The Mean Value Theorem — Question 8

Question 8 - Solution

We are given: $$f(x) = \sin(x)$$

(a) MVT Conditions:

- $f(x)$ is continuous on $[0, \pi]$: ✓ Sine is continuous everywhere.

- $f(x)$ is differentiable on $(0, \pi)$: ✓ Sine is differentiable everywhere.

✅ MVT applies.

(b) Apply MVT:

Average rate of change: $$f(\pi) = \sin(\pi) = 0, \quad f(0) = \sin(0) = 0$$ $$\frac{f(\pi) - f(0)}{\pi - 0} = \frac{0 - 0}{\pi} = 0$$

We want: $$f'(c) = \cos(c) = 0 \Rightarrow c = \frac{\pi}{2}$$

$$\boxed{c = \frac{\pi}{2}}$$

Graph of $f(x) = \sin(x)$ on $[0, \pi]$: