Question

What is the rate of change for f(x)=-2cos4x-3 on the interval from x=0 to x=pi over 4

Answer 1

Answer:

$\frac{16}{\pi}$

Step-by-step explanation:

The average rate of a function $f$ on interval $x=a$ to $x=b$ is:

$\frac{f(b)-f(a)}{b-a}$.

$a=0$ and $b=\frac{\pi}{4}$ here for this problem where $f(x)=-2\cos(4x)-3$.

So we will need to evaluate $f$ for $x=0 \text{ and } b=\frac{\pi}{4}$.

To find $f(0)$ we will replace $x$ in $f(x)=-2\cos(4x)-3$ with $0$:

$f(0)=-2\cos(4(0))-3$

$f(0)=-2\cos(0)-3$

$f(0)=-2(1)-3$

$f(0)=-2-3$

$f(0)=-5$

To find $f(\frac{\pi}{4})$ we will replace $x$ in $f(x)=-2\cos(4x)-3$ with $\frac{\pi}{4}$:

$f(\frac{\pi}{4})=-2\cos(4(\frac{\pi}{4}))-3$

$f(\frac{\pi}{4})=-2\cos(\pi)-3$

$f(\frac{\pi}{4})=-2(-1)-3$

$f(\frac{\pi}{4})=2-3$

$f(\frac{\pi}{4})=-1$

So now we need to compute the change in $f$ over the change in $x$:

$\frac{-1-(-5)}{\frac{\pi}{4}-0}$

$\frac{-1+5}{\frac{\pi}{4}}$

Dividing by $\frac{\pi}{4}$ is the same as multiplying by $\frac{4}{\pi}$:

$(-1+5)\cdot \frac{4}{\pi}$

$4 \cdot \frac{4}{\pi}$

$\frac{16}{\pi}$