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# Find the average rate of change of the function over the given interval. Compare this average rate of change with the instantaneous rates of change at the endpoints of the interval. f(t)=3t+5, [1, 2]

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$\hspace*{5mm}$ We need to find the average rate of change of the function

$\color{#4257b2}{f(t)=3 t+5, \ \ [1,2]}$

\begin{align*} \frac{\Delta f(t)}{\Delta t} &=\frac{f(2)-f(1)}{2-1} \\ &=\frac{((3(2)+5)-(3(1)+5))}{1} \\ &=(6+5)-(3+5) \\ &=11-8 \end{align*}

$\hspace*{5mm}$From this we have,

\begin{align*} \frac{\Delta f}{\Delta t} &=11-8 \\ &=3 \end{align*}

$\color{#c34632}{\text{The average rate of change is in given interval}}$

\begin{align*} \lim _{x \rightarrow \infty} \frac{\Delta f(t)}{\Delta t} &=\frac{d}{d t}(f(t)) \\ &=f^{\prime}(t) \end{align*}

$\hspace*{5mm}$Differentiate the given function with respect to t is

\begin{align*} f^{\prime}(t) &=\frac{d}{d t}(3 t+5) \\ &=3 \end{align*}

$\color{#c34632}{f^{\prime}(t)_{t=1}=3\text{ and }f^{\prime}(t)_{t=2}=3}$

$\hspace*{5mm}$The instantaneous rate of change at the end intervals is same and is equal to $3 .$It can be seen that the average rate of change and the instantaneous rate of change at the end points are same.

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