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 \[f\left( x \right) = 3{x^5} + 5{x^4} - 20{x^3}.\]
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.
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 \[f\left( x \right) = x{e^{ - 2x}}.\]
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.
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:
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:
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
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.
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:
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:
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:
where \(\mu\) is the mean value of the distribution, \(\sigma\) is its standard deviation. Next, we assume that \(\mu = 0.\) Calculate the first derivative:
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 \[f\left( x \right) = \sqrt[3]{{{x^2}\left( {x + 1} \right)}}.\]
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:
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):
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:
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: