Math 1 Quiz: Comparing Function Growth

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Comparing Function GrowthQuestion 1 of 9

A piecewise function $h(x)$ equals $2x + 1$ for $x \leq 3$ and $x^2 - 2x + 7$ for $x > 3$ . Compared to the linear function $j(x) = 4x - 5$ , which statement about their growth over $[1, 6]$ is most accurate?

Function

j

grows faster on

[1, 3]

but

h

grows faster on

(3, 6]

due to quadratic acceleration

Function

h

grows faster on

[3, 6]

because the quadratic piece accelerates beyond

j

's rate

Function

j

grows faster throughout because linear functions maintain constant positive slope

The functions have identical growth patterns because both are continuous and differentiable everywhere

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Math 1 Quiz

Math 1 Quiz: Comparing Function Growth

Practice Comparing Function Growth in Math 1 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 Comparing Function Growth, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 1.

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

Function $j$ grows faster on $[1, 3]$ but $h$ grows faster on $(3, 6]$ due to quadratic acceleration (correct answer)
Function $h$ grows faster on $[3, 6]$ because the quadratic piece accelerates beyond $j$ 's rate
Function $j$ grows faster throughout because linear functions maintain constant positive slope
The functions have identical growth patterns because both are continuous and differentiable everywhere

Explanation: When analyzing piecewise functions, you need to examine each piece separately and understand how growth rates change across different intervals. Let's compare the growth rates by looking at the derivatives (slopes). For

h(x)

, when

x \leq 3

, we have

h(x) = 2x + 1

with slope 2. When

x > 3

, we have

h(x) = x^2 - 2x + 7

with derivative

h'(x) = 2x - 2

. For

j(x) = 4x - 5

, the slope is constantly 4. On the interval

[1, 3]

, function

h

has slope 2 while

j

has slope 4, so

j

grows faster. On the interval

(3, 6]

, function

h

's slope is

2x - 2

. At

x = 3

, this gives slope

2(3) - 2 = 4

, matching

j

's slope. But as

x

increases beyond 3,

h

's slope becomes

2(4) - 2 = 6

x = 4

, then

2(6) - 2 = 10

x = 6

. The quadratic piece accelerates, eventually outpacing

j

's constant growth rate. Answer A correctly identifies both phases:

j

grows faster initially, then

h

overtakes it due to quadratic acceleration. Answer B ignores the

[1, 3]

portion where

j

grows faster. Answer C incorrectly assumes constant slope always wins, missing that quadratic functions can accelerate past linear ones. Answer D is wrong because the functions aren't even differentiable at

x = 3

(the derivatives from left and right don't match), and their growth patterns clearly differ. Study tip: When comparing function growth rates, always check derivatives and remember that quadratic functions accelerate over time while linear functions maintain constant rates.

Question 2

Function $f$ is defined by $f(x) = 2x^3 - 5x^2 + 3x$ , and function $g$ has the property that $g'(x) = 6x^2 + 4x - 1$ for all $x$ . Over the interval $[3, 5]$ , which statement best describes the relative growth rates of these functions?

Function $f$ grows faster than $g$ throughout the entire interval
Function $g$ grows faster than $f$ throughout the entire interval (correct answer)
The functions have identical growth rates at exactly two points in the interval
Neither function consistently grows faster; their relative growth changes within the interval

Explanation: To compare growth rates, we need to compare the derivatives. We have

f'(x) = 6x^2 - 10x + 3

and

g'(x) = 6x^2 + 4x - 1

. The difference is

g'(x) - f'(x) = 14x - 4

. For

x \in [3,5]

, we have

14x - 4 \geq 14(3) - 4 = 38 > 0

, so

g'(x) > f'(x)

throughout the interval. Choice A incorrectly reverses the comparison. Choice C is wrong because the derivatives are equal when

14x - 4 = 0

, or

x = 2/7

, which is not in our interval. Choice D is incorrect because the relative growth is consistent.

Question 3

Function $h(x) = \sqrt{x + 4}$ is compared to function $k(x)$ over the interval $[5, 12]$ . If the average rate of change of $k$ over this interval is $\frac{3}{7}$ , and $k$ is linear, which function grows faster over this interval?

Function $h$ grows faster because its instantaneous rate exceeds $\frac{3}{7}$ throughout the interval
Function $k$ grows faster because linear functions always outpace square root functions
Function $k$ grows faster because $\frac{3}{7} > \frac{1}{6}$ , which bounds $h'(x)$ on this interval (correct answer)
The functions grow at identical rates because both have constant average rate $\frac{3}{7}$

Explanation: For

h(x) = \sqrt{x + 4}

, we have

h'(x) = \frac{1}{2\sqrt{x + 4}}

. On

[5,12]

, this ranges from

\frac{1}{2\sqrt{16}} = \frac{1}{8}

\frac{1}{2\sqrt{9}} = \frac{1}{6}

. Since

k

is linear, its instantaneous rate equals its average rate of

\frac{3}{7}

everywhere. Since

\frac{3}{7} \approx 0.429 > \frac{1}{6} \approx 0.167

, function

k

grows faster. Choice A is wrong because

h'(x) < \frac{3}{7}

throughout. Choice B gives an incorrect general rule. Choice D is wrong because

h

's average rate is

\frac{\sqrt{16} - \sqrt{9}}{7} = \frac{1}{7} \neq \frac{3}{7}

Question 4

Function $p(x) = 2^x$ and function $q(x) = x^3 + 4x$ are compared over the interval $[0, 2]$ . A student claims that $p(x)$ grows faster because exponential functions always grow faster than polynomial functions. What is the most accurate assessment of this claim?

The claim is correct because $p(2) > q(2)$ , confirming exponential dominance over the given interval
The claim is incorrect because $q(x)$ actually grows faster over $[0, 2]$ despite the general exponential dominance principle (correct answer)
The claim is partially correct because while exponential functions eventually dominate, the interval $[0, 2]$ is too short to observe this
The claim is incorrect because polynomial functions of degree 3 always grow faster than exponential functions with base 2

Explanation:

p(0) = 1

p(2) = 4

, change = 3.

q(0) = 0

q(2) = 16

, change = 16. Over

[0,2]

q(x)

grows faster (16 > 3) despite exponential functions eventually dominating. Choice A incorrectly compares endpoint values rather than growth. Choice C misses that

q

actually grows faster here. Choice D makes a false general statement about polynomial vs exponential growth.

Question 5

Two rational functions $f(x) = \frac{x^2 + 3x + 2}{x + 1}$ and $g(x) = \frac{2x^2 - x + 3}{x + 2}$ are compared over $[1, 4]$ . Which grows faster on this interval?

Function $f$ grows faster with constant slope after simplification
Function $g$ grows faster due to higher leading coefficient
Function $f$ grows faster due to smaller denominator effects
Function $g$ grows faster with average rate exceeding $f$ (correct answer)

Explanation: Function

f(x) = \frac{x^2 + 3x + 2}{x + 1} = \frac{(x+1)(x+2)}{x+1} = x + 2

for

x \neq -1

, so

f

has constant rate of change 1. Over

[1,4]

f

increases from 3 to 6, a total change of 3. For

g(x)

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

and

g(4) = \frac{32-4+3}{6} = \frac{31}{6}

. The total change for

g

\frac{31}{6} - \frac{4}{3} = \frac{31-8}{6} = \frac{23}{6} \approx 3.83

. Since

g

's average rate is

\frac{23/6}{3} \approx 1.28 > 1

, function

g

grows faster.

Question 6

Function $m(x) = \ln(x^2 + 1)$ is compared to function $n(x) = \frac{x^3}{12} + 2$ over the interval $[2, 5]$ . Which function demonstrates faster growth, and why?

Function $m$ grows faster because logarithmic functions accelerate rapidly for large arguments like $x^2 + 1$
Function $n$ grows faster because cubic polynomials eventually dominate logarithmic functions in growth rate
Function $m$ grows faster because its derivative $\frac{2x}{x^2 + 1}$ exceeds $n'(x) = \frac{x^2}{4}$ on this interval
Function $n$ grows faster because $\frac{x^2}{4} > \frac{2x}{x^2 + 1}$ for all $x$ in the given interval (correct answer)

Explanation: We have

m'(x) = \frac{2x}{x^2 + 1}

and

n'(x) = \frac{x^2}{4}

. To compare these on

[2,5]

, we check when

\frac{x^2}{4} > \frac{2x}{x^2 + 1}

. Cross-multiplying (both sides positive):

x^2(x^2 + 1) > 8x

, so

x^4 + x^2 > 8x

, or

x^3 + x > 8

(dividing by

x > 0

). At

x = 2

8 + 2 = 10 > 8

✓. The inequality holds throughout

[2,5]

, so

n

grows faster. Choice A incorrectly describes logarithmic growth. Choice B gives the right answer but wrong reasoning (this isn't about eventual dominance). Choice C incorrectly claims

m'(x) > n'(x)

Question 7

Function $a(x) = \sqrt{x + 3}$ and function $b(x) = 0.5x + 2$ are defined for $x \geq 0$ . Over the interval $[1, 6]$ , which analysis of their relative growth is correct?

Function $a$ grows faster because square root functions eventually outgrow linear functions with fractional coefficients
Function $b$ grows faster because linear functions maintain constant positive growth while root functions decelerate
Function $a$ grows faster because its total change over the interval exceeds that of function $b$
Function $b$ grows faster because its average rate of change over $[1, 6]$ exceeds that of function $a$ (correct answer)

Explanation: For

a

a(1) = \sqrt{4} = 2

a(6) = \sqrt{9} = 3

. Average rate =

(3-2)/(6-1) = 0.2

. For

b

b(1) = 2.5

b(6) = 5

. Average rate =

(5-2.5)/(6-1) = 0.5

. Since 0.5 > 0.2,

b

grows faster. Choice A makes incorrect claims about eventual behavior. Choice B gives correct reasoning but wrong conclusion about which grows faster. Choice C is wrong:

a

changes by 1,

b

changes by 2.5.

Question 8

Function $h(x) = x^2 - 4x + 5$ and function $k(x) = -x^2 + 6x - 1$ are compared over the interval $[2, 4]$ . Which statement correctly describes their growth comparison?

Function $h$ grows faster because it opens upward while $k$ opens downward, ensuring positive versus negative growth
Function $k$ grows faster because its vertex is higher than $h$ 's vertex, indicating greater overall increase
Function $h$ grows faster because its average rate of change over the interval exceeds that of $k$ (correct answer)
Function $k$ grows faster because its rate of change at $x = 3$ is greater than $h$ 's rate at the same point

Explanation:

h(2) = 1

h(4) = 5

, so average rate =

(5-1)/(4-2) = 2

k(2) = 7

k(4) = -1

, so average rate =

(-1-7)/(4-2) = -4

. Since

h

increases while

k

decreases,

h

grows faster. Choice A incorrectly assumes parabola direction determines growth over any interval. Choice B confuses vertex height with growth rate. Choice D incorrectly uses instantaneous rates at one point rather than overall growth.

Question 9

Function $u(x) = \log_2(x + 1)$ and function $v$ satisfies $v(1) = 0$ , $v(3) = 4$ , and $v(5) = 6$ . Assuming $v$ is linear, which function grows faster over the interval $[1, 5]$ ?

Function $u$ grows faster because logarithmic functions increase without bound while linear functions have constant rates
Function $v$ grows faster because its constant rate of change exceeds the average rate of change of $u$ (correct answer)
Function $u$ grows faster because its average rate of change over $[1, 5]$ exceeds that of $v$
Function $v$ grows faster because linear functions always outpace logarithmic functions over finite intervals

Explanation: For

u

u(1) = \log_2(2) = 1

u(5) = \log_2(6) \approx 2.58

. Average rate =

(2.58-1)/(5-1) \approx 0.4

. For

v

: since it's linear through

(1,0)

and

(5,6)

, average rate =

6/(5-1) = 1.5

. Since 1.5 > 0.4,

v

grows faster. Choice A makes incorrect generalizations about function behavior. Choice C reverses the comparison. Choice D makes a false general statement.