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Which function has the greater average rate of change over the interval 5x0-5 \leq x \leq 0 :the translation of the function f(x)=x3f(x)=\sqrt[3]{x} to the right 1 unit and up 2 units, or the function r(x)=x3+3r(x)=\sqrt[3]{x}+3 ?

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We let the translated function be gg so that it is given by:

g(x)=x13+2g(x)=\sqrt[3]{x-1}+2

Using the functions,

g(5)=513+20.18g(0)=013+2=1h(5)=53+31.29h(0)=03+3=3\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.

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

g(0)g(5)0(5)=10.180(5)0.16\dfrac{g(0)- g(-5)}{0 -(-5)}=\dfrac{1-0.18}{0-(-5)}\approx 0.16

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

h(0)h(5)0(5)=31.290(5)0.34\dfrac{h(0)-h(-5)}{0 -(-5)}=\dfrac{3-1.29}{0-(-5)}\approx 0.34

Since 0.16<0.340.16<0.34, then h(x)\color{#c34632}{h(x)} has the greater average rate of change.

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