The Mean Value Theorem — Question 7

Question 7 - Solution

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

(a) MVT Conditions:

- $f(x)$ is continuous on $[1, 4]$: The function is a rational function with no discontinuities on this interval.

- $f(x)$ is differentiable on $(1, 4)$: Differentiable as a quotient of differentiable functions, and denominator never zero.

✅ MVT applies.

(b) Compute the average rate of change:

$$f(1) = \frac{1}{2}, \quad f(4) = \frac{4}{5}$$ $$\frac{f(4) - f(1)}{4 - 1} = \frac{\frac{4}{5} - \frac{1}{2}}{3} = \frac{\frac{8 - 5}{10}}{3} = \frac{3}{10} \cdot \frac{1}{3} = \frac{1}{10}$$

Find derivative:

$$f'(x) = \frac{(x + 1)(1) - x(1)}{(x + 1)^2} = \frac{1}{(x + 1)^2}$$

Set: $$\frac{1}{(c + 1)^2} = \frac{1}{10} \Rightarrow (c + 1)^2 = 10 \Rightarrow c + 1 = \sqrt{10} \Rightarrow c = \sqrt{10} - 1$$

Since $\sqrt{10} \approx 3.16$, we find $c \approx 2.16 \in (1, 4)$

$$\boxed{c = \sqrt{10} - 1}$$

Graph of $f(x)$ and Secant Line: