The Shape of a Graph, Part II — Question 5

Question 5 - Solution

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

First derivative

Using the quotient rule, $$f'(x) = \frac{(x^2 + 1) - 2x^2}{(x^2 + 1)^2} = \frac{1 - x^2}{(x^2 + 1)^2}.$$

Second derivative

Differentiating $f'(x)$, $$f''(x) = \frac{-2x(3 - x^2)}{(x^2 + 1)^3}.$$

(a) Concavity

Setting $f''(x) = 0$ gives $$x = -\sqrt{3}, \quad x = 0, \quad x = \sqrt{3}.$$

For $x < -\sqrt{3}$, the second derivative is negative, so the graph is concave down.

For $-\sqrt{3} < x < 0$, the second derivative is positive, so the graph is concave up.

For $0 < x < \sqrt{3}$, the second derivative is negative, so the graph is concave down.

For $x > \sqrt{3}$, the second derivative is positive, so the graph is concave up.

Concave up on $$\boxed{(-\sqrt{3}, 0) \cup (\sqrt{3}, \infty)}$$

Concave down on $$\boxed{(-\infty, -\sqrt{3}) \cup (0, \sqrt{3})}$$

(b) Inflection points

Because the concavity changes at each critical value, inflection points occur at all three values.

$$f(0) = 0, \qquad f(\pm\sqrt{3}) = \pm\frac{\sqrt{3}}{4}$$

Inflection points at $$\boxed{ \left(-\sqrt{3}, -\frac{\sqrt{3}}{4}\right), \quad (0, 0), \quad \left(\sqrt{3}, \frac{\sqrt{3}}{4}\right) }$$

Graph of $f(x)$