Comparing Function Growth - Math 1

1. 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?
   Answer: Function $g$ grows faster with average rate exceeding $f$ 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 $g$ is $\frac{31}{6} - \frac{4}{3} = \frac{31-8}{6} = \frac{23}{6} \approx 3.83$. Since $g$'s average rate is $\frac{23/6}{3} \approx 1.28 > 1$, function $g$ grows faster.
2. 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?
   Answer: The claim is incorrect because $q(x)$ actually grows faster over $[0, 2]$ despite the general exponential dominance principle $p(0) = 1$, $p(2) = 4$, 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.
3. 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?
   Answer: Function $h$ grows faster because its average rate of change over the interval exceeds that of $k$ $h(2) = 1$, $h(4) = 5$, 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.
4. 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?
   Answer: Function $j$ grows faster on $[1, 3]$ but $h$ grows faster on $(3, 6]$ due to quadratic acceleration 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 $2(4) - 2 = 6$ at $x = 4$, then $2(6) - 2 = 10$ at $x = 6$. 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.
5. 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?
   Answer: Function $k$ grows faster because $\frac{3}{7} > \frac{1}{6}$, which bounds $h'(x)$ on this interval For $h(x) = \sqrt{x + 4}$, we have $h'(x) = \frac{1}{2\sqrt{x + 4}}$. On $[5,12]$, this ranges from $\frac{1}{2\sqrt{16}} = \frac{1}{8}$ to $\frac{1}{2\sqrt{9}} = \frac{1}{6}$. 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 $\frac{\sqrt{16} - \sqrt{9}}{7} = \frac{1}{7} \neq \frac{3}{7}$.
6. 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?
   Answer: Function $g$ grows faster than $f$ throughout the entire interval 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.
7. 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 $[1, 5]$?
   Answer: Function $v$ grows faster because its constant rate of change exceeds the average rate of change of $u$ 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.
8. 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?
   Answer: Function $b$ grows faster because its average rate of change over $[1, 6]$ exceeds that of function $a$ 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.
9. 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?
   Answer: Function $n$ grows faster because $\frac{x^2}{4} > \frac{2x}{x^2 + 1}$ for all $x$ in the given interval 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 $m'(x) > n'(x)$.

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