How to find concavity and inflection points from the second derivative

Expert intro: Concavity questions reward patience. Once the second derivative is simplified, the rest is really a sign chart problem with a little interpretation at the end.

Detailed walkthrough

We are given

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

We want the intervals of concavity and the inflection points.

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1) First derivative

Use the quotient rule:

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

Simplify:

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

2) Second derivative

Differentiate again. A simplified form is

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

3) Critical values for concavity

Set the numerator equal to zero:

$2x(x^2-3)=0$

So

$x=0,\quad x=\pm\sqrt{3}.$

The denominator $(x^2+1)^3$ is always positive, so the sign of $f''(x)$ depends only on

$2x(x^2-3).$

4) Sign chart

Check the intervals:

• $(-\infty,-\sqrt{3})$
• $(-\sqrt{3},0)$
• $(0,\sqrt{3})$
• $(\sqrt{3},\infty)$

The sign of $f''(x)$ gives:

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

5) Inflection points

Inflection points occur where concavity changes. Evaluate $f(x)$ at the candidate $x$-values:

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

So the inflection points are

$\left(-\sqrt{3},-\frac{\sqrt{3}}{4}\right),\ (0,0),\ \left(\sqrt{3},\frac{\sqrt{3}}{4}\right).$

💡 Pitfall guide

A common trap is stopping after finding where $f''(x)=0$. That is not enough by itself. You still need a sign change in concavity. Another mistake is forgetting that the denominator is always positive here, so it never changes the sign of $f''(x)$. That makes the numerator the only thing that matters.

🔄 Real-world variant

If the function were $f(x)=\frac{x}{x^2+a^2}$ with $a>0$, the same strategy would work, but the algebraic details of $f''(x)$ would change. For a polynomial instead of a rational function, the denominator issue disappears, but the sign-chart method for concavity stays the same. If you only need intervals of concavity and not inflection points, the sign of $f''$ is enough.

🔍 Related terms

second derivative, inflection point, concavity