Finding Absolute Extrema — Question 3

Question 3 - 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 $f'(x) = 0$: $$\frac{1 - x^2}{(x^2 + 1)^2} = 0 \Rightarrow 1 - x^2 = 0 \Rightarrow x = \pm 1$$

So, critical points: $x = -1$, $x = 1$

(b) Evaluate at endpoints and critical points:

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

(c) Conclusion:

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

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

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