Inflection Points
Solved Problems
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
\[f\left( { - 1} \right) = 3 \cdot {\left( { - 1} \right)^5} + 5 \cdot {\left( { - 1} \right)^4} - 20 \cdot {\left( { - 1} \right)^3} = 22;\]
\[f\left( {0} \right) = 3 \cdot {0^5} + 5 \cdot {0^4} - 20 \cdot {0^3} = 0;\]
\[f\left( {2} \right) = 3 \cdot {2^5} + 5 \cdot {2^4} - 20 \cdot {2^3} = 16,\]
we obtain the following answer:
\[3 \text{ inflection points: } \left( { - 1,22} \right), \left( {0,0} \right), \left( {2,16} \right).\]
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.
Figure 6.
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 \(7.\)
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.
Figure 8. 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.
Figure 9. 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.
Figure 10. 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
\[f\left( { \pm \frac{{\sqrt 3 }}{3}} \right) = \frac{{{{\left( { \pm \frac{{\sqrt 3 }}{3}} \right)}^2}}}{{1 + {{\left( { \pm \frac{{\sqrt 3 }}{3}} \right)}^2}}} = \frac{{\frac{1}{3}}}{{1 + \frac{1}{3}}} = \frac{{\frac{1}{3}}}{{\frac{4}{3}}} = \frac{1}{4},\]
we get the following answer:
\[\text{two inflection points: } \left( { - \frac{{\sqrt 3 }}{3},\frac{1}{4}} \right), \left( {\frac{{\sqrt 3 }}{3},\frac{1}{4}} \right).\]
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 \(11.\)
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.
We start with calculating the first derivative:
\[f'\left( x \right) = \left( {\sqrt[3]{{{x^2}\left( {x + 1} \right)}}} \right)^\prime = \left[ {{{\left( {{x^2}\left( {x + 1} \right)} \right)}^{\frac{1}{3}}}} \right]^\prime = \frac{1}{3}{\left( {{x^2}\left( {x + 1} \right)} \right)^{ - \frac{2}{3}}} \cdot {\left( {{x^2}\left( {x + 1} \right)} \right)^\prime } = \frac{{2{x^2} + 2x + {x^2}}}{{3\sqrt[3]{{{{\left( {{x^2}\left( {x + 1} \right)} \right)}^2}}}}} = \frac{{3{x^2} + 2x}}{{3\sqrt[3]{{{{\left( {{x^2}\left( {x + 1} \right)} \right)}^2}}}}} = \frac{{\cancel{x}\left( {3x + 2} \right)}}{{3\cancel{x}\sqrt[3]{{x{{\left( {x + 1} \right)}^2}}}}} = \frac{{3x + 2}}{{3\sqrt[3]{{x{{\left( {x + 1} \right)}^2}}}}}.\]
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 \(12\)).
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.
Figure 13.
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 \(14.\)
Figure 14.
