Answer
The proof is below.
Work Step by Step
Let, $f(x)$ be the function of consideration. The rate of change would be $\frac{d}{d x} f(x)=f^{\prime}(x)$
The average value of $f^{\prime}(x)$ over an interval, say $[a, b]$, is given by $\frac{\int_{a}^{b} f^{\prime}(x) d x}{b-a}$ The quantity can be written simply as $\frac{-f(a)+f(b)}{-a+b}$
If we were to talk about the average rate of change of $\mathrm{f}(\mathrm{x})$ over the interval $b-a,$ we write
Average rate of change $=\frac{\text { Change in } \mathrm{f}(\mathrm{x})}{\text { Change in } \mathrm{x}}=\frac{-f(a)+f(b)}{-a+b}$
We can verify that both are the same quantity .
