>Applications of the Derivative>What Are Inflection Points of a Function?

What Are Inflection Points of a Function?

An inflection point shows where a graph changes from being concave to convex—or the opposite. This point is also where the graph is increasing or decreasing the fastest. You find it by solving the equation $f^{″} (x) = 0$ .

Rule

Inflection Points

You find the inflection points of a function by solving the equation

f^{″} (x) = 0

and using a sign chart.

The sign chart of the second derivative $f^{″} (x)$ shows where the graph of the main function $f (x)$ is convex and concave. It also shows where the graph has inflection points. Furthermore it shows where $f^{″} (x)$ is above and below the $x$ -axis.

Example 1

Find the inflection point of the function

f (x) = 3 x^{3} + 2 x^{2} - 4 x + 3

You know you need the second derivative $f^{″} (x)$ to find the inflection point, so you differentiate the function $f (x)$ twice:

\begin{array}{l} f^{'} (x) & = 9 x^{2} + 4 x - 4 \\ f^{″} (x) & = 18 x + 4 \end{array}

You then solve the equation $f^{″} (x) = 0$ : $\begin{array}{l} 18 x + 4 & = 0 \\ 18 x & = - 4 | \div 18 \\ x & = - \frac{4}{18} \\ x & = - \frac{2}{9} \end{array}$

You then find the $y$ -value of the inflection point by inserting $x = - \frac{2}{9}$ into the main function $f (x) = 3 x^{3} + 2 x^{2} - 4 x + 3$ . You then get

\begin{array}{l} f (- \frac{2}{9}) & = 3 {(- \frac{2}{9})}^{3} + 2 {(- \frac{2}{9})}^{2} \\ - 4 (- \frac{2}{9}) + 3 \\ \approx 4 \end{array}

\begin{array}{l} f (- \frac{2}{9}) & = 3 {(- \frac{2}{9})}^{3} + 2 {(- \frac{2}{9})}^{2} - 4 (- \frac{2}{9}) + 3 \\ \approx 4 \end{array}

The inflection point of

f (x)

is thus the point

(- \frac{2}{9}, 4)

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