# Inflection Points

## Related calculator: Inflection Points and Concavity Calculator

**Definition.** Point $$$c$$$ is an **inflection point** of function $$$y={f{{\left({x}\right)}}}$$$ if function at this point changes direction of concavity (i.e. from concave upward becomes concave downward or from concave downward becomes concave upward).

It appears that $$${f{''}}{\left({x}\right)}$$$ plays same role in finding inflection points as played $$${f{'}}{\left({x}\right)}$$$ in finding extrema.

**Fact.** If point $$${c}$$$ is inflection point of the function $$${y}={f{{\left({x}\right)}}}$$$ then either $$${f{''}}{\left({c}\right)}={0}$$$ or $$${f{''}}{\left({c}\right)}$$$ doesn't exist.

Converse is not true. Consider function $$${f{{\left({x}\right)}}}={{x}}^{{4}}$$$. $$${f{''}}{\left({x}\right)}={12}{{x}}^{{2}}$$$, so $$${f{''}}{\left({0}\right)}={0}$$$, but $$${x}={0}$$$ is not an inflection point.

But we know that when $$${f{''}}{\left({x}\right)}>{0}$$$ function is concave upward, and when $$${f{''}}{\left({x}\right)}<{0}$$$ function is concave downward.

So, point will be point of inflection if second derivative changes sign at this point.

This rule is very similar to First Derivative Test for extrema.

**Procedure for Finding Inflection Points**:

- Find points where $$${f{''}}{\left({x}\right)}=0$$$ or $$${f{''}}{\left({x}\right)}$$$ doesn't exist.
- Test all points in the following way: if second derivative changes sign at point then it is point of inflection. Otherwise, it is not a point of inflection.

Now it is clear why $$${x}={0}$$$ is not point of inflection for $$${f{{\left({x}\right)}}}={{x}}^{{4}}$$$. That's because $$${f{''}}{\left({x}\right)}={12}{{x}}^{{2}}\ge{0}$$$ and second derivative doesn't change sign at $$${x}={0}$$$.

**Example 1.** Find points of inflection of $$${f{{\left({x}\right)}}}={{x}}^{{\frac{{1}}{{3}}}}$$$.

We have that $$${f{'}}{\left({x}\right)}=\frac{{1}}{{3}}{{x}}^{{-\frac{{2}}{{3}}}}$$$ and $$${f{''}}{\left({x}\right)}=-\frac{{2}}{{9}}{{x}}^{{-\frac{{5}}{{3}}}}$$$.

Second derivative is never zero, but it doesn't exist at $$${x}={0}$$$.

Since $$${f{''}}{\left({x}\right)}>{0}$$$ for $$${x}<{0}$$$ and $$${f{''}}{\left({x}\right)}<{0}$$$ for $$${x}>{0}$$$ then second derivative changes sign at $$${x}={0}$$$.

This means that $$${x}={0}$$$ is point of inflection.

**Example 2**. Find points of inflection of $$${f{{\left({x}\right)}}}={\sin{{\left({x}\right)}}}$$$.

Since $$${f{'}}{\left({x}\right)}=-{\cos{{\left({x}\right)}}}$$$ and $$${f{''}}{\left({x}\right)}=-{\sin{{\left({x}\right)}}}$$$. then points of inflection are points where $$${\sin{{\left({x}\right)}}}={0}$$$ (clearly at these points second derivative changes sign).

Thus, points of the form $$${x}=\pi{k},{k}\in\mathbb{Z}$$$ are points of inflection.

This means that $$${\sin{{\left({x}\right)}}}$$$ has infinite number of points of inflection.

**Example 3.** Find points of inflection of $$${f{{\left({x}\right)}}}={3}{{x}}^{{5}}-{5}{{x}}^{{4}}+{5}{x}-{7}$$$.

We have that $$${f{'}}{\left({x}\right)}={15}{{x}}^{{4}}-{20}{{x}}^{{3}}+{5}$$$ and $$${f{''}}{\left({x}\right)}={60}{{x}}^{{3}}-{60}{{x}}^{{2}}={60}{{x}}^{{2}}{\left({x}-{1}\right)}$$$.

Derivative is defined for all $$${x}$$$ so there are no such points where derivative doesn't exist.

So, inflection points are among roots of equation $$${f{''}}{\left({x}\right)}={0}$$$.

Equation $$${60}{{x}}^{{2}}{\left({x}-{1}\right)}={0}$$$ gives two roots: $$${x}={0}$$$ and $$${x}={1}$$$.

Now use method of intervals to where second derivative is positive and where negative.

Number line is divided by points $$${x}={0}$$$ and $$${x}={1}$$$ on 3 intervals: $$${\left(-\infty,{0}\right)}\cup{\left({0},{1}\right)}\cup{\left({1},\infty\right)}$$$.

First interval is $$${\left(-\infty,{0}\right)}$$$: here $$${{x}}^{{2}}>{0}$$$ and $$${x}-{1}<{0}$$$ so $$${f{''}}{\left({x}\right)}<{0}$$$.

Second interval is $$${\left({0},{1}\right)}$$$: here $$${{x}}^{{2}}>{0}$$$ and $$${x}-{1}<{0}$$$ so $$${f{''}}{\left({x}\right)}<{0}$$$.

First interval is $$${\left({1},\infty\right)}$$$: here $$${{x}}^{{2}}>{0}$$$ and $$${x}-{1}>{0}$$$ so $$${f{''}}{\left({x}\right)}>{0}$$$.

We see that derivative doesn't change sign at $$${x}={0}$$$ and changes sign from negative to positive at $$${x}={1}$$$.

This means that the only point of inflection is $$${x}={1}$$$.

As in case with extrema we can formulate criterion of finding inflection points using higher-order derivatives.

**Fact.** Suppose that for function $$${y}={f{{\left({x}\right)}}}$$$ there is point $$${c}$$$ such that $$${f{''}}{\left({c}\right)}={0}$$$. Also suppose $$${k}$$$ $$$\left({k}>{2}\right)$$$ is the smallest number such that $$${{f}}^{{{\left({k}\right)}}}{\left({c}\right)}\ne{0}$$$. If $$${k}$$$ is odd then $$${c}$$$ is point of inflection, if $$${k}$$$ is even then there is no point of inflection.

Let's solve Example 3 using above fact.

We have that $$${f{'''}}{\left({x}\right)}={180}{{x}}^{{2}}-{120}{x}$$$.

Since $$${f{'''}}{\left({1}\right)}={60}\ne{0}$$$ and 3 is odd number, then $$${x}={1}$$$ is point of inflection.

Since $$${f{'''}}{\left({0}\right)}={0}$$$, then third derivative is inconclusive for $$${x}={0}$$$.

Find fourth derivative $$${{f}}^{{{\left({4}\right)}}}{\left({x}\right)}={360}{x}-{120}$$$.

Since $$${{f}}^{{{\left({4}\right)}}}{\left({0}\right)}=-{120}\ne{0}$$$ and 4 is even number, then $$${x}={0}$$$ is not a point of inflection.