Q. 80 - Precalculus Enhanced with Graphing Utilities

Question

Find the average rate of change of f form $0$to$\frac{\mathrm{\pi }}{2}$.

$f\left(x\right)=\cos \left(2x\right)$

Step-by-Step Solution

Answer

The average rate of change of $f\left(x\right)=\cos \left(2x\right)$from $0$to $\frac{\mathrm{\pi }}{2}$is $0.$

Step 1. Given data

The given function is

$f\left(x\right)=\cos \left(2x\right)$

The given interval is$\left(0,\frac{\mathrm{\pi }}{2}\right)$

Step 2. The average rate of change

differentiate function

$$\begin{gathered} f\left(x\right)=\cos \left(2x\right) \\ f\text{'}\left(x\right)=-2\sin \left(2x\right) \end{gathered}$$

Consider  $(a,b)=\left(0,\frac{\mathrm{\pi }}{2}\right)$

The average rate of change

$$\begin{gathered} \frac{f\text{'}\left(b\right)-f\text{'}\left(a\right)}{b-a}=\frac{f\text{'}\left({\displaystyle \frac{\mathrm{\pi }}{2}}\right)-f\text{'}\left(0\right)}{\frac{\mathrm{\pi }}{2}-0} \\ =\frac{-2\sin \left(2\left(\frac{\mathrm{\pi }}{2}\right)\right)+2\sin \left(2\right(0\left)\right))}{\frac{\mathrm{\pi }}{2}-0} \\ ==\frac{-2\sin \left(\mathrm{\pi }\right)+2\sin \left(0\right)}{\frac{\mathrm{\pi }}{2}} \\ =\frac{-0+0}{\frac{\mathrm{\pi }}{2}} \\ =0 \end{gathered}$$