Calculus

Applications of the Derivative

Applications of Derivative Logo

Local Extrema of Functions

Solved Problems

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Example 1

Find the local extrema of the function \[f\left( x \right) = {e^{ - {x^2}}}.\]

Example 2

Find the local (relative) extrema of the function \[f\left( x \right) = - {x^2} + 4x - 3.\]

Example 3

Find the local extrema of the function \[f\left( x \right) = {x^3} - 12x.\]

Example 4

Find the local extrema of the cubic function \[f\left( x \right) = {x^3} - 3{x^2} - 9x + 2.\]

Example 5

Using the second derivative test, find the local extrema of the function \[f\left( x \right) = {x^3} - 9{x^2} + 24x - 7.\]

Example 6

Find the local extrema of the function \[f\left( x \right) = {x^{\frac{1}{x}}}.\]

Example 7

Find the local extrema of the function \[f\left( x \right) = {x^2}{e^{ - x}}.\]

Example 8

Find the local extrema points of the function \[f\left( x \right) = \left( {x - a} \right){e^x},\] where \(a\) is an arbitrary real number.

Example 9

Find the local extrema of the function \[f\left( x \right) = x + \cos 2x.\]

Example 10

Determine the local extrema of the function \[f\left( x \right) = \sin x - \cos x\] on \(\left[ {0,2\pi } \right].\)

Example 11

Determine the local extrema of the function \[f\left( x \right) = {x^2}\ln x.\]

Example 12

Find the local extrema of the function \[f\left( x \right) = x\ln x.\]

Example 13

Find the local extrema of the function

\[f\left( x \right) = {x^4} - 8{x^3} + 22{x^2} - 24x + 1.\]

Example 14

Find the local extrema of the function \[f\left( x \right) = \frac{{{x^2}}}{{{x^4} + 16}}.\]

Example 1.

Find the local extrema of the function \[f\left( x \right) = {e^{ - {x^2}}}.\]

Solution.

Sign chart for the first derivative of f(x)=exp(-x^2).
Figure 4.

The function is defined and differentiable for all \(x \in \mathbb{R}.\) We use the first derivative test. The derivative is given by

\[f^\prime\left( x \right) = \left( {{e^{ - {x^2}}}} \right)^\prime = {e^{ - {x^2}}} \cdot \left( { - {x^2}} \right)^\prime = - 2x{e^{ - {x^2}}}.\]

Then we get

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

The derivative changes its sign as shown in the sign chart above.

Hence, the function has a maximum at \(x = 0.\) The maximum value is

\[{f_{\max }} = f\left( 0 \right) = {e^{ - {0^2}}} = {e^0} = 1.\]

Example 2.

Find the local (relative) extrema of the function \[f\left( x \right) = - {x^2} + 4x - 3.\]

Solution.

This function is differentiable everywhere on the set \(\mathbb{R}.\) Consequently, the extrema of the function are contained among its stationary points. Solve the equation \(f'\left( x \right) = 0:\)

\[f'\left( x \right) = \left( { - {x^2} + 4x - 3} \right)^\prime = - 2x + 4,\]
\[f'\left( x \right) = 0,\;\; \Rightarrow - 2x + 4 = 0,\;\; \Rightarrow x = 2.\]

The function has one stationary point \(x = 2.\) Determine the sign of the derivative to the left and right of the point \(x = 2.\) The derivative is positive for \(x \lt 2\) and negative for \(x \gt 2\). Thus, when passing through the point \(x = 2\), the derivative changes sign from plus to minus. By the first derivative test, this means that \(x = 2\) is a maximum point.

The maximum value (that is the value of the function at the maximum point) is equal to

\[{f_{\max }} = f\left( 2 \right) = - {2^2} + 4 \cdot 2 - 3 = 1.\]

Example 3.

Find the local extrema of the function \[f\left( x \right) = {x^3} - 12x.\]

Solution.

Determine the critical points. The first derivative is given by

\[f^\prime\left( x \right) = \left( {{x^3} - 12x} \right)^\prime = 3{x^2} - 12.\]

It is equal to zero at the following points:

\[f^\prime\left( x \right) = 0,\;\; \Rightarrow 3{x^2} - 12 = 0,\;\; \Rightarrow {x^2} = 4,\;\; \Rightarrow {x_1} = - 2,{x_2} = 2.\]

These two points are critical since the function is defined and continuous over all \(x.\) The derivative also exists for all \(x,\) so there are no other critical points.

We use the Second Derivative Test:

\[f^{\prime\prime}\left( x \right) = \left( {3{x^2} - 12} \right)^\prime = 6x,\]
\[f^{\prime\prime}\left( { - 2} \right) = 6 \cdot \left( { - 2} \right) = -12 \lt 0,\]
\[f^{\prime\prime}\left( {2} \right) = 6 \cdot 2 = 12 \gt 0.\]

Hence, \(x = -2\) is a point of local maximum, and \(x = 2\) is a point of local minimum.

Calculate the \(y-\)values for these points:

\[{f_{\max }} = f\left( { - 2} \right) = {\left( { - 2} \right)^3} - 12 \cdot \left( { - 2} \right) = 16,\]
\[{f_{\min }} = f\left( 2 \right) = {2^3} - 12 \cdot 2 = - 16.\]

Answer:

\[\text{local max:}\,\left( { -2, 16} \right);\,\text{local min:}\,\left( {2, -16} \right).\]

Example 4.

Find the local extrema of the cubic function \[f\left( x \right) = {x^3} - 3{x^2} - 9x + 2.\]

Solution.

The function is differentiable on the whole set of real numbers. Therefore, the extremum points are contained among the stationary points (where the derivative is equal to zero).

We find these points:

\[f'\left( x \right) = 0,\;\; \Rightarrow \left( {{x^3} - 3{x^2} - 9x + 2} \right) = 0,\;\; \Rightarrow 3{x^2} - 6x - 9 = 0,\;\; \Rightarrow {x^2} - 2x - 3 = 0,\;\; \Rightarrow {x_1} = - 1,\;{x_2} = 2.\]

Substituting test values of \(x\), we determine the sign of the derivative \(f'\left( x \right) = 3{x^2} - 6x - 9\) in the corresponding intervals (Figure \(5\)).

Signs of derivative of f(x)=x^3-3x^2-9x+2.
Figure 5.

As seen, when passing through the point \(x = - 1\), the derivative changes sign from plus to minus. By the first derivative test, this point is a local maximum point. Similarly, we establish that \(x = 2\) is a local minimum point.

We now determine the maximum and minimum values of the function:

\[{f_{\max }} = f\left( { - 1} \right) = {\left( { - 1} \right)^3} - 3 \cdot {\left( { - 1} \right)^2} - 9 \cdot \left( { - 1} \right) + 2 = 7,\]
\[{f_{\min }} = f\left( 2 \right) = {2^3} - 3 \cdot {2^2} - 9 \cdot 2 + 2 = - 20.\]

Example 5.

Using the second derivative test, find the local extrema of the function \[f\left( x \right) = {x^3} - 9{x^2} + 24x - 7.\]

Solution.

The function is defined for all \(x.\) Take the first derivative and determine the critical points:

\[f^\prime\left( x \right) = \left( {{x^3} - 9{x^2} + 24x - 7} \right)^\prime = 3{x^2} - 18x + 24.\]
\[f^\prime\left( x \right) = 0,\;\; \Rightarrow 3{x^2} - 18x + 24 = 0,\;\; \Rightarrow 3\left( {{x^2} - 6x + 8} \right) = 0,\;\; \Rightarrow 3\left( {x - 2} \right)\left( {x - 4} \right) = 0,\;\; \Rightarrow {x_1} = 2,{x_2} = 4.\]

The second derivative is given by

\[f^{\prime\prime}\left( x \right) = \left( {3{x^2} - 18x + 24} \right)^\prime = 6x - 18.\]

Determine the sign of the \(2\)nd derivative at the critical points:

\[f^{\prime\prime}\left( {{x_1}} \right) = f^{\prime\prime}\left( 2 \right) = 6 \cdot 2 - 18 = - 6 \lt 0.\]
\[f^{\prime\prime}\left( {{x_2}} \right) = f^{\prime\prime}\left( 4 \right) = 6 \cdot 4 - 18 = 6 \gt 0.\]

Hence, the point \({x_1} = 2\) is a local maximum, and the point \({x_2} = 4\) is a local minimum.

Compute the \(y-\)coordinates:

\[f\left( {{x_1}} \right) = {2^3} - 9 \cdot {2^2} + 24 \cdot 2 - 7 = 13,\]
\[f\left( {{x_2}} \right) = {4^3} - 9 \cdot {4^2} + 24 \cdot 4 - 7 = 9.\]

The answer is

\[\text{local max:}\,\left( {2,13} \right);\,\text{local min:}\,\left( {4,9} \right).\]

Example 6.

Find the local extrema of the function \[f\left( x \right) = {x^{\frac{1}{x}}}.\]

Solution.

This function belongs to the family of power-exponential functions. In general, they have the form of \(y = g{\left( x \right)^{h\left( x \right)}}.\) It is usually assumed that the domain of power-exponential functions satisfies the condition \(g\left( x \right) \gt 0.\) (In some special cases, the base \(g\left( x \right)\) can be negative − for example, if \(h = {\frac{1}{3}}.\)) In our case we suppose that \(x \gt 0.\) It follows from here that the function takes only positive values.

Determine the derivative:

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

Note: The derivative of the function \(f\left( x \right) = {x^{\frac{1}{x}}}\) can also be found using logarithmic differentiation.

Calculate the critical points:

\[f'\left( x \right) = 0,\;\; \Rightarrow {x^{\frac{1}{x} - 2}}\left( {1 - \ln x} \right) = 0,\;\; \Rightarrow 1 - \ln x = 0,\;\; \Rightarrow \ln x = 1,\;\; \Rightarrow x = e.\]

In the left neighborhood of \(x = e,\) the derivative is positive, and in the right neighborhood it is negative. Consequently, the function attains a local maximum at the point \(x = e.\) The maximum value is equal

\[{f_{\max }} = f\left( {x = e} \right) = {e^{\frac{1}{e}}} \approx 1,445.\]

Example 7.

Find the local extrema of the function \[f\left( x \right) = {x^2}{e^{ - x}}.\]

Solution.

The function is defined and differentiable on the whole set \(\mathbb{R}.\) Calculate its derivative:

\[f'\left( x \right) = {\left( {{x^2}{e^{ - x}}} \right)^\prime } = {\left( {{x^2}} \right)^\prime }{e^{ - x}} + {x^2}{\left( {{e^{ - x}}} \right)^\prime } = 2x{e^{ - x}} - {x^2}{e^{ - x}} = x{e^{ - x}}\left( {2 - x} \right).\]

Find the roots of the equation \(f'\left( x \right) = 0:\)

\[x{e^{ - x}}\left( {2 - x} \right) = 0,\;\; \Rightarrow {x_1} - 0,\;{x_2} = 2.\]

When passing through these points, the derivative changes sign as shown above in Figure \(6.\)

Signs of derivative of f(x)=x^2*exp(-x).
Figure 6.

Hence, at the point \(x = 0,\) the function has a minimum, and at the point \(x = 0\) it has a maximum. The minimum and maximum values, respectively, are equal to:

\[{f_{\min }} = f\left( 0 \right) = {0^2}{e^{ - 0}} = 0,\]
\[{f_{\max }} = f\left( 2 \right) = {2^2}{e^{ - 2}} = \frac{4}{{{e^2}}} \approx 0,541.\]

Example 8.

Find the local extrema points of the function \[f\left( x \right) = \left( {x - a} \right){e^x},\] where \(a\) is an arbitrary real number.

Solution.

Sign chart for the first derivative of f(x)=(x-a)exp(x).
Figure 7.

Calculate the derivative using the product rule:

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

Determine the critical points:

\[f^\prime\left( x \right) = 0,\;\; \Rightarrow \left( {x - a + 1} \right){e^x} = 0,\;\; \Rightarrow x - a + 1 = 0,\;\; \Rightarrow x = a - 1.\]

It follows from the sign chart (see above) that the point \(x = a-1\) is a point of local minimum. The minimum value is

\[\require{cancel} {f_{\min }} = f\left( {a - 1} \right) = \left( {\cancel{a} - 1 - \cancel{a}} \right){e^{a - 1}} = - {e^{a - 1}}.\]

Hence, we obtain the following answer:

\[\text{local min:}\,\left( {a - 1, - {e^{a - 1}}} \right).\]

Example 9.

Find the local extrema of the function \[f\left( x \right) = x + \cos 2x.\]

Solution.

The function is differentiable on the whole set \(\mathbb{R}.\) Calculate the derivative and determine the stationary points:

\[f'\left( x \right) = \left( {x + \cos 2x} \right)^\prime = 1 - 2\sin 2x,\]
\[f'\left( x \right) = 0,\;\; \Rightarrow 1 - 2\sin 2x = 0,\;\; \Rightarrow \sin 2x = \frac{1}{2}.\]

The resulting trigonometric equation has two branches of solutions:

\[1)\;{2x = {\frac{\pi }{6}} + 2\pi n,} \Rightarrow {x_1} = {\frac{\pi }{{12}}} + \pi n, n \in \mathbb{Z};\]
\[2)\;{2x = {\frac{5\pi }{6}} + 2\pi k,} \Rightarrow {x_2} = {\frac{5\pi }{{12}}} + \pi k, k \in \mathbb{Z}.\]

To determine the type of the extrema, we use the second derivative test. The second derivative is given by

\[f^{\prime\prime}\left( x \right) = \left( {1 - 2\sin 2x} \right)^\prime = - 4\cos 2x.\]

For the first set of solutions \(\left( {2x = {\frac{\pi }{6}} + 2\pi n,\,n \in \mathbb{Z}} \right),\) the second derivative is negative. Therefore, this set corresponds to a maximum point. The second set of solutions \(\left( {2x = {\frac{5\pi }{6}} + 2\pi k,}\right.\) \(\left.{k \in \mathbb{Z}} \right)\) corresponds to a positive second derivative. Here we have a minimum point. Compute the values of the function at these maximum and minimum points:

\[{f_{\max}} = f\left( {\frac{\pi }{{12}} + \pi n} \right) = \frac{\pi }{{12}} + \pi n + \cos \left[ {2\left( {\frac{\pi }{{12}} + \pi n} \right)} \right] = \frac{\pi }{{12}} + \pi n + \cos \left( {\frac{\pi }{6} + 2\pi n} \right) = \frac{\pi }{{12}} + \frac{{\sqrt 3 }}{2} + \pi n\; \approx 1,13 + 3,14n,\;\;n \in Z;\]
\[{f_{\min}} = f\left( {\frac{5\pi }{{12}} + \pi k} \right) = \frac{5\pi }{{12}} + \pi k + \cos \left[ {2\left( {\frac{5\pi }{{12}} + \pi k} \right)} \right] = \frac{5\pi }{{12}} + \pi k + \cos \left( {\frac{5\pi }{6} + 2\pi k} \right) = \frac{5\pi }{{12}} - \frac{{\sqrt 3 }}{2} + \pi k\; \approx 0,44 + 3,14k,\;\;k \in Z.\]

This function is the sum of a linear function and a cosine function. Such a combination results in that the maxima and minima linearly increase with the numbers \(n\) and \(k\) as shown in Figure \(8\).

Sum of linear function and cosine
Figure 8.

Example 10.

Determine the local extrema of the function \[f\left( x \right) = \sin x - \cos x\] on \(\left[ {0,2\pi } \right].\)

Solution.

We use the \(2\)nd Derivative Test. First we find the critical points on the given interval:

\[f^\prime\left( x \right) = \left( {\sin x - \cos x} \right)^\prime = \cos x + \sin x.\]
\[f^\prime\left( x \right) = 0,\;\; \Rightarrow \cos x + \sin x = 0,\;\; \Rightarrow \cot x + 1 = 0,\;\; \Rightarrow \cot x = - 1, \Rightarrow {x_1} = \frac{{3\pi }}{4},\;{x_2} = \frac{{7\pi }}{4}.\]

The function is continuous and differentiable over all \(x\), so it has no other singular or critical points.

Differentiate once more:

\[f^{\prime\prime}\left( x \right) = \left( {\cos x + \sin x} \right)^\prime = - \sin x + \cos x.\]

Determine the sign of the \(2\)nd derivative at the critical points we found above:

\[f^{\prime\prime}\left( {{x_1}} \right) = f^{\prime\prime}\left( {\frac{{3\pi }}{4}} \right) = - \sin \frac{{3\pi }}{4} + \cos \frac{{3\pi }}{4} = - \frac{{\sqrt 2 }}{2} + \left( { - \frac{{\sqrt 2 }}{2}} \right) = - \sqrt 2 \lt 0;\]
\[f^{\prime\prime}\left( {{x_2}} \right) = f^{\prime\prime}\left( {\frac{{7\pi }}{4}} \right) = - \sin \frac{{7\pi }}{4} + \cos \frac{{7\pi }}{4} = - \left( { - \frac{{\sqrt 2 }}{2}} \right) + \frac{{\sqrt 2 }}{2} = \sqrt 2 \gt 0.\]

We see that \({x_1} = \frac{{3\pi }}{4}\) is a local maximum point, and \({x_2} = \frac{{7\pi }}{4}\) is a local minimum point.

Calculate the (y-)values:

\[{f_{\max }} = f\left( {\frac{{3\pi }}{4}} \right) = \sin \frac{{3\pi }}{4} - \cos \frac{{3\pi }}{4} = \frac{{\sqrt 2 }}{2} - \left( { - \frac{{\sqrt 2 }}{2}} \right) = \sqrt 2 ;\]
\[{f_{\min }} = f\left( {\frac{{7\pi }}{4}} \right) = \sin \frac{{7\pi }}{4} - \cos \frac{{7\pi }}{4} = - \frac{{\sqrt 2 }}{2} - \frac{{\sqrt 2 }}{2} = - \sqrt 2 .\]

Answer:

\[\text{local max:}\,\left( {\frac{{3\pi }}{4},\sqrt 2 } \right);\,\text{local min:}\,\left( {\frac{{7\pi }}{4}, - \sqrt 2 } \right).\]

Example 11.

Determine the local extrema of the function \[f\left( x \right) = {x^2}\ln x.\]

Solution.

The given function is defined and differentiable for \(x \gt 0.\) Find the derivative and determine the stationary points of the function in the given interval of \(x:\)

\[f'\left( x \right) = {\left( {{x^2}\ln x} \right)^\prime } = {\left( {{x^2}} \right)^\prime }\ln x + {x^2}{\left( {\ln x} \right)^\prime } = 2x \cdot \ln x + {x^2} \cdot \frac{1}{x} = 2x\ln x + x = x\left( {2\ln x + 1} \right) = 0.\]

The root \(x = 0\) is outside of the domain. Therefore, we consider the second solution:

\[2\ln x + 1 = 0,\;\; \Rightarrow \ln x = - \frac{1}{2},\;\; \Rightarrow x = {e^{ -\frac{1}{2}}} = \frac{1}{{\sqrt e }}.\]

In the left neighborhood of the point \(x = {\frac{1}{{\sqrt e }}}\) the derivative is negative, and in the right neighborhood it is positive. Consequently, the function has a minimum at this point. Its value is equal to

\[{f_{\min }} = f\left( {\frac{1}{{\sqrt e }}} \right) = {\left( {\frac{1}{{\sqrt e }}} \right)^2}\ln \frac{1}{{\sqrt e }} = \frac{1}{e} \cdot \left( { - \frac{1}{2}} \right) = - \frac{1}{{2e}}.\]

Example 12.

Find the local extrema of the function \[f\left( x \right) = x\ln x.\]

Solution.

Sign chart for the first derivative of f(x)=x ln(x).
Figure 9.

The domain of the function is \(x \gt 0.\) We start with the first derivative:

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

Find the critical points:

\[f^\prime\left( x \right) = 0,\;\; \Rightarrow \ln x + 1 = 0,\;\; \Rightarrow \ln x = - 1,\;\; \Rightarrow x = {e^{ - 1}} = \frac{1}{e}.\]

A sign chart for \(f^\prime\left( x \right)\) is given above. Hence, by the first derivative test, the function has a local minimum at \( x = \frac{1}{e}.\) The minimum value is

\[f\left( {\frac{1}{e}} \right) = \frac{1}{e}\ln \frac{1}{e} = \frac{1}{e} \cdot \left( { - 1} \right) = - \frac{1}{e}.\]

Answer:

\[\text{local min:}\,\left( {\frac{1}{e}, - \frac{1}{e}} \right).\]

Example 13.

Find the local extrema of the function

\[f\left( x \right) = {x^4} - 8{x^3} + 22{x^2} - 24x + 1.\]

Solution.

The function is defined and differentiable on the whole set \(\mathbb{R}.\) Consequently, the local extrema of the function are contained among its stationary points. Calculate the first derivative:

\[f'\left( x \right) = {\left( {{x^4} - 8{x^3} + 22{x^2} - 24x + 1} \right)^\prime } = 4{x^3} - 24{x^2} + 44x - 24 = 4\left( {{x^3} - 6{x^2} + 11x - 6} \right).\]

Find the stationary points:

\[f'\left( x \right) = 0,\;\; \Rightarrow 4\left( {{x^3} - 6{x^2} + 11x - 6} \right) = 0,\; \Rightarrow {x^3} - 6{x^2} + 11x - 6 = 0.\]

By substitution, we find that \(x = 1\) is a root of the cubic equation. Dividing the cubic function by \(x - 1,\) we can factor it as follows:

\[{x^3} - 6{x^2} + 11x - 6 = 0,\;\; \Rightarrow \left( {x - 1} \right)\left( {{x^2} - 5x + 6} \right) = 0,\;\; \Rightarrow \left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right) = 0,\; \Rightarrow {x_1} = 1,\;{x_2} = 2,\;\;{x_3} = 3.\]

So we have found three "candidates" for the extrema of the function. To check these points, we use the second derivative test. The second derivative is written as

\[f^{\prime\prime}\left( x \right) = {\left[ {4\left( {{x^3} - 6{x^2} + 11x - 6} \right)} \right]^\prime } = 4\left( {3{x^2} - 12x + 11} \right).\]

Substituting the coordinates of the stationary points, we find the values of the second derivative:

\[f^{\prime\prime}\left( 1 \right) = 4\left( {3 \cdot {1^2} - 12 \cdot 1 + 11} \right) = 8,\]
\[f^{\prime\prime}\left( 2 \right) = 4\left( {3 \cdot {2^2} - 12 \cdot 2 + 11} \right) = - 4,\]
\[f^{\prime\prime}\left( 3 \right) = 4\left( {3 \cdot {3^2} - 12 \cdot 3 + 11} \right) = 8.\]

It follows that \({x_1} = 1\) and \({x_3} = 3\) are local minimum points, and \({x_2} = 2\) is a local maximum point. The function has the following values at these points:

\[{f_{\min }} = f\left( 1 \right) = {1^4} - 8 \cdot {1^3} + 22 \cdot {1^2} - 24 \cdot 1 + 1 = - 8,\]
\[{f_{\min }} = f\left( 3 \right) = {3^4} - 8 \cdot {3^3} + 22 \cdot {3^2} - 24 \cdot 3 + 1 = - 8,\]
\[{f_{\max }} = f\left( 2 \right) = {2^4} - 8 \cdot {2^3} + 22 \cdot {2^2} - 24 \cdot 2 + 1 = - 7.\]

Example 14.

Find the local extrema of the function \[f\left( x \right) = \frac{{{x^2}}}{{{x^4} + 16}}.\]

Solution.

This function is defined and differentiable for all \(x \in \mathbb{R}.\) Find the derivative and calculate the stationary points of the function:

\[f'\left( x \right) = \left( {\frac{{{x^2}}}{{{x^4} + 16}}} \right)^\prime = \frac{{{{\left( {{x^2}} \right)}^\prime }\left( {{x^4} + 16} \right) - {x^2}{{\left( {{x^4} + 16} \right)}^\prime }}}{{{{\left( {{x^4} + 16} \right)}^2}}} = \frac{{2x \cdot \left( {{x^4} + 16} \right) - {x^2} \cdot 4{x^3}}}{{{{\left( {{x^4} + 16} \right)}^2}}} = \frac{{32x - 2{x^5}}}{{{{\left( {{x^4} + 16} \right)}^2}}} = \frac{{2x\left( {16 - {x^4}} \right)}}{{{{\left( {{x^4} + 16} \right)}^2}}}.\]

Then

\[ f'\left( x \right) = 0,\;\;\Rightarrow \frac{{2x\left( {16 - {x^4}} \right)}}{{{{\left( {{x^4} + 16} \right)}^2}}} = 0,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {2x\left( {16 - {x^4}} \right) = 0}\\ {{{\left( {{x^4} + 16} \right)}^2} \ne 0} \end{array}} \right.,\;\; \Rightarrow x\left( {4 - {x^2}} \right)\left( {4 + {x^2}} \right) = 0,\;\; \Rightarrow x\left( {2 - x} \right)\left( {2 + x} \right) = 0,\;\; \Rightarrow {x_1} = 0,\;{x_{2,3}} = \pm 2.\]

Determine the sign of the derivative by the interval method (Figure \(10\)).

Signs of derivative of f(x)=x^2/(x^2+16).
Figure 10.

According to the first derivative test, we have maximums at \(x = -2\) and \(x = 2.\) Since the function is even, the values of the function at these points are the same:

\[{f_{\max }} = f\left( { \pm 2} \right) = \frac{{{2^2}}}{{{2^4} + 16}} = \frac{4}{{32}} = \frac{1}{8}.\]

At \(x = 0\), the function has a local minimum equal

\[{f_{\min }} = f\left( {0} \right) = \frac{{{0^2}}}{{{0^4} + 16}} = 0.\]
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