The Mean Value Theorem — Question 9

Question 9 - Solution

We are given: $$f(x) = \frac{1}{x}$$

(a) Conditions for MVT:

- $f(x)$ is continuous on $[1, 3]$ ✓ - $f(x)$ is differentiable on $(1, 3)$ ✓

✅ Therefore, MVT applies.

(b) Apply MVT:

$$f(3) = \frac{1}{3}, \quad f(1) = 1$$ $$\frac{f(3) - f(1)}{3 - 1} = \frac{\frac{1}{3} - 1}{2} = \frac{-\frac{2}{3}}{2} = -\frac{1}{3}$$

Now compute derivative: $$f'(x) = \frac{d}{dx} \left( \frac{1}{x} \right) = -\frac{1}{x^2}$$ $$f'(c) = -\frac{1}{c^2} = -\frac{1}{3} \Rightarrow c^2 = 3 \Rightarrow c = \sqrt{3}$$

$$\boxed{c = \sqrt{3} \in (1, 3)}$$

Graph of $f(x) = \frac{1}{x}$ on $[1, 3]$: