# Calculus

## Applications of the Derivative # Inflection Points

• ### Definition of an Inflection Point

Consider a function $$y = f\left( x \right),$$ which is continuous at a point $${x_0}.$$ The function $$f\left( x \right)$$ can have a finite or infinite derivative $$f’\left( {{x_0}} \right)$$ at this point. If, when passing through $${x_0}$$, the function changes the direction of convexity, i.e. there exists a number $$\delta \gt 0$$ such that the function is convex upward on one of the intervals $$\left( {{x_0} – \delta ,{x_0}} \right)$$ or $$\left( {{x_0},{x_0} + \delta } \right)$$, and is convex downward on the other, then $${x_0}$$ is called a point of inflection of the function $$y = f\left( x \right).$$

The geometric meaning of an inflection point is that the graph of the function $$f\left( x \right)$$ 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

${f^{\prime\prime}\left( {{x_0}} \right) \lt 0\;\;\text{or}\;\;\;}\kern-0.3pt{f^{\prime\prime}\left( {{x_0}} \right) \lt 0\;\forall \;x \in \left( {{x_0} – \delta ,{x_0} + \delta } \right).}$

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

### 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 ${f\left( x \right) \text{ = }}\kern0pt{ {x^4} – 12{x^3} + 48{x^2} + 12x + 1.}$

### Example 7

Find the inflection points of the function $$f\left( x \right) = 3{x^5} + 5{x^4} – 20{x^3}.$$

### Example 8

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

### Example 9

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

### Example 10

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

### Example 11

Find the inflection points of the function $$f\left( x \right) = x + {x^{\large{\frac{5}{3}}\normalsize}}.$$

### Example 12

Find the points of inflection of the function $$f\left( x \right) = {e^{\sin x}}.$$

### Example 13

For what values of $$a$$ and $$b$$ the point $$\left( { – 1,2} \right)$$ is an inflection point of the graph of the function $$y\left( x \right) = a{x^3} + b{x^2}?$$

### Example 14

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

### Example 15

Find the inflection points of the function $$f\left( x \right) = {x^2} – \large{\frac{1}{{{x^2}}}}\normalsize.$$

### Example 16

Find the points of inflection of the function $$f\left( x \right) = \large{\frac{{{x^2}}}{{1 + {x^2}}}}\normalsize.$$

### Example 17

Find the inflection points of a Gaussian function.

### Example 18

Find the points of inflection of the function $f\left( x \right) = \sqrt[\large 3\normalsize]{{{x^2}\left( {x + 1} \right)}}.$

### Example 19

Find the inflection points of the curve defined by the parametric equations: $x = {t^2},\;\;y = t + {t^3}.$

### Example 20

Show that the graph of the function $$y = {\large\frac{{x + 1}}{{{x^2} + 1}}\normalsize}$$ has three points of inflection lying on one straight line.

### 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:

${f’\left( x \right) = {\left( {\sin x} \right)^\prime } = \cos x,\;\;\;}\kern-0.3pt {f^{\prime\prime}\left( x \right) = {\left( {\cos x} \right)^\prime } = – \sin x,\;\;\;}\kern-0.3pt {f^{\prime\prime\prime}\left( x \right) = {\left( { – \sin x} \right)^\prime } = – \cos x.}$

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

${f^{\prime\prime}\left( 0 \right) = – \sin 0 = 0,\;\;\;}\kern-0.3pt {f^{\prime\prime\prime}\left( 0 \right) = – \cos 0 = – 1.}$

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:

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

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

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 ${f\left( x \right) \text{ = }}\kern0pt{ {x^4} – 12{x^3} + 48{x^2} + 12x + 1.}$

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.

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Problems 1-6
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Problems 7-20