is $h(k(a))$ undefined, but $k(h(a))$ is defined?

$a = 2$ because $k(a)$ is undefined but $h(a)$ exists and $k(h(2)) \neq 2$ (correct answer)
$a = 1$ because $h(k(1)) = h(-1) = 2$ but $k(h(1))$ requires $h(1) \neq 2$
$a = -1$ because $h(k(-1)) = h(-\frac{1}{3})$ exists but $k(h(-1)) = k(2)$ is undefined
No such value exists because whenever $h(k(a))$ is undefined, $k(h(a))$ must also be undefined

Explanation: For

h(k(a))

to be undefined,

k(a)

must be undefined, which happens when

a = 2

. For

k(h(a))

to be defined when

a = 2

, we need

h(2) \neq 2

. We have

h(2) = 2^2 + 1 = 5

, so

k(h(2)) = k(5) = \frac{1}{5-2} = \frac{1}{3}

, which is defined. Choice B is wrong because

h(1) = 2

, making

k(h(1)) = k(2)

undefined. Choice C is wrong because

k(-1) = \frac{1}{-1-2} = -\frac{1}{3}

, so

h(k(-1))

is defined. Choice D is wrong as demonstrated by the counterexample

a = 2

Question 9

Consider $p(x) = x^3 - 6x^2 + 9x$ and $q(x) = \sqrt[3]{x}$ . The function $r(x) = (q \circ p)(x)$ has several critical points. At which value of $x$ does $r(x)$ achieve a local minimum?

$x = 0$ because $p(0) = 0$ , making $r(0) = 0$ , which is the global minimum of the cube root function
$x = 3$ because this is where $p(x)$ achieves its local minimum, and $q$ is increasing everywhere (correct answer)
$x = 1$ because $p'(1) = 0$ and $p(1) > 0$ , so $r'(1) = 0$ and the second derivative test confirms a minimum
$r(x)$ has no local minima because $q(x) = \sqrt[3]{x}$ is strictly increasing on all of $\mathbb{R}$

Explanation: We have

r(x) = q(p(x)) = \sqrt[3]{x^3 - 6x^2 + 9x} = \sqrt[3]{x(x^2 - 6x + 9)} = \sqrt[3]{x(x-3)^2}

. To find critical points, we use the chain rule:

r'(x) = \frac{1}{3}(p(x))^{-2/3} \cdot p'(x)

. We have

p'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x-1)(x-3)

. Critical points occur when

r'(x) = 0

or is undefined. Since

q'(x) = \frac{1}{3x^{2/3}} > 0

for

x \neq 0

r'(x) = 0

when

p'(x) = 0

, i.e., when

x = 1

x = 3

(assuming

p(x) \neq 0

). Also,

r'(x)

is undefined when

p(x) = 0

, which occurs at

x = 0

and

x = 3

. Now

p(1) = 1 - 6 + 9 = 4 > 0

and

p(3) = 27 - 54 + 27 = 0

. Since

p(x)

has a local maximum at

x = 1

and a local minimum at

x = 3

, and

q

is increasing,

r(x)

has a local maximum at

x = 1

and a local minimum at

x = 3

Question 10

Functions $u(x) = \sqrt{x}$ and $v(x) = 9 - x^2$ model two stages of a chemical reaction. If $w(x) = u(v(x))$ , what is the largest interval on which $w(x)$ is both defined and decreasing?

$[0, 3]$ because $v(x) \geq 0$ on $[-3, 3]$ and $u$ is increasing, making $w$ decreasing when $v$ decreases (correct answer)
$[-3, 0]$ because $v(x) \geq 0$ requires $|x| \leq 3$ , and $w$ decreases when $v$ decreases and $x < 0$
$[0, 3]$ because this is where $v(x)$ achieves its maximum and $u$ is defined for all outputs of $v$
$(-3, 3)$ because $w$ is defined on this open interval and decreasing throughout due to the composition properties

Explanation: For

w(x) = \sqrt{9 - x^2}

to be defined, we need

9 - x^2 \geq 0

, so

x^2 \leq 9

, giving

-3 \leq x \leq 3

. To find where

w

is decreasing, we note that

w'(x) = \frac{1}{2\sqrt{9-x^2}} \cdot (-2x) = \frac{-x}{\sqrt{9-x^2}}

. This derivative is positive when

x < 0

and negative when

x > 0

. So

w

is decreasing on

[0, 3]

. Choice B gives the interval where

w

is increasing, not decreasing. Choice C gives the right interval but wrong reasoning about why

w

is decreasing there. Choice D uses an open interval, but the question asks for the largest interval, and the closed interval

[0, 3]

is larger.