Finding Absolute Extrema — Question 10

Question 10 - Solution

We are given: $$f(x) = \frac{x}{x^2 + 1}$$

(a) Find critical points:

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

Set $f'(x) = 0$: $$1 - x^2 = 0 \Rightarrow x = \pm 1$$

(b) Evaluate $f(x)$:

$$f(-2) = \frac{-2}{4 + 1} = -\frac{2}{5}, \quad f(-1) = \frac{-1}{1 + 1} = -\frac{1}{2}, \quad f(1) = \frac{1}{1 + 1} = \frac{1}{2}, \quad f(2) = \frac{2}{4 + 1} = \frac{2}{5}$$

(c) Absolute extrema:

• Absolute maximum: $\boxed{f(1) = \frac{1}{2}}$

• Absolute minimum: $\boxed{f(-1) = -\frac{1}{2}}$

Graph of $f(x) = \frac{x}{x^2 + 1}$ on $[-2, 2]$: