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Question

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

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We 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.

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