Concavity and inflection point of a rational function

Problem

Study the concavity, convexity and inflection points of the function $f(x) = \frac{2x+1}{x^2}.$

Solution

First derivative. $f'(x) = -\frac{2}{x^2}-\frac{2}{x^3} = -\frac{2x+2}{x^3}.$ It vanishes for $x=-1$ (a stationary point, to be classified by studying the sign of $f'$).

Second derivative. $f''(x) = \frac{4}{x^3}+\frac{6}{x^4} = \frac{4x+6}{x^4}, \qquad f''=0 \iff x = -\frac{3}{2}.$

Checking the change of sign. Since $x^4>0$, the sign of $f''$ is that of $4x+6$:

• for $x < -\tfrac{3}{2}$ we have $f''<0$: the function is concave;
• for $-\tfrac{3}{2} < x < 0$ we have $f''>0$: the function is convex.

The concavity changes: the point $x=-\dfrac{3}{2}$ is an $\boxed{\text{inflection point}}$.

Links

Topics: Curve sketching
Concepts: Concavity and convexity · Second derivative · Inflection point
Functions: Rational function
Skills: Differentiating · Sketching a function
Exercise type: Curve sketching