# Inflection Points

## Solved Problems

Click or tap a problem to see the solution.

### 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 $f\left( x \right) = x + {x^{\frac{5}{3}}}.$

### 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 $f\left( x \right) = \frac{{{x^3}}}{{1 - {x^2}}}.$

### Example 15

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

### Example 16

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

### Example 17

Find the inflection points of a Gaussian function.

### Example 18

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

### 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 $y = {\frac{{x + 1}}{{{x^2} + 1}}}$ has three points of inflection lying on one straight line.

### Example 7.

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

Solution.

Compute the derivatives:

$f^\prime\left( x \right) = \left( {3{x^5} + 5{x^4} - 20{x^3}} \right)^\prime = 15{x^4} + 20{x^3} - 60{x^2};$
$f^{\prime\prime}\left( x \right) = \left( {15{x^4} + 20{x^3} - 60{x^2}} \right)^\prime = 60{x^3} + 60{x^2} - 120x = 60x\left( {{x^2} + x - 2} \right) = 60x\left( {x + 1} \right)\left( {x - 2} \right).$

The second derivatives has zeros at $$x = -1,$$ $$x = 0,$$ $$x = 2.$$

We use the third derivative test to identify inflection points.

$f^{\prime\prime\prime}\left( x \right) = \left( {60{x^3} + 60{x^2} - 120x} \right)^\prime = 180{x^2} + 120x - 120 = 60\left( {3{x^2} + 2x - 2} \right).$

Compute the values of the third derivative at these points:

$f^{\prime\prime\prime}\left( { - 1} \right) = 60 \cdot \left( {3 \cdot {{\left( { - 1} \right)}^2} + 2 \cdot \left( { - 1} \right) - 2} \right) = - 60 \ne 0;$
$f^{\prime\prime\prime}\left( {0} \right) = 60 \cdot \left( {3 \cdot {{0}^2} + 2 \cdot 0 - 2} \right) = - 120 \ne 0;$
$f^{\prime\prime\prime}\left( {2} \right) = 60 \cdot \left( {3 \cdot {{2}^2} + 2 \cdot 2 - 2} \right) = 840 \ne 0.$

We have checked that $$f^{\prime\prime\prime}\left( x \right) \ne 0$$ at all three points: $$x = -1,$$ $$x = 0$$ and $$x = 2.$$ Hence, these are inflection points.

Given that

$f\left( { - 1} \right) = 3 \cdot {\left( { - 1} \right)^5} + 5 \cdot {\left( { - 1} \right)^4} - 20 \cdot {\left( { - 1} \right)^3} = 22;$
$f\left( {0} \right) = 3 \cdot {0^5} + 5 \cdot {0^4} - 20 \cdot {0^3} = 0;$
$f\left( {2} \right) = 3 \cdot {2^5} + 5 \cdot {2^4} - 20 \cdot {2^3} = 16,$

$3 \text{ inflection points: } \left( { - 1,22} \right), \left( {0,0} \right), \left( {2,16} \right).$

### Example 8.

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

Solution.

Determine the second derivative:

$f'\left( x \right) = \left( {{x^2}\ln x} \right)^\prime = {\left( {{x^2}} \right)^\prime }\ln x + {x^2}{\left( {\ln x} \right)^\prime } = 2x\ln x + {x^2} \cdot \frac{1}{x} = 2x\ln x + x = x\left( {2\ln x + 1} \right);$
$f^{\prime\prime}\left( x \right) = \left[ {x\left( {2\ln x + 1} \right)} \right]^\prime = x'\left( {2\ln x + 1} \right) + x{\left( {2\ln x + 1} \right)^\prime } = 2\ln x + 1 + x \cdot \frac{2}{x} = 2\ln x + 3.$

Solve the equation $$f^{\prime\prime}\left( x \right) = 0:$$

$\Rightarrow 2\ln x + 3,\;\; \Rightarrow \ln x = - \frac{3}{2},\;\; \Rightarrow x = {e^{ - \frac{3}{2}}} = \frac{1}{{\sqrt {{e^3}} }}.$

The second derivative is a monotonically increasing function. Therefore, the second derivative changes sign at the point found above. Hence, this point is the point of inflection.

### Example 9.

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

Solution.

Calculate the first and second derivatives:

$f^\prime\left( x \right) = \left( {x{e^{ - 2x}}} \right)^\prime = x^\prime \cdot {e^{ - 2x}} + x \cdot \left( {{e^{ - 2x}}} \right)^\prime = 1 \cdot {e^{ - 2x}} + x \cdot {e^{ - 2x}} \cdot \left( { - 2} \right) = {e^{ - 2x}} - 2x{e^{ - 2x}} = \left( {1 - 2x} \right){e^{ - 2x}};$
$f^{\prime\prime}\left( x \right) = \left[ {\left( {1 - 2x} \right){e^{ - 2x}}} \right]^\prime = \left( {1 - 2x} \right)^\prime \cdot {e^{ - 2x}} + \left( {1 - 2x} \right) \cdot \left( {{e^{ - 2x}}} \right)^\prime = - 2 \cdot {e^{ - 2x}} + \left( {1 - 2x} \right) \cdot {e^{ - 2x}} \cdot \left( { - 2} \right) = - 2{e^{ - 2x}} - \left( {2 - 4x} \right){e^{ - 2x}} = \left( {4x - 4} \right){e^{ - 2x}}.$

Solve the equation $$f^{\prime\prime}\left( x \right) = 0:$$

$f^{\prime\prime}\left( x \right) = 0,\;\; \Rightarrow \left( {4x - 4} \right){e^{ - 2x}} = 0,\;\; \Rightarrow 4x - 4 = 0,\;\; \Rightarrow x = 1.$

Since $${e^{ - 2x}}$$ is always positive, then $$f^{\prime\prime}\left( x \right)$$ is negative to the left of $$1$$ and positive to the right of $$1.$$ Thus, $$x = 1$$ is a point of inflection.

$f\left( 1 \right) = 1 \cdot {e^{ - 2 \cdot 1}} = {e^{ - 2}} = \frac{1}{{{e^2}}}.$

Answer: $$\left( {1,\frac{1}{{{e^2}}}} \right).$$

### Example 10.

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

Solution.

Compute the derivatives:

$f'\left( x \right) = {\left( {{e^{ - {x^2}}}} \right)^\prime } = {e^{ - {x^2}}} \cdot {\left( { - {x^2}} \right)^\prime } = - 2x{e^{ - {x^2}}};$
$f^{\prime\prime}\left( x \right) = {\left( { - 2x{e^{ - {x^2}}}} \right)^\prime } = {\left( { - 2x} \right)^\prime }{e^{ - {x^2}}} + \left( { - 2x} \right){\left( {{e^{ - {x^2}}}} \right)^\prime } = - 2{e^{ - {x^2}}} - 2x{e^{ - {x^2}}} \cdot \left( { - 2x} \right) = - 2{e^{ - {x^2}}} + 4{x^2}{e^{ - {x^2}}} = 4{e^{ - {x^2}}}\left( {{x^2} - \frac{1}{2}} \right).$

The second derivative is equal to zero at the following points:

$f^{\prime\prime}\left( x \right) = 0,\;\; \Rightarrow 4{e^{ - {x^2}}}\left( {{x^2} - \frac{1}{2}} \right) = 0,\;\; \Rightarrow {x^2} - \frac{1}{2} = 0,\;\; \Rightarrow {x_{1,2}} = \pm \frac{1}{{\sqrt 2 }}.$

When passing through the points $${x_1} = -{\frac{1}{{\sqrt 2 }}},$$ $${x_2} = {\frac{1}{{\sqrt 2 }}},$$ the second derivative changes its sign (Figure $$6$$). Hence, these points are points of inflection.

### Example 11.

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

Solution.

The function is defined for all $$x \in \mathbb{R}.$$ Take the first and second derivatives:

$f^\prime\left( x \right) = \left( {x + {x^{\frac{5}{3}}}} \right)^\prime = 1 + \frac{5}{3}{x^{\frac{2}{3}}};$
$f^{\prime\prime}\left( x \right) = \left( {1 + \frac{5}{3}{x^{\frac{2}{3}}}} \right)^\prime = 0 + \frac{5}{3} \cdot \frac{2}{3}{x^{ - \frac{1}{3}}} = \frac{{10}}{{9\sqrt[3]{x}}}.$

We see that the $$2$$nd derivative is undefined at $$x = 0.$$ For $$x \lt 0,$$ the second derivative is negative. For $$x \gt 0,$$ it is positive.

Hence, the graph of $$f\left( x \right)$$ has an inflection point at $$x = 0.$$

Compute the $$y-$$value at this point:

$f\left( 0 \right) = 0 + {0^{\frac{5}{3}}} = 0.$

So, the inflection point is $$\left( {0,0} \right).$$

### Example 12.

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

Solution.

Determine the derivatives of the given function:

$f'\left( x \right) = \left( {{e^{\sin x}}} \right)^\prime = {e^{\sin x}}{\left( {\sin x} \right)^\prime} = {e^{\sin x}}\cos x;$
$f^{\prime\prime}\left( x \right) = \left( {{e^{\sin x}}\cos x} \right)^\prime = {\left( {{e^{\sin x}}} \right)^\prime }\cos x + {e^{\sin x}}{\left( {\cos x} \right)^\prime} = {e^{\sin x}}{\cos ^2}x - {e^{\sin x}}\sin x = {e^{\sin x}}\left( {{{\cos }^2}x - \sin x} \right).$

The inflection points must satisfy the equation $$f^{\prime\prime}\left( x \right) = 0.$$ Then

$f^{\prime\prime}\left( x \right) = 0,\;\; \Rightarrow {e^{\sin x}}\left( {{{\cos }^2}x - \sin x} \right) = 0,\;\; \Rightarrow {\cos ^2}x - \sin x = 0,\;\; \Rightarrow 1 - {\sin ^2}x - \sin x = 0,\;\; \Rightarrow {\sin ^2}x + \sin x - 1 = 0.$

This trigonometric equation from school mathematics can be solved by replacing $$\sin x = t.$$ Consequently,

${t^2} + t - 1 = 0,\;\; \Rightarrow D = 1 - 4 \cdot \left( { - 1} \right) = 5,\;\; \Rightarrow {t_{1,2}} = \frac{{ - 1 \pm \sqrt 5 }}{2}.$

The first root $${t_1}$$ with a minus sign in the numerator must be discarded since $${t_1} \lt -1.$$ Then the equation has the following unique solution:

$t = \frac{{ - 1 + \sqrt 5 }}{2} = \frac{{\sqrt 5 - 1}}{2},\;\; \Rightarrow \sin x = t = \frac{{\sqrt 5 - 1}}{2},\;\; \Rightarrow {x_n} = {\left( { - 1} \right)^n}\arcsin \frac{{\sqrt 5 - 1}}{2},\;\;n \in Z.$

Thus, we have an infinite number of points at which the second derivative is zero. There are two solutions within the interval $$\left[ {0,2\pi } \right],$$ which are equal

${x_1} = \arcsin \frac{{\sqrt 5 - 1}}{2} \approx 0,66;\;\;{x_2} = \pi - \arcsin \frac{{\sqrt 5 - 1}}{2} \approx 2,48.$

The remaining solutions $${x_n}$$ are formed owing to the periodicity of the sine and hence the entire function $$f\left( x \right).$$

Both solutions $${x_1}$$ and $${x_2}$$ correspond to the right root of the quadratic equation. It is clear that when passing through the root (in the case of two distinct real roots), a quadratic function changes its sign.

Therefore, the second derivative of the form

$f^{\prime\prime}\left( x \right) = {e^{\sin x}}\left( {{{\cos }^2}x - \sin x} \right) = {e^{\sin x}}\left( { - {\sin^2}x - \sin x + 1} \right)$

also changes sign when passing through the points $${x_n}.$$ Hence, according to the first sufficient criterion, it follows that all these points $${x_n}$$ are points of inflection. A schematic view of the function on the interval $$\left[ {0,2\pi } \right]$$ is shown in Figure $$7.$$

### 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 $y\left( x \right) = a{x^3} + b{x^2}?$

Solution.

The first and second derivatives of the given cubic function are expressed by the formulas:

$y'\left( x \right) = \left( {a{x^3} + b{x^2}} \right)^\prime = 3a{x^2} + 2bx,$
$y^{\prime\prime}\left( x \right) = \left( {3a{x^2} + 2bx} \right)^\prime = 6ax + 2b.$

At an inflection point, the second derivative is zero, that is the first equation is as follows:

$6ax + 2b = 0\;\;\text{or}\;\;3ax + b = 0.$

The second equation is the equation of the function. As a result, we obtain the following system for determining $$a$$ and $$b:$$

$\left\{ \begin{array}{l} 3ax + b = 0\\ a{x^3} + b{x^2} = y \end{array} \right..$

Substitute the known coordinates of the inflection point of the graph of the function:

$\left\{ \begin{array}{l} 3a \cdot \left( { - 1} \right) + b = 0\\ a \cdot {\left( { - 1} \right)^3} + b \cdot {\left( { - 1} \right)^2} = y \end{array} \right.,\;\; \Rightarrow \left\{ \begin{array}{l} b - 3a = 0\\ b - a = 2 \end{array} \right.,\;\; \Rightarrow \left\{ \begin{array}{l} 2a = 2\\ b = a + 2 \end{array} \right.,\;\; \Rightarrow \left\{ \begin{array}{l} a = 1\\ b = 3 \end{array} \right..$

Using the second sufficient condition, we verify that the inflection point is found correctly. Check that the third derivative is not equal to zero:

$y^{\prime\prime\prime}\left( x \right) = \left( {6ax + 2b} \right)^\prime = 6a \ne 0\;\;\text{at}\;\;a = 1.$

Consequently, the given point $$\left( { - 1,2} \right)$$ is the inflection point of the graph of the function $$y\left( x \right) = -{x^3} + 2{x^2}.$$

### Example 14.

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

Solution.

The function is defined for all $$x$$ except the points $$x = \pm 1$$ where it has discontinuities:

Calculate the first derivative using the quotient rule:

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

Continue differentiating to find the second derivative:

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

Solve the equation $$f^{\prime\prime}\left( x \right) = 0:$$

$f^{\prime\prime}\left( x \right) = 0,\;\; \Rightarrow \frac{{2x\left( {{x^2} + 3} \right)}}{{{{\left( {1 - {x^2}} \right)}^3}}} = 0,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {2x\left( {{x^2} + 3} \right) = 0}\\ {{{\left( {1 - {x^2}} \right)}^3} \ne 0} \end{array}} \right.,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {x = 0}\\ {x \ne \pm 1} \end{array}} \right., \Rightarrow x = 0.$

Draw a sign chart for $$f^{\prime\prime}\left( x \right)$$ (see above). We also indicate the points $$x = \pm 1$$ on the number line as the function may change concavity around these points.

To determine the $$+$$ and $$-$$ signs for the second derivative, we just test any point in each region:

$f^{\prime\prime}\left( { - 2} \right) = \frac{{2 \cdot \left( { - 2} \right)\left( {{{\left( { - 2} \right)}^3} + 3} \right)}}{{{{\left( {1 - {{\left( { - 2} \right)}^2}} \right)}^3}}} = 1.037 \gt 0;$
$f^{\prime\prime}\left( { -0.5} \right) = \frac{{2 \cdot \left( { -0.5} \right)\left( {{{\left( { -0.5} \right)}^3} + 3} \right)}}{{{{\left( {1 - {{\left( { - 0.5} \right)}^2}} \right)}^3}}} = -7.704 \lt 0;$
$f^{\prime\prime}\left( { 0.5} \right) = \frac{{2 \cdot \left( { 0.5} \right)\left( {{{\left( { 0.5} \right)}^3} + 3} \right)}}{{{{\left( {1 - {{\left( {0.5} \right)}^2}} \right)}^3}}} = 7.704 \gt 0;$
$f^{\prime\prime}\left( { 2} \right) = \frac{{2 \cdot 2 \cdot \left( {{{2}^3} + 3} \right)}}{{{{\left( {1 - {{2}^2}} \right)}^3}}} = -1.037 \lt 0.$

Note that both points $$x = \pm 1$$ are not inflection points because the function itself is undefined at these points.

As to $$x = 0,$$ we see that the second derivative changes sign around it. Hence, it is a point of inflection of the given function.

Calculate the $$y-$$coordinate:

$f\left( 0 \right) = \frac{{{0^3}}}{{1 - {0^2}}} = 0.$

So, the point of inflection is at $$\left( {0,0} \right).$$

### Example 15.

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

Solution.

Calculate the first and second derivatives:

$f^\prime\left( x \right) = \left( {{x^2} - \frac{1}{{{x^2}}}} \right)^\prime = 2x + \frac{2}{{{x^3}}} ;$
$f^{\prime\prime}\left( x \right) = \left( {2x + \frac{2}{{{x^3}}}} \right)^\prime = 2 - \frac{6}{{{x^4}}} = \frac{{2\left( {{x^4} - 3} \right)}}{{{x^4}}}.$

Solve the equation $$f^{\prime\prime}\left( x \right) = 0:$$

$f^{\prime\prime}\left( x \right) = 0,\;\; \Rightarrow \frac{{2\left( {{x^4} - 3} \right)}}{{{x^4}}} = 0,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {2\left( {{x^4} - 3} \right) = 0}\\ {{x^4} \ne 0} \end{array}} \right.,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {x = \pm \sqrt[4]{3}}\\ {x \ne 0} \end{array}} \right.,\;\; \Rightarrow x = \pm \sqrt[4]{3} \approx \pm 1.316$

Draw a sign chart for $$f^{\prime\prime}\left( x \right)$$ and test a point in each region:

$f^{\prime\prime}\left( { - 2} \right) = \frac{{2\left( {{{\left( { - 2} \right)}^4} - 3} \right)}}{{{{\left( { - 2} \right)}^4}}} = 1.625 \gt 0;$
$f^{\prime\prime}\left( { - 1} \right) = \frac{{2\left( {{{\left( { - 1} \right)}^4} - 3} \right)}}{{{{\left( { - 1} \right)}^4}}} = -4 \lt 0.$

As the function is even, we get

$f^{\prime\prime}\left( {1} \right) = -4 \lt 0;$
$f^{\prime\prime}\left( {2} \right) = 1.625 \gt 0.$

Hence, we have two inflection points: $${x_{1,2}} = \pm \sqrt[4]{3}.$$ Calculate the $$y-$$coordinates:

$f\left( { \pm \sqrt[4]{3}} \right) = {\left( { \pm \sqrt[4]{3}} \right)^2} - \frac{1}{{{{\left( { \pm \sqrt[4]{3}} \right)}^2}}} = \sqrt 3 - \frac{1}{{\sqrt 3 }} = \frac{{3 - 1}}{{\sqrt 3 }} = \frac{2}{{\sqrt 3 }}.$
$\text{Answer: } \left( { - \sqrt[4]{3},\frac{2}{{\sqrt 3 }}} \right), \left( {\sqrt[4]{3},\frac{2}{{\sqrt 3 }}} \right).$

### Example 16.

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

Solution.

Take the first derivative:

$f^\prime\left( x \right) = \left( {\frac{{{x^2}}}{{1 + {x^2}}}} \right)^\prime = \frac{{2x \cdot \left( {1 + {x^2}} \right) - {x^2} \cdot 2x}}{{{{\left( {1 + {x^2}} \right)}^2}}} = \frac{{2x + \cancel{2{x^3}} - \cancel{2{x^3}}}}{{{{\left( {1 + {x^2}} \right)}^2}}} = \frac{{2x}}{{{{\left( {1 + {x^2}} \right)}^2}}}.$

The second derivative is given by

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

Determine the roots of the second derivative.

$f^{\prime\prime}\left( x \right) = 0,\;\; \Rightarrow \frac{{2 - 6{x^2}}}{{{{\left( {1 + {x^2}} \right)}^3}}} = 0,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {2 - 6{x^2} = 0}\\ {{{\left( {1 + {x^2}} \right)}^3} \ne 0} \end{array}} \right.,\;\; \Rightarrow {x^2} = \frac{1}{3},\;\; \Rightarrow x = \pm \frac{{\sqrt 3 }}{3} \approx \pm 0.58$

We test the values of the second derivative in each region to draw the sign chart.

$f^{\prime\prime}\left( { - 1} \right) = \frac{{2 - 6 \cdot {{\left( { - 1} \right)}^2}}}{{{{\left( {1 + {{\left( { - 1} \right)}^2}} \right)}^3}}} = \frac{{ - 4}}{8} = - \frac{1}{2} \lt 0;$
$f^{\prime\prime}\left( {0} \right) = \frac{{2 - 6 \cdot {0^2}}}{{{{\left( {1 + {{0^2}}} \right)}^3}}} = \frac{{2}}{1} = 2 \gt 0.$

$$f^{\prime\prime}\left( {1} \right) = f^{\prime\prime}\left( {-1} \right)$$ as the function is even.

So, the $$2$$nd derivative changes sign around both points $${x_{1,2}} = \pm \frac{{\sqrt 3 }}{3}.$$ Hence, these points are inflection points.

Given that

$f\left( { \pm \frac{{\sqrt 3 }}{3}} \right) = \frac{{{{\left( { \pm \frac{{\sqrt 3 }}{3}} \right)}^2}}}{{1 + {{\left( { \pm \frac{{\sqrt 3 }}{3}} \right)}^2}}} = \frac{{\frac{1}{3}}}{{1 + \frac{1}{3}}} = \frac{{\frac{1}{3}}}{{\frac{4}{3}}} = \frac{1}{4},$

$\text{two inflection points: } \left( { - \frac{{\sqrt 3 }}{3},\frac{1}{4}} \right), \left( {\frac{{\sqrt 3 }}{3},\frac{1}{4}} \right).$

### Example 17.

Find the inflection points of a Gaussian function.

Solution.

A Gaussian function or the probability density function of normal distribution is defined by the formula

$f\left( x \right) = \frac{1}{{\sigma \sqrt {2\pi } }}{e^{ - \frac{{{{\left( {x - \mu } \right)}^2}}}{{2{\sigma ^2}}}}},$

where $$\mu$$ is the mean value of the distribution, $$\sigma$$ is its standard deviation. Next, we assume that $$\mu = 0.$$ Calculate the first derivative:

$f'\left( x \right) = \left( {\frac{1}{{\sigma \sqrt {2\pi } }}{e^{ - \frac{{{x^2}}}{{2{\sigma ^2}}}}}} \right)^\prime = \frac{1}{{\sigma \sqrt {2\pi } }}{e^{ - \frac{{{x^2}}}{{2{\sigma ^2}}}}} \cdot {\left( { - \frac{{{x^2}}}{{2{\sigma ^2}}}} \right)^\prime } = \frac{1}{{\sigma \sqrt {2\pi } }}{e^{ - \frac{{{x^2}}}{{2{\sigma ^2}}}}} \cdot \left( { - \frac{{2x}}{{2{\sigma ^2}}}} \right) = - \frac{x}{{{\sigma ^3}\sqrt {2\pi } }}{e^{ - \frac{{{x^2}}}{{2{\sigma ^2}}}}}.$

Differentiating once more, we find the second derivative:

$f^{\prime\prime}\left( x \right) = \left( { - \frac{x}{{{\sigma ^3}\sqrt {2\pi } }}{e^{ - \frac{{{x^2}}}{{2{\sigma ^2}}}}}} \right)^\prime = {\left( { - \frac{x}{{{\sigma ^3}\sqrt {2\pi } }}} \right)^\prime }{e^{ - \frac{{{x^2}}}{{2{\sigma ^2}}}}} - \frac{x}{{{\sigma ^3}\sqrt {2\pi } }}{\left( {{e^{ - \frac{{{x^2}}}{{2{\sigma ^2}}}}}} \right)^\prime } = \left( { - \frac{1}{{{\sigma ^3}\sqrt {2\pi } }}} \right){e^{ - \frac{{{x^2}}}{{2{\sigma ^2}}}}} - \frac{x}{{{\sigma ^3}\sqrt {2\pi } }}\left( {{e^{ - \frac{{{x^2}}}{{2{\sigma ^2}}}}}} \right) \cdot \left( { - \frac{x}{{{\sigma ^2}}}} \right) = - \frac{1}{{{\sigma ^3}\sqrt {2\pi } }}{e^{ - \frac{{{x^2}}}{{2{\sigma ^2}}}}} + \frac{{{x^2}}}{{{\sigma ^5}\sqrt {2\pi } }}{e^{ - \frac{{{x^2}}}{{2{\sigma ^2}}}}} = \frac{1}{{{\sigma ^3}\sqrt {2\pi } }}{e^{ - \frac{{{x^2}}}{{2{\sigma ^2}}}}}\left( {\frac{{{x^2}}}{{{\sigma ^2}}} - 1} \right).$

The roots of the second derivative are given by

$f^{\prime\prime}\left( x \right) = 0,\;\; \Rightarrow \frac{1}{{{\sigma ^3}\sqrt {2\pi } }}{e^{ - \frac{{{x^2}}}{{2{\sigma ^2}}}}}\left( {\frac{{{x^2}}}{{{\sigma ^2}}} - 1} \right) = 0,\;\; \Rightarrow \frac{{{x^2}}}{{{\sigma ^2}}} - 1 = 0,\;\; \Rightarrow {x^2} = {\sigma ^2},\;\; \Rightarrow {x_{1,2}} = \pm \sigma .$

The roots of the second derivative are given by

$f^{\prime\prime}\left( x \right) = 0,\;\; \Rightarrow \frac{1}{{{\sigma ^3}\sqrt {2\pi } }}{e^{ - \frac{{{x^2}}}{{2{\sigma ^2}}}}}\left( {\frac{{{x^2}}}{{{\sigma ^2}}} - 1} \right) = 0,\;\; \Rightarrow \frac{{{x^2}}}{{{\sigma ^2}}} - 1 = 0,\;\; \Rightarrow {x^2} = {\sigma ^2},\;\; \Rightarrow {x_{1,2}} = \pm \sigma .$

Since the sign of the second derivative is determined by the quadratic expression $${x^2} - {\sigma ^2},$$ it is clear that when passing through the points $$x = \pm \sigma,$$ the second derivative will change its sign. Therefore, according to the first sufficient condition, the points $$x = \pm \sigma$$ are the inflection points. A schematic view of the Gaussian function for $$\sigma = 1$$ and $$\sigma = 2$$ is given in Figure $$11.$$

### Example 18.

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

Solution.

$f'\left( x \right) = \left( {\sqrt[3]{{{x^2}\left( {x + 1} \right)}}} \right)^\prime = \left[ {{{\left( {{x^2}\left( {x + 1} \right)} \right)}^{\frac{1}{3}}}} \right]^\prime = \frac{1}{3}{\left( {{x^2}\left( {x + 1} \right)} \right)^{ - \frac{2}{3}}} \cdot {\left( {{x^2}\left( {x + 1} \right)} \right)^\prime } = \frac{{2{x^2} + 2x + {x^2}}}{{3\sqrt[3]{{{{\left( {{x^2}\left( {x + 1} \right)} \right)}^2}}}}} = \frac{{3{x^2} + 2x}}{{3\sqrt[3]{{{{\left( {{x^2}\left( {x + 1} \right)} \right)}^2}}}}} = \frac{{\cancel{x}\left( {3x + 2} \right)}}{{3\cancel{x}\sqrt[3]{{x{{\left( {x + 1} \right)}^2}}}}} = \frac{{3x + 2}}{{3\sqrt[3]{{x{{\left( {x + 1} \right)}^2}}}}}.$

Differentiating once more, we determine the second derivative:

$f^{\prime\prime}\left( x \right) = \left( {\frac{{3x + 2}}{{3\sqrt[3]{{x{{\left( {x + 1} \right)}^2}}}}}} \right)^\prime = - \frac{2}{{9\sqrt[3]{{{x^4}{{\left( {x + 1} \right)}^5}}}}}.$

It can be seen that the second derivative is nowhere zero. However, points of inflection may be at $$x = - 1$$ and $$x = 0,$$ where the denominator is zero and the second derivative does not exist. Examine the sign of the second derivative in the vicinity of these points (Figure $$12$$).

Thus, when passing through the point $$x = - 1,$$ the second derivative changes sign from plus to minus. Therefore, this point is a point of inflection. Another point $$x = 0$$ is not an inflection point since the second derivative does not change sign around it.

### Example 19.

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

Solution.

The first derivative $${y'_x}$$ of a parametric function is determined by the formula

${y'_x} = \frac{{{y'_t}}}{{{x'_t}}}.$

Substituting the given expressions $$x\left( t \right),$$ $$x\left( t \right),$$ we obtain:

${y'_x} = \frac{{{y'_t}}}{{{x'_t}}} = \frac{{{{\left( {t + {t^3}} \right)}^\prime }}}{{{{\left( {{t^2}} \right)}^\prime }}} = \frac{{1 + 3{t^2}}}{{2t}}.$

Similarly, we can find the second derivative $${y^{\prime\prime}_{xx}}:$$

${y^{\prime\prime}_{xx}} = \frac{{{{\left( {{y'_x}} \right)}'_t}}}{{{x'_t}}} = \frac{{{{\left( {\frac{{1 + 3{t^2}}}{{2t}}} \right)}^\prime }}}{{{{\left( {{t^2}} \right)}^\prime }}} = \frac{{\frac{{6t \cdot 2t - \left( {1 + 3{t^2}} \right) \cdot 2}}{{4{t^2}}}}}{{2t}} = \frac{{12{t^2} - 2 - 6{t^2}}}{{8{t^3}}} = \frac{{6{t^2} - 2}}{{8{t^3}}} = \frac{{3{t^2} - 1}}{{4{t^3}}}.$

Calculate the value of the parameter $$t,$$ at which the second derivative is equal to zero:

${y^{\prime\prime}_{xx}} = 0,\;\; \Rightarrow \frac{{3{t^2} - 1}}{{4{t^3}}} = 0,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {3{t^2} - 1 = 0}\\ {4{t^3} = 0} \end{array}} \right.,\;\; \Rightarrow {t^2} = \frac{1}{3},\;\; \Rightarrow {t_{1,2}} = \pm \frac{1}{{\sqrt 3 }}.$

To find out whether the found values of $${t_1},$$ $${t_2}$$ correspond to points of inflection, we calculate the third derivative (that is, we apply the second sufficient condition for inflection points):

${y^{\prime\prime\prime}_{xxx}} = \frac{{{{\left( {{y^{\prime\prime}_{xx}}} \right)}'_t}}}{{{x'_t}}} = \frac{{{{\left( {\frac{{3{t^2} - 1}}{{4{t^3}}}} \right)}^\prime }}}{{{{\left( {{t^2}} \right)}^\prime }}} = \frac{{\frac{{6t \cdot 4{t^2} - \left( {3{t^2} - 1} \right) \cdot 12{t^2}}}{{16{t^2}}}}}{{2t}} = \frac{{24{t^4} - 36{t^4} + 12{t^2}}}{{32{t^7}}} = \frac{{12{t^2} - 12{t^4}}}{{32{t^7}}} = \frac{{3 - 3{t^2}}}{{8{t^5}}}.$

After substitution of the values of $${t_{1,2}} = \pm {\frac{1}{{\sqrt 3 }}}$$ we see that the third derivative is not equal to zero. Therefore, by the second sufficient criterion, the curve has the points of inflection at $${t_{1,2}} = \pm {\frac{1}{{\sqrt 3 }}}.$$

Calculate the $$x$$ and $$y$$ coordinates of the inflection points:

$x\left( { - \frac{1}{{\sqrt 3 }}} \right) = {\left( { - \frac{1}{{\sqrt 3 }}} \right)^2} = \frac{1}{3} \approx 0,33;$
$y\left( { - \frac{1}{{\sqrt 3 }}} \right) = - \frac{1}{{\sqrt 3 }} + {\left( { - \frac{1}{{\sqrt 3 }}} \right)^2} = - \frac{1}{{\sqrt 3 }} - \frac{1}{{3\sqrt 3 }} = \frac{{ - 3 - 1}}{{3\sqrt 3 }} = - \frac{4}{{3\sqrt 3 }} \approx - 0,77;$
$x\left( { \frac{1}{{\sqrt 3 }}} \right) = {\left( {\frac{1}{{\sqrt 3 }}} \right)^2} = \frac{1}{3} \approx 0,33;$
$y\left( {\frac{1}{{\sqrt 3 }}} \right) = \frac{1}{{\sqrt 3 }} + {\left( {\frac{1}{{\sqrt 3 }}} \right)^2} = \frac{1}{{\sqrt 3 }} + \frac{1}{{3\sqrt 3 }} = \frac{{3 + 1}}{{3\sqrt 3 }} = \frac{4}{{3\sqrt 3 }} \approx 0,77.$

Figure $$13$$ schematically shows the given curve together with its inflection points. As can be seen, it consists of two branches that are symmetric about the $$x$$-axis.

### Example 20.

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

Solution.

Following the general scheme, we find successively the first and second derivative:

$y'\left( x \right) = {\left( {\frac{{x + 1}}{{{x^2} + 1}}} \right)^\prime } = \frac{{{x^2} + 1 - 2x\left( {x + 1} \right)}}{{{{\left( {{x^2} + 1} \right)}^2}}} = \frac{{{x^2} + 1 - 2{x^2} - 2x}}{{{{\left( {{x^2} + 1} \right)}^2}}} = \frac{{ - {x^2} - 2x + 1}}{{{{\left( {{x^2} + 1} \right)}^2}}};$
$y^{\prime\prime}\left( x \right) = {\left( {\frac{{ - {x^2} - 2x + 1}}{{{{\left( {{x^2} + 1} \right)}^2}}}} \right)^\prime } = \frac{{2\left( {{x^3} + {3{x^2}} - {3x} - {1}} \right)}}{{{{\left( {{x^2} + 1} \right)}^3}}}.$

The cubic polynomial in the numerator is factored as follows:

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

We factor the quadratic function as well:

${x^2} + 4x + 1 = 0,\;\; \Rightarrow D = 16 - 4 = 12,\;\; \Rightarrow {x_{1,2}} = \frac{{ - 4 \pm \sqrt {12} }}{2} = - 2 \pm \sqrt 3 .$

Consequently, the second derivative of the original function is written as

$y^{\prime\prime}\left( x \right) = \frac{{2\left( {{x^3} + 3{x^2} - 3x - 1} \right)}}{{{{\left( {{x^2} + 1} \right)}^3}}} = \frac{{2\left( {x - 1} \right) \left( {x + 2 + \sqrt 3 } \right)\left( {x + 2 - \sqrt 3 } \right)}}{{{{\left( {{x^2} + 1} \right)}^3}}}$

Thus, the second derivative has three roots:

${x_1} = 1,\;\;{x_2} = - 2 - \sqrt 3 ,\;\;{x_3} = - 2 + \sqrt 3 ,$

and the derivative changes sign when passing through each of these points. This means that the identified points are the points of inflection.

Calculate the corresponding $$y$$-coordinates:

${y_1} = y\left( 1 \right) = \frac{{1 + 1}}{{{1^2} + 1}} = 1;$
${y_2} = y\left( { - 2 - \sqrt 3 } \right) = \frac{{ - 2 - \sqrt 3 + 1}}{{{{\left( { - 2 - \sqrt 3 } \right)}^2} + 1}} = \frac{{ - 1 - \sqrt 3 }}{{4 + 4\sqrt 3 + 3 + 1}} = \frac{{ - 1 - \sqrt 3 }}{{8 + 4\sqrt 3 }} = \frac{{ - 1 - \sqrt 3 }}{{4\left( {2 + \sqrt 3 } \right)}} = \frac{{\left( { - 1 - \sqrt 3 } \right)\left( {2 - \sqrt 3 } \right)}}{{4\left( {2 + \sqrt 3 } \right)\left( {2 - \sqrt 3 } \right)}} = \frac{{ - 2 - 2\sqrt 3 + \sqrt 3 + 3}}{{4\left( {4 - 3} \right)}} = \frac{{1 - \sqrt 3 }}{4};$
${y_3} = y\left( { - 2 + \sqrt 3 } \right) = \frac{{ - 2 + \sqrt 3 + 1}}{{{{\left( { - 2 + \sqrt 3 } \right)}^2} + 1}} = \frac{{ - 1 + \sqrt 3 }}{{4 - 4\sqrt 3 + 3 + 1}} = \frac{{ - 1 + \sqrt 3 }}{{8 - 4\sqrt 3 }} = \frac{{ - 1 + \sqrt 3 }}{{4\left( {2 - \sqrt 3 } \right)}} = \frac{{\left( { - 1 + \sqrt 3 } \right)\left( {2 + \sqrt 3 } \right)}}{{4\left( {2 - \sqrt 3 } \right)\left( {2 + \sqrt 3 } \right)}} = \frac{{ - 2 + 2\sqrt 3 - \sqrt 3 + 3}}{{4\left( {4 - 3} \right)}} = \frac{{1 + \sqrt 3 }}{4}.$

So the points of inflection have the following coordinates:

$\left( {1,1} \right),\;\;\left( { - 2 - \sqrt 3 ,\frac{{1 - \sqrt 3 }}{4}} \right),\;\;\left( { - 2 + \sqrt 3 ,\frac{{1 + \sqrt 3 }}{4}} \right).$

Make sure that these three points lie on one straight line. In this case, the following proportion is valid:

$\frac{{{y_3} - {y_2}}}{{{y_3} - {y_1}}} = \frac{{{x_3} - {x_2}}}{{{x_3} - {x_1}}}.$

Substitute the known coordinates:

$\frac{{\frac{{1 + \sqrt 3 }}{4} - \frac{{1 - \sqrt 3 }}{4}}}{{\frac{{1 + \sqrt 3 }}{4} - 1}} = \frac{{ - 2 + \sqrt 3 - \left( { - 2 - \sqrt 3 } \right)}}{{ - 2 + \sqrt 3 - 1}},\;\; \Rightarrow \frac{{\frac{{\cancel{1} + \sqrt 3 - \cancel{1} + \sqrt 3 }}{4}}}{{\frac{{1 + \sqrt 3 - 4}}{4}}} = \frac{{ - \cancel{2} + \sqrt 3 + \cancel{2} + \sqrt 3 }}{{\sqrt 3 - 3}},\;\; \Rightarrow \frac{{2\sqrt 3 }}{{\sqrt 3 - 3}} = \frac{{2\sqrt 3 }}{{\sqrt 3 - 3}}.$

This proves that the inflection points are located on the same straight line. A schematic graph of the function is shown in Figure $$14.$$