Compare the average rate of change of the function over the given interval with the instantaneous rates of change at the endpoints of the interval.
$g(x)=x^2+e^x,[0,1]$ 98, $h(x)=x^3-2 e^x, \quad[-1,0]$

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Quizlet Admin · Accepted Answer

The average rate of change of $g(x)$ in the interval of $[a,b]$ is $\dfrac{g(b)-g(a)}{b-a}$.
So, in our case the average rate of change is
\begin{align*}
\dfrac{g(1)-g(0)}{1-0}&=\dfrac{1^2+e^1-(0^2+e^0)}{1}\
&=1+e-1\
&={\color{Mahogany}e\approx 2.71828}\
\end{align*}
The instantaneous rate of change is the derivative of the function.
To find the derivative of $g(x)$ we will use some {\color{Blue}Basic Differentiation Rules}:
\begin{enumerate}[(1)]
\item The Sum and Difference rule: $\dfrac{d}{dx}[f(x)\pm g(x)]=\dfrac{d}{dx}[f(x)]\pm \dfrac{d}{dx}[g(x)]$
\item The Power Rule: $\dfrac{d}{dx}[x^n]=nx^{n-1}$
\item Derivative of the Natural Exponential Function: $\dfrac{d}{dx}[e^x]=e^x$
\end{enumerate}
\begin{align*}
g'(x)&\overset{(1)}{=}\dfrac{d}{dx}[x^2]+\dfrac{d}{dx}[e^x]\
&\overset{(2),(3)}{=}2x+e^x
\end{align*}
So, the instantaneous rates of change at the endpoints of the interval are:
\begin{align*}
{\color{OliveGreen}g'(0)}&{\color{OliveGreen}=2\cdot 0 +e^0=1}\
{\color{OliveGreen}g'(1)}&{\color{OliveGreen}=2\cdot 1+e^1=2+e\approx 4.71828}
\end{align*}
Let's compare the average rate of change with the instantaneous rates of change at the endpoints of the interval.
\begin{align*}
{\color{OliveGreen}1} 	extless {\color{Mahogany}e} 	extless {\color{OliveGreen}2+e}
\end{align*}

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