The Shape of a Graph, Part II — Question 3

Question 3 - Solution

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

First derivative

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

Second derivative

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

(a) Concavity

Setting $f''(x) = 0$ yields $$1 - 3x^2 = 0 \Rightarrow x = \pm \frac{1}{\sqrt{3}}.$$

For $|x| < \frac{1}{\sqrt{3}}$, the second derivative is positive, so the graph is concave up.

For $|x| > \frac{1}{\sqrt{3}}$, the second derivative is negative, so the graph is concave down.

Concave up on $$\boxed{\left(-\frac{1}{\sqrt{3}},\,\frac{1}{\sqrt{3}}\right)}$$

Concave down on $$\boxed{\left(-\infty,-\frac{1}{\sqrt{3}}\right) \cup \left(\frac{1}{\sqrt{3}},\infty\right)}$$

(b) Inflection points

Because the concavity changes at $x = \pm \frac{1}{\sqrt{3}}$, inflection points occur there.

$$f\!\left(\pm \frac{1}{\sqrt{3}}\right) = \frac{1}{4}$$

Inflection points at $$\boxed{\left(-\frac{1}{\sqrt{3}}, \frac{1}{4}\right) \quad \text{and} \quad \left(\frac{1}{\sqrt{3}}, \frac{1}{4}\right)}$$

Graph of $f(x)$