## Related questions with answers

Which function has the greater average rate of change over the interval $-5 \leq x \leq 0$ :the translation of the function $f(x)=\sqrt[3]{x}$ to the right 1 unit and up 2 units, or the function $r(x)=\sqrt[3]{x}+3$ ?

Solutions

VerifiedWe let the translated function be $g$ so that it is given by:

$g(x)=\sqrt[3]{x-1}+2$

Using the functions,

$\begin{align*} g(-5)&= \sqrt[3]{-5-1}+2\approx 0.18\\ g(0)&= \sqrt[3]{0-1}+2=1\\ h(-5)&= \sqrt[3]{-5}+3\approx 1.29\\ h(0)&= \sqrt[3]{0}+3=3 \end{align*}$

Find the average rate of change of each function over the given interval.

$\textbf{\color{#4257b2}{For $g(x)$:}}$

$\dfrac{g(0)- g(-5)}{0 -(-5)}=\dfrac{1-0.18}{0-(-5)}\approx 0.16$

$\textbf{\color{#4257b2}{For $h(x)$:}}$

$\dfrac{h(0)-h(-5)}{0 -(-5)}=\dfrac{3-1.29}{0-(-5)}\approx 0.34$

Since $0.16<0.34$, then $\color{#c34632}{h(x)}$ has the greater average rate of change.

Let the translated function of $f(x)$ is $g(x)$ such that

$\begin{align*} f(x)=&\;\sqrt[3]{x} &&\text{Given function}\\ g(x)=&\;\sqrt[3]{x-1}+2 &&\text{Translated function}\\ r(x)=&\;\sqrt[3]{x}+3 \end{align*}$

The domain of the function is

$\begin{align*} -5 \leq x \leq 0 \end{align*}$

The range of the functions are

$\begin{align*} g(0)=&\;\sqrt[3]{0-1}+2\\ =&\;1\\ r(0)=&\;\sqrt[3]{0}+3\\ =&\;3\\ g(-5)=&\;\sqrt[3]{-5-1}+2\\ \approx&\;0.18\\ r(-5)=&\;\sqrt[3]{-5}+3\\ \approx &\;1.29 \end{align*}$

The average rate of change is

$\begin{align*} \dfrac{g(-5)-g(0)}{-5-0}=&\;\dfrac{0.18-1}{-5-0}\\ \approx&\; 0.16 && \text{$g(x)$}\\ \dfrac{r(-5)-r(0)}{-5-0}=&\;\dfrac{0.29-3}{-5-0}\\ \approx&\; 0.34 && \text{$r(x)$}\\ r(x) >&\; g(x) &&\text{Average rate of change of functions} \end{align*}$

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