# Calculus

## Applications of the Derivative # Inflection Points

## Definition of an Inflection Point

Consider a function y = f (x), which is continuous at a point x0. The function f (x) can have a finite or infinite derivative f '(x0) at this point. If, when passing through x0, the function changes the direction of convexity, i.e. there exists a number δ > 0 such that the function is convex upward on one of the intervals (x0δ, x0) or (x0, x0 + δ), and is convex downward on the other, then x0 is called a point of inflection of the function y = f (x).

The geometric meaning of an inflection point is that the graph of the function f (x) passes from one side of the tangent line to the other at this point, i.e. the curve and the tangent line intersect (see Figure 1).

Another interesting feature of an inflection point is that the graph of the function $$f\left( x \right)$$ in the vicinity of the inflection point $${x_0}$$ is located within a pair of the vertical angles formed by the tangent and normal (Figure $$2$$).

## Necessary Condition for an Inflection Point (Second Derivative Test)

If $${x_0}$$ is a point of inflection of the function $$f\left( x \right)$$, and this function has a second derivative in some neighborhood of $${x_0},$$ which is continuous at the point $${x_0}$$ itself, then

$f^{\prime\prime}\left( {{x_0}} \right) = 0.$

### Proof.

Suppose that the second derivative at the inflection point $${x_0}$$ is not zero: $$f^{\prime\prime}\left( {{x_0}} \right) \ne 0.$$ Since it is continuous at $${x_0},$$ then there exists a $$\delta$$-neighborhood of the point $${x_0}$$ where the second derivative preserves its sign, that is

$f^{\prime\prime}\left( {{x_0}} \right) \lt 0\;\;\text{or}\;\;f^{\prime\prime}\left( {{x_0}} \right) \lt 0\;\forall \;x \in \left( {{x_0} - \delta ,{x_0} + \delta } \right).$

In this case, the function is either strictly convex upward (when $$f^{\prime\prime}\left( x \right) \lt 0$$) or strictly convex downward (when $$f^{\prime\prime}\left( x \right) \gt 0$$). But then the point $${x_0}$$ is not an inflection point. Hence, the assumption is wrong and the second derivative of the inflection point must be equal to zero.

## First Sufficient Condition for an Inflection Point (Second Derivative Test)

If the function $$f\left( x \right)$$ is continuous and differentiable at a point $${x_0},$$ has a second derivative $$f^{\prime\prime}\left( {{x_0}} \right)$$ in some deleted $$\delta$$-neighborhood of the point $${x_0}$$ and if the second derivative changes sign when passing through the point $${x_0},$$ then $${x_0}$$ is a point of inflection of the function $$f\left( x \right).$$

### Proof.

Suppose, for example, that the second derivative $$f^{\prime\prime}\left( x \right)$$ changes sign from plus to minus when passing through the point $${x_0}.$$ Hence, in the left $$\delta$$-neighborhood $$\left( {{x_0} - \delta ,{x_0}} \right),$$ the inequality $$f^{\prime\prime}\left( x \right) \gt 0,$$ holds, and in the right $$\delta$$-neighborhood $$\left( {{x_0},{x_0} + \delta } \right),$$ the inequality $$f^{\prime\prime}\left( x \right) \lt 0$$ is valid.

In this case, according to the sufficient conditions for convexity, the function $$f\left( x \right)$$ is convex downward in the left $$\delta$$-neighborhood of the point $${x_0}$$ and is convex upward in the right $$\delta$$-neighborhood.

Consequently, the function changes the direction of convexity at the point $${x_0},$$ that is by definition, $${x_0}$$ is a point of inflection.

## Second Sufficient Condition for an Inflection Point (Third Derivative Test)

Let $$f^{\prime\prime}\left( {{x_0}} \right) = 0,$$ $$f^{\prime\prime\prime}\left( {{x_0}} \right) \ne 0.$$ Then $${x_0}$$ is a point of inflection of the function $$f\left( x \right).$$

### Proof.

As $$f^{\prime\prime\prime}\left( {{x_0}} \right) \ne 0,$$ the second derivative is either strictly increasing at $${x_0}$$ (if $$f^{\prime\prime\prime}\left( {{x_0}} \right) \gt 0$$) or strictly decreasing at this point (if $$f^{\prime\prime\prime}\left( {{x_0}} \right) \lt 0$$). Because $$f^{\prime\prime}\left( {{x_0}} \right) = 0,$$ then the second derivative for some $$\delta \gt 0$$ has different signs in the left and right $$\delta$$-neighborhood of $${x_0}.$$ Hence, on the basis of the previous theorem, it follows that $${x_0}$$ is a point of inflection of the function $$f\left( x \right).$$

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Determine whether the point $$x = 0$$ is an inflection point of the function $f\left( x \right) = \sin x.$

### Example 2

Find the points of inflection of the function $f\left( x \right) = {x^3} - 3{x^2} - 1.$

### Example 3

Find the points of inflection of the function $f\left( x \right) = 2{x^3} + 6{x^2} - 5x + 1.$

### Example 4

Find the inflection points of the function $f\left( x \right) = {x^4} - 6{x^2}.$

### Example 5

Find the points of inflection of the function $f\left( x \right) = 2{x^4} - 3x.$

### Example 6

Find the points of inflection of the function

$f\left( x \right) = {x^4} - 12{x^3} + 48{x^2} + 12x + 1.$

### Example 1.

Determine whether the point $$x = 0$$ is an inflection point of the function $f\left( x \right) = \sin x.$

Solution.

We use the second sufficient condition for the existence of an inflection point. Find the derivatives of the sine function up to the third order:

$f'\left( x \right) = {\left( {\sin x} \right)^\prime } = \cos x,\;\;f^{\prime\prime}\left( x \right) = \left( {\cos x} \right)^\prime = - \sin x,\;\;f^{\prime\prime\prime}\left( x \right) = \left( { - \sin x} \right)^\prime = - \cos x.$

At the point $$x = 0,$$ the second and third derivatives have the following values:

$f^{\prime\prime}\left( 0 \right) = - \sin 0 = 0,\;\;f^{\prime\prime\prime}\left( 0 \right) = - \cos 0 = - 1.$

Thus, we have $$f^{\prime\prime}\left( {{x_0}} \right) = 0,$$ $$f^{\prime\prime\prime}\left( {{x_0}} \right) \ne 0.$$ Hence, by the second sufficient condition, the point $$x = 0$$ is a point of inflection.

### Example 2.

Find the points of inflection of the function $f\left( x \right) = {x^3} - 3{x^2} - 1.$

Solution.

Compute the first and second derivatives:

$f^\prime\left( x \right) = \left( {{x^3} - 3{x^2} - 1} \right)^\prime = 3{x^2} - 6x;$
$f^{\prime\prime}\left( x \right) = \left( {3{x^2} - 6x} \right)^\prime = 6x - 6.$

We see that $$f^{\prime\prime}\left( x \right) = 0$$ at $$x = 1.$$ The function changes concavity as shown in figure above. Since

$f\left( 1 \right) = {1^3} - 3 \cdot {1^2} - 1 = - 3,$

the inflection point is at $$\left( {1, - 3} \right).$$

### Example 3.

Find the points of inflection of the function $f\left( x \right) = 2{x^3} + 6{x^2} - 5x + 1.$

Solution.

We differentiate this function twice to get the second derivative:

$f^\prime\left( x \right) = \left( {2{x^3} + 6{x^2} - 5x + 1} \right)^\prime = 6{x^2} + 12x - 5;$
$f^{\prime\prime}\left( x \right) = \left( {6{x^2} + 12x - 5} \right)^\prime = 12x + 12.$

Clearly that $$f^{\prime\prime}\left( x \right)$$ exists for all $$x.$$ Determine the points where it is equal to zero:

$f^{\prime\prime}\left( x \right) = 0,\;\; \Rightarrow 12x + 12 = 0,\;\; \Rightarrow x = - 1.$

The function $$f\left( x \right)$$ is concave down $$\left( {f^{\prime\prime} \lt 0} \right)$$ for $$x \lt -1$$ and it is concave up $$\left( {f^{\prime\prime} \gt 0} \right)$$ for $$x \gt -1.$$ Therefore, $$x = -1$$ is an inflection point.

Calculate the corresponding $$y-$$coordinate:

$f\left( { - 1} \right) = 2 \cdot {\left( { - 1} \right)^3} + 6 \cdot {\left( { - 1} \right)^2} - 5 \cdot \left( { - 1} \right) + 1 = - 2 + 6 + 5 + 1 = 10.$

So, the inflection point is $$\left( {-1, 10} \right).$$

### Example 4.

Find the inflection points of the function $f\left( x \right) = {x^4} - 6{x^2}.$

Solution.

Compute the first derivative:

$f^\prime\left( x \right) = \left( {{x^4} - 6{x^2}} \right)^\prime = 4{x^3} - 12x.$

The second derivative is

$f^{\prime\prime}\left( x \right) = \left( {4{x^3} - 12x} \right)^\prime = 12{x^2} - 12 = 12\left( {{x^2} - 1} \right).$

Find the roots of the second derivative:

$f^{\prime\prime}\left( x \right) = 0,\;\; \Rightarrow 12\left( {{x^2} - 1} \right) = 0,\;\; \Rightarrow {x_1} = - 1,{x_2} = 1.$

We need to determine where the second derivative changes sign. Draw a sign chart for $$f^{\prime\prime}\left( x \right)$$ (see above).

Clearly, the concavity changes at both points, $$x = -1$$ and $$x = 1.$$ Hence, these points are points of inflection.

We can easily calculate their $$y-$$coordinates:

$f\left( { - 1} \right) = \left( { - 1} \right)^4 - 6 \cdot \left( { - 1} \right)^2 = - 5;$
$f\left( 1 \right) = {1^4} - 6 \cdot {1^2} = - 5.$

So, the inflection points are $$\left( {-1,-5} \right)$$ and $$\left( {1,-5} \right).$$

### Example 5.

Find the points of inflection of the function $f\left( x \right) = 2{x^4} - 3x.$

Solution.

We compute the derivatives and draw a sign chart for the second derivative.

$f^\prime\left( x \right) = \left( {2{x^4} - 3x} \right)^\prime = 8{x^3} - 3;$
$f^{\prime\prime}\left( x \right) = \left( {8{x^3} - 3} \right)^\prime = 24{x^2}.$

We see that the concavity does not change at $$x = 0.$$ Consequently, $$x = 0$$ is not a point of inflection.

The second derivative is a continuous function defined over all $$x$$. Therefore, we conclude that $$f\left( x \right)$$ has no inflection points.

### Example 6.

Find the points of inflection of the function

$f\left( x \right) = {x^4} - 12{x^3} + 48{x^2} + 12x + 1.$

Solution.

Find the derivatives:

$f'\left( x \right) = {\left( {{x^4} - 12{x^3} + 48{x^2} + 12x + 1} \right)^\prime } = 4{x^3} - 36{x^2} + 96x + 12 = 4\left( {{x^3} - 9{x^2} + 24x + 3} \right);$
$f^{\prime\prime}\left( x \right) = \left( {4\left( {{x^3} - 9{x^2} + 24x + 3} \right)} \right)^\prime = 4\left( {3{x^2} - 18x + 24} \right) = 12\left( {{x^2} - 6x + 8} \right).$

Calculate the roots of the second derivative:

$f^{\prime\prime}\left( x \right) = 0,\;\; \Rightarrow 12\left( {{x^2} - 6x + 8} \right) = 0,\;\; \Rightarrow {x^2} - 6x + 8 = 0,\;\; \Rightarrow {x_1} = 2,\;{x_2} = 4.$

In this case it is convenient to use the second sufficient condition for the existence of an inflection point. The third derivative is written as

$f^{\prime\prime\prime}\left( x \right) = {\left( {12\left( {{x^2} - 6x + 8} \right)} \right)^\prime } = 12\left( {2x - 6} \right) = 24\left( {x - 3} \right).$

From this we immediately see that the third derivative is not zero at the points $${x_1} = 2$$ and $${x_2} = 4.$$ Therefore, these points are points of inflection.

See more problems on Page 2.