Calculus

Applications of the Derivative

Applications of Derivative Logo

Global Extrema of Functions

Solved Problems

Example 11.

Find the global extrema of \[f\left( x \right) = \sqrt {5 - 4x} \] in the interval \(\left[ { - 1,1} \right].\)

Solution.

Global maximum and minimum points of the root function f(x)=sqrt(5-4x).
Figure 12.

First we determine the domain of the function:

\[5 - 4x \ge 0,\;\; \Rightarrow 4x \le 5,\;\; \Rightarrow x \le \frac{5}{4},\]

so the given function is defined on the interval \(\left( { - \infty ,\frac{5}{4}} \right].\)

Find the derivative:

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

We see that the first derivative is nowhere zero. It is undefined at \(x = \frac{5}{4},\) but this point does not belong to the interval \(\left[ { - 1,1} \right].\) Therefore, we must calculate the function value only at the endpoints of the interval:

\[f\left( { - 1} \right) = \sqrt {5 - 4 \cdot \left( { - 1} \right)} = \sqrt 9 = 3,\]
\[f\left( 1 \right) = \sqrt {5 - 4 \cdot 1} = \sqrt 1 = 1.\]

We get the following answer:

\[\text{global max:}\,\left( { -1, 3} \right);\,\text{global min:}\,\left( {1, 1} \right).\]

Example 12.

Find the global maximum and global minimum of \[f\left( x \right) = {2^x}\] in the interval \(\left[ { - 1,3} \right].\)

Solution.

Global maximum and minimum points of the exponential function f(x)=2^x.
Figure 13.

Take the derivative:

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

The derivative of the function exists at all \(x,\) but is never equal to 0. Therefore, the function has no critical points.

Hence, we must test the function values only at the endpoints:

\[f\left( { - 1} \right) = 2^{ - 1} = \frac{1}{2},\]
\[f\left( 3 \right) = {2^3} = 8.\]

Thus, the global maximum is \(8\) at \(x = 3,\) and global minimum is \(\frac{1}{2}\) at \(x = -1.\)

Example 13.

Find the global extrema of the function \[f\left( x \right) = \sqrt[3]{{{{\left( {x - 2} \right)}^2}}}\] in \(\left[ {1,4} \right].\)

Solution.

Global extrema points of the function f(x)=sqrt[3]((x-2)^2).
Figure 14.

The function is defined over all \(x.\) Find the critical points:

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

We see that the first derivative is nowhere zero, but it does not exist at \(x = 2.\) Hence, this point is critical.

Clearly, we must evaluate the function at \(x = 1,2,4.\)

\[f\left( 1 \right) = \sqrt[3]{{{{\left( {1 - 2} \right)}^2}}} = \sqrt[3]{1} = 1,\]
\[f\left( 2 \right) = \sqrt[3]{{{{\left( {2 - 2} \right)}^2}}} = \sqrt[3]{0} = 0,\]
\[f\left( 4 \right) = \sqrt[3]{{{{\left( {4 - 2} \right)}^2}}} = \sqrt[3]{4}.\]

Thus, the global maximum is \(\sqrt[3]{4}\) at \(x = 4,\) and the global minimum is \(0\) at \(x = 2.\)

Example 14.

Find the global maximum and global minimum of \[f\left( x \right) = \frac{{{x^2}}}{{{2^x}}}\] in the interval \(\left[ { - 1,3} \right].\)

Solution.

We see that the function is defined and differentiable for all \(x \in \mathbb{R}.\) Therefore, all critical points can be found from the condition \(f'\left( x \right) = 0:\)

\[f'\left( x \right) = \left( {\frac{{{x^2}}}{{{2^x}}}} \right)^\prime = \frac{{{{\left( {{x^2}} \right)}^\prime }{2^x} - {x^2}{{\left( {{2^x}} \right)}^\prime }}}{{{{\left( {{2^x}} \right)}^2}}} = \frac{{2x \cdot {2^x} - {x^2} \cdot {2^x}\ln 2}}{{{{\left( {{2^x}} \right)}^2}}} = \frac{{\cancel{2^x}x\left( {2 - x\ln 2} \right)}}{{\cancel{2^x} \cdot {2^x}}} = \frac{{x\left( {2 - x\ln 2} \right)}}{{{2^x}}};\]
\[f'\left( x \right) = 0,\;\; \Rightarrow \frac{{x\left( {2 - x\ln 2} \right)}}{{{2^x}}} = 0,\;\; \Rightarrow x\left( {2 - x\ln 2} \right) = 0,\;\; \Rightarrow {x_1} = 0,\;{x_2} = \frac{2}{{\ln 2}} \approx 2,885.\]

Found \(2\) points of local extremum: \({x_1} = 0\) and \({x_2} = \frac{2}{{\ln 2}}.\)

Now we can calculate the values of the function at these points and at the endpoints of the interval and determine the maximum and minimum values of the function on the given interval:

\[f\left( 0 \right) = \frac{{{0^2}}}{{{2^0}}} = 0;\;\;\;f\left( {\frac{2}{{\ln 2}}} \right) = \frac{{{{\left( {\frac{2}{{\ln 2}}} \right)}^2}}}{{{2^{\frac{2}{{\ln 2}}}}}} \approx \frac{{{{\left( {2,885} \right)}^2}}}{{{2^{2,885}}}} \approx 1,127;\;\;f\left( { - 1} \right) = \frac{{{{\left( { - 1} \right)}^2}}}{{{2^{ - 1}}}} = 2;\;\;f\left( 3 \right) = \frac{{{3^2}}}{{{2^3}}} = 1,125.\]

Consequently, the maximum value of the function is equal to \(2\) when \(x = -1,\) the minimum value is equal to \(0\) at \(x = 0.\)

Example 15.

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

Solution.

Here, the function is not restricted by any interval, so we will consider its change over the domain \(\left( { - \infty ,\infty } \right).\)

Compute the derivative:

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

Solve the equation \(f^\prime\left( c \right) = 0:\)

\[ - 2c{e^{ - {c^2}}} = 0,\;\; \Rightarrow c = 0.\]

The point \(c = 0\) is critical. The function has no other critical points as the derivative exists everywhere.

The derivative changes sign from positive to negative when passing through the point \(c = 0.\) Therefore, this is a global maximum point. The \(y-\)coordinate at \(x = 0\) is

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

The function has no global minimum.

Answer:

\[\text{global max:}\,\left( {0,1} \right).\]

Example 16.

Find the global extrema of \[f\left( x \right) = {x^2}{e^{ - x}}\] in the interval \(\left[ { - 1,1} \right].\)

Solution.

Global extrema points of the function f(x)=x^2 exp(-x).
Figure 15.

Take the derivative:

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

Determine the critical points:

\[f^\prime\left( c \right) = 0,\;\; \Rightarrow c{e^{ - c}}\left( {2 - c} \right) = 0,\;\; \Rightarrow {c_1} = 0,\;{c_2} = 2.\]

The second critical point \({c_2} = 2\) lies outside the interval \(\left[ { - 1,1} \right],\) so we calculate the function values at \({c_1} = 0\) and at the endpoints \(-1\) and \(1:\)

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

As you can see, the global maximum is equal to \(e\) at \(x = -1,\) and the global minimum is \(0\) at \(x = 0.\)

Example 17.

Find the global maximum and global minimum of \[f\left( x \right) = \sqrt[x]{x}\] in the interval \(\left[ {2,3} \right].\)

Solution.

This function is defined and differentiable for \(x \gt 0.\) Its derivative can be found using logarithmic differentiation:

\[y = \sqrt[x]{x} = {x^{\frac{1}{x}}},\;\; \Rightarrow \ln y = \ln {x^{\frac{1}{x}}},\;\; \Rightarrow \ln y = \frac{1}{x}\ln x,\;\; \Rightarrow {\left( {\ln y} \right)^\prime } = {\left( {\frac{1}{x}\ln x} \right)^\prime },\;\; \Rightarrow \frac{{y'}}{y} = {\left( {\frac{1}{x}} \right)^\prime }\ln x + \frac{1}{x}{\left( {\ln x} \right)^\prime },\;\; \Rightarrow \frac{{y'}}{y} = \frac{1}{{{x^2}}}\left( {1 - \ln x} \right),\;\; \Rightarrow y' = {x^{\frac{1}{x}}} \cdot \frac{1}{{{x^2}}}\left( {1 - \ln x} \right),\;\; \Rightarrow y' = {x^{\frac{1}{x} - 2}}\left( {1 - \ln x} \right).\]

Equating the derivative to zero, we find the critical (stationary) points of the function:

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

Now we compute the value of the function at the critical point \(x = e\) (that falls in the given interval) and at the endpoints of the interval: at \(x = 2\) and \(x = 3:\)

\[f\left( e \right) = e^{\frac{1}{e}} \approx 1,445;\;\; f\left( 2 \right) = 2^{\frac{1}{2}} = \sqrt 2 \approx 1,414;\;\; f\left( 3 \right) = 3^{\frac{1}{3}} = \sqrt[3]{3} \approx 1,442.\]

Thus, the minimum value of the function is \(f\left( 2 \right) \approx 1,414,\) and the maximum values respectively is \(f\left( e \right) \approx 1,445.\)

Example 18.

Find the global maximum and global minimum of the function \[f\left( x \right) = \sin x - x\] in the interval \(\left[ { - \pi ,\pi } \right].\)

Solution.

Note that the function is defined and differentiable over all \(x.\) Take the derivative:

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

Solving the equation \(f^\prime\left( c \right) = 0,\) we get the critical points:

\[f^\prime\left( c \right) = 0,\;\; \Rightarrow \cos c - 1 = 0,\;\; \Rightarrow \cos c = 1,\;\; \Rightarrow c = 0.\]

Hence, the function has one critical point in the interval \(\left[ { - \pi ,\pi } \right].\)

Compute \(f\left( x \right)\) at the critical point and at the endpoints of the given interval:

\[f\left( { - \pi } \right) = \sin \left( { - \pi } \right) - \left( { - \pi } \right) = 0 + \pi = \pi ,\]
\[f\left( \pi \right) = \sin \pi - \pi = 0 - \pi = - \pi ,\]
\[f\left( 0 \right) = \sin 0 - 0 = 0.\]

Comparing the data we find that the global maximum is attained at \(x = -\pi\) and equal to \(\pi.\) Respectively, the global minimum occurs at \(x = \pi\) and equal to \(-\pi.\)

Example 19.

Find the global maximum and minimum of the function in the given interval: \[f\left( x \right) = {\cos ^2}x - 2\sin x,\] where \(x \in \left[ {0,2\pi } \right].\)

Solution.

The derivative of the function is given by

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

Find the critical points by setting \(f'\left( x \right) = 0:\)

\[f'\left( x \right) = 0,\;\; \Rightarrow - 2\cos x\left( {1 + \sin x} \right) = 0.\]
\[1)\;2\cos x = 0,\;\;\Rightarrow \cos x = 0,\;\;\Rightarrow x = {\frac{\pi }{2}} + \pi n, n \in \mathbb{Z};\]
\[2)\;1 + \sin x = 0,\;\;\Rightarrow \sin x = - 1,\;\; \Rightarrow x = {\frac{{3\pi }}{2}} + 2\pi k, k \in \mathbb{Z}.\]

The following critical points fall in the interval \(\left[ {0,2\pi } \right]:\)

\[x = \frac{\pi }{2},\;\;x = \frac{{3\pi }}{2}.\]

Now it is sufficient to calculate the values of the function at these points and at the endpoints of the interval to determine the global maximum and minimum:

\[f\left( {\frac{\pi }{2}} \right) = {\cos ^2}\frac{\pi }{2} - 2\sin \frac{\pi }{2} = 0 - 2 = - 2;\]
\[f\left( {\frac{{3\pi }}{2}} \right) = {\cos ^2}\left( {\frac{{3\pi }}{2}} \right) - 2\sin \frac{{3\pi }}{2} = 0 + 2 = 2;\]
\[f\left( 0 \right) = {\cos^2}0 - 2\sin 0 = 1 - 0 = 1.\]

The function has a period of \(2\pi.\) Therefore, \(f\left( {2\pi } \right) = f\left( 0 \right) = 1.\)

So, the maximum value of the function on the interval \(\left[ {0,2\pi } \right]\) is equal to \(2\) at \(x = {\frac{{3\pi }}{2}},\) and the minimum value is equal to \(-2\) at \(x = {\frac{{\pi }}{2}}.\)

Example 20.

Find the global maximum and minimum of the function in the given interval: \[f\left( x \right) = \arctan \frac{{1 - x}}{{1 + x}},\] where \(x \in \left[ {0,1} \right].\)

Solution.

Obviously, the function is defined everywhere on the real axis except the point \(x = -1,\) which does not fall into the given interval. Find the derivative:

\[f'\left( x \right) = {\left( {\arctan \frac{{1 - x}}{{1 + x}}} \right)^\prime } = \frac{1}{{1 + {{\left( {\frac{{1 - x}}{{1 + x}}} \right)}^2}}} \cdot {\left( {\frac{{1 - x}}{{1 + x}}} \right)^\prime } = \frac{{ - \left( {1 + x} \right) - \left( {1 - x} \right)}}{{\color{maroon}{1} + \cancel{\color{red}{2x}} + \color{blue}{x^2} + \color{maroon}{1} - \cancel{\color{red}{2x}} + \color{blue}{x^2}}} = \frac{{ - \color{green}{1} - \cancel{\color{red}{x}} - \color{green}{1} + \cancel{\color{red}{x}}}}{{2\left( {1 + {x^2}} \right)}} = - \frac{1}{{1 + {x^2}}}.\]

As it can be seen, the derivative is nowhere zero. Therefore the function contains no other critical points besides \(x = -1.\) Hence, the function is monotonically decreasing (given that the derivative is negative everywhere in the domain). In this case, the maximum and minimum values of the function are attained at the endpoints of the interval:

\[f\left( 0 \right) = \arctan \frac{{1 - 0}}{{1 + 0}} = \arctan 1 = \frac{\pi }{4};\;\;f\left( 1 \right) = \arctan \frac{{1 - 1}}{{1 + 1}} = \arctan 0 = 0.\]

The minimum value is equal to \(f\left( 1 \right) = 0,\) the maximum value is \(f\left( 0 \right) = {\frac{\pi }{4}}.\)

Example 21.

Find the global maximum and minimum of the function in the given interval: \[f\left( x \right) = \left| {\sin x - \frac{{\sqrt 3 }}{2}} \right|,\] where \(x \in \left[ {0,\pi } \right].\)

Solution.

As the function \(f\left( x \right)\) is everywhere non-negative, the minimum value is \(0\) and it is attained in the interval \(\left[ {0,\pi } \right]\) at the following critical points:

\[\sin x - \frac{{\sqrt 3 }}{2} = 0,\;\; \Rightarrow \sin x = \frac{{\sqrt 3 }}{2},\;\; \Rightarrow {x_1} = \frac{\pi }{3},\;\;{x_2} = \frac{{2\pi }}{3}.\]

To determine the maximum value, we calculate the derivative of the function:

\[f'\left( x \right) = \left( {\left| {\sin x - \frac{{\sqrt 3 }}{2}} \right|} \right)^\prime = \frac{{\left| {\sin x - \frac{{\sqrt 3 }}{2}} \right|}}{{\sin x - \frac{{\sqrt 3 }}{2}}} \cdot {\left( {\sin x - \frac{{\sqrt 3 }}{2}} \right)^\prime } = \frac{{\left| {\sin x - \frac{{\sqrt 3 }}{2}} \right|}}{{\sin x - \frac{{\sqrt 3 }}{2}}} \cdot \cos x.\]

The critical points \(x = {\frac{\pi }{3}},\) \(x = {\frac{2\pi }{3}},\) that satisfy the equation \({\sin x = {\frac{{\sqrt 3 }}{2}}},\) have already been found above. Therefore, we consider only the solution

\[\cos x = 0,\;\; \Rightarrow x = \frac{\pi }{2}.\]

Thus, the global maximum is attained either at the point \(x = {\frac{\pi }{2}},\) or at the endpoints of the interval, i.e. at \(x = 0\) or \(x = \pi.\) Calculations lead to the following results:

\[f\left( {\frac{\pi }{2}} \right) = \left| {\sin \frac{\pi }{2} - \frac{{\sqrt 3 }}{2}} \right| = \left| {1 - \frac{{\sqrt 3 }}{2}} \right| \approx 0,732;\]
\[f\left( 0 \right) = \left| {\sin 0 - \frac{{\sqrt 3 }}{2}} \right| = \frac{{\sqrt 3 }}{2} \approx 1,732;\]
\[f\left( \pi \right) = \left| {\sin \pi - \frac{{\sqrt 3 }}{2}} \right| = \frac{{\sqrt 3 }}{2} \approx 1,732.\]

Hence, the maximum value is \({\frac{{\sqrt 3 }}{2}} \approx 1,732\) and it is attained at two points: \(x = 0,\) \(x = \pi.\)

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