Calculus

Applications of the Derivative

Applications of Derivative Logo

Global Extrema of Functions

Solved Problems

Example 1.

Find the global maximum and minimum of the function in the given interval: \[f\left( x \right) = {x^2} - 2x + 5,\] where \(x \in \left[ { - 1,4} \right].\)

Solution.

The function is defined and differentiable for all \(x \in \mathbb{R}.\) Determine its stationary points:

\[f'\left( x \right) = 0,\;\; \Rightarrow {\left( {{x^2} - 2x + 5} \right)^\prime } = 0,\;\; \Rightarrow 2x - 2 = 0,\;\; \Rightarrow x = 1.\]

This local extremum point belongs to the interval \(\left( { - 1,4} \right).\) We compute the values of the function at \(x = 1\) and at the endpoints of the interval:

\[f\left( 1 \right) = {1^2} - 2 \cdot 1 + 5 = 4,\;\;f\left( { - 1} \right) = {\left( { - 1} \right)^2} - 2 \cdot \left( { - 1} \right) + 5 = 8,\;\;f\left( 4 \right) = {4^2} - 2 \cdot 4 + 5 = 13.\]

Consequently, the maximum value of the function is equal \(f\left( 4 \right) = 13,\) and the minimum value is \(f\left( 1 \right) = 4.\)

Example 2.

Calculate the difference \(d\) between the global maximum and global minimum values of \[f\left( x \right) = {x^2} - 4x + 6\] in the interval \(\left[ { - 2,4} \right].\)

Solution.

Find the derivative:

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

Solve the equation \(f^\prime\left( c \right) = 0\) to determine the critical points:

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

We must evaluate \(f\left( x \right)\) at the critical point \(x = 2\) and at the endpoints \(x = -2,\) \(x = 4:\)

\[f\left( { - 2} \right) = {\left( { - 2} \right)^2} - 4 \cdot \left( { - 2} \right) + 6 = 18,\]
\[f\left( 2 \right) = {2^2} - 4 \cdot 2 + 6 = 2,\]
\[f\left( 4 \right) = {4^2} - 4 \cdot 4 + 6 = 6.\]

Thus, the global maximum value is 18 and the global minimum value is 2, so the difference \(d\) is equal to

\[d = {f_{\max }} - {f_{\min }} = 18 - 2 = 16.\]

Example 3.

Find the global extrema of the function \[f\left( x \right) = {x^3} - 6{x^2} - 15x + 100\] in the interval \(\left[ { - 3,6} \right].\)

Solution.

Global extrema of the cubic function f(x)=x^3-6x^2-15x+100.
Figure 7.

Write the derivative of the function:

\[f^\prime\left( x \right) = \left( {{x^3} - 6{x^2} - 15x + 100} \right)^\prime = 3{x^2} - 12x - 15.\]

Determine the critical points by solving the equation \(f^\prime\left( c \right) = 0:\)

\[3{c^2} - 12c - 15 = 0,\;\; \Rightarrow 3\left( {{c^2} - 4c - 5} \right) = 0,\;\; \Rightarrow 3\left( {c + 1} \right)\left( {c - 5} \right) = 0,\;\; \Rightarrow {c_1} = - 1,\;{c_2} = 5.\]

Now we evaluate the function at these points and at the endpoints of the interval:

\[f\left( { - 3} \right) = {\left( { - 3} \right)^3} - 6 \cdot {\left( { - 3} \right)^2} - 15 \cdot \left( { - 3} \right) + 100 = 64,\]
\[f\left( { - 1} \right) = {\left( { - 1} \right)^3} - 6 \cdot {\left( { - 1} \right)^2} - 15 \cdot \left( { - 1} \right) + 100 = 108,\]
\[f\left( {5} \right) = {{5}^3} - 6 \cdot {{5} ^2} - 15 \cdot {5} + 100 = 0,\]
\[f\left( {6} \right) = {{6}^3} - 6 \cdot {{6} ^2} - 15 \cdot {6} + 100 = 10.\]

We see that \(\left( {5,0} \right)\) is the global minimum point and \(\left( {-1,108} \right)\) is the global maximum point.

Example 4.

Find the global maximum and minimum of the function on the given interval: \[f\left( x \right) = x + \frac{2}{x},\] where \(x \in \left[ {0.5,2} \right].\)

Solution.

This function is not defined at \(x = 0,\) but this point is not included in the given interval. Differentiating, we find the points of local extremum:

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

As you can see, only the point \(x = \sqrt 2\) falls in the interval \(\left[ {0.5,2} \right].\) Calculate the values of the function at the extremum point \(x = \sqrt 2\) and at the boundary points of the interval:

\[f\left( {\sqrt 2 } \right) = \sqrt 2 + \frac{2}{{\sqrt 2 }} = 2\sqrt 2 \approx 2.83;\;\;f\left( {0.5} \right) = 0.5 + \frac{2}{{0.5}} = 4.5;\;\;f\left( 2 \right) = 2 + \frac{2}{2} = 3.\]

So the maximum value of the function in this interval is equal to \(4.5\) at the point \(x = 0.5,\) and the minimum value is \(2.83\) at \(x = \sqrt 2.\)

Example 5.

Find the global maximum and global minimum of the function \[f\left( x \right) = {x^2} - \frac{{16}}{x}\] in the interval \(\left[ { - 4, - 1} \right].\)

Solution.

Note that the domain of \(f\left( x \right)\) does not contain \(x = 0,\) and this point is not in the interval \(\left[ { - 4, - 1} \right].\)

Compute the derivative:

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

Solve the equation \(f^\prime\left( c \right) = 0\) to determine the critical points:

\[f^\prime\left( c \right) = 0,\;\; \Rightarrow \frac{{2{c^3} + 16}}{{{c^2}}} = 0,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {2{c^3} + 16 = 0}\\ {{c^2} \ne 0} \end{array}} \right.,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {{c^3} = - 8}\\ {c \ne 0} \end{array}} \right.,\;\; \Rightarrow c = - 2.\]

Calculate the function values at the critical point and at the boundaries of the interval:

\[f\left( { - 4} \right) = {\left( { - 4} \right)^2} - \frac{{16}}{{\left( { - 4} \right)}} = 20,\]
\[f\left( { - 2} \right) = {\left( { - 2} \right)^2} - \frac{{16}}{{\left( { - 2} \right)}} = 12,\]
\[f\left( { - 1} \right) = {\left( { - 1} \right)^2} - \frac{{16}}{{\left( { - 1} \right)}} = 17.\]

Thus, the global (absolute) minimum occurs at \(\left( { - 2,12} \right),\) and the global maximum is attained at \(\left( { - 4,20} \right).\)

Example 6.

Find the global extrema of the function \[f\left( x \right) = 3{x^4} - 6{x^2} + 2\] in the interval \(\left[ { - 2,2} \right].\)

Solution.

This function is defined and differentiable on the whole real axis. In this case, all local extrema can be found from the equation \(f'\left( x \right) = 0:\)

\[f'\left( x \right) = {\left( {3{x^4} - 6{x^2} + 2} \right)^\prime } = 12{x^3} - 12x = 12x\left( {{x^2} - 1} \right) = 12x\left( {x - 1} \right)\left( {x + 1} \right);\]
\[f'\left( x \right) = 0,\;\; \Rightarrow 12x\left( {x - 1} \right)\left( {x + 1} \right) = 0,\;\; \Rightarrow {x_1} = 0,\;{x_2} = - 1,\;{x_3} = 1.\]

As it can be seen, the function has three local extrema and all these points fall in the given interval \(\left[ { - 2,2} \right].\) Calculate the values of the function at the points of extremum and at the endpoints of the interval:

\[f\left( 0 \right) = 3 \cdot {0^4} - 6 \cdot {0^2} + 2 = 2;\;\;f\left( { - 1} \right) = 3 \cdot {\left( { - 1} \right)^4} - 6 \cdot {\left( { - 1} \right)^2} + 2 = - 1;\;\;f\left( { - 2} \right) = 3 \cdot {\left( { - 2} \right)^4} - 6 \cdot {\left( { - 2} \right)^2} + 2 = 26.\]

Since the function is even, we can write:

\[f\left( 1 \right) = f\left( { - 1} \right) = - 1,\;\;f\left( 2 \right) = f\left( { - 2} \right) = 26.\]

Thus, the function has the minimum value \(-1\) at two points: at \(x = -1\) and \(x = 1.\) The maximum value \(26\) is also attained at two points: at \(x = -2\) and \(x = 2.\) A schematic graph of the function is given in Figure \(8.\)

Global maximum and minimum points of the quartic function y=3x^4-6x^2+2.
Figure 8.

Example 7.

Find the global and global minimum of the function \[f\left( x \right) = {x^4} - 8{x^2} + 3\] within the interval \(\left[ { - 1,2} \right].\)

Solution.

Global maximum and minimum points of the quartic function y=x^4-8x^2+3.
Figure 9.

The function \(f\left( x \right)\) is differentiable everywhere. Its derivative is given by

\[f^\prime\left( x \right) = \left( {{x^4} - 8{x^2} + 3} \right)^\prime = 4{x^3} - 16x.\]

So, the function has the following critical points

\[f^\prime\left( c \right) = 0,\;\; \Rightarrow 4{c^3} - 16c = 0,\;\; \Rightarrow 4c\left( {{c^2} - 4} \right) = 0,\;\; \Rightarrow 4c\left( {c - 2} \right)\left( {c + 2} \right) = 0,\;\; \Rightarrow {c_1} = - 2,\;{c_2} = 0,\;{c_3} = 2.\]

Evaluate the function value at the critical points and at the endpoints of the interval \(\left[ { {- 1},{2}} \right]:\)

\[f\left( { - 1} \right) = {\left( { - 1} \right)^4} - 8 \cdot {\left( { - 1} \right)^2} + 3 = - 4,\]
\[f\left( 0 \right) = {0^4} - 8 \cdot {0^2} + 3 = 3,\]
\[f\left( 2 \right) = {2^4} - 8 \cdot {2^2} + 3 = - 13.\]

The maximum value is \(3\) at \(x = 0,\) and the minimum value is \(-13\) at \(x = 2.\)

Example 8.

Find the global extrema of the function \[f\left( x \right) = {x^4} - 2{x^2} - 1\] on \(\left[ {0, + \infty } \right).\)

Solution.

Global extrema points of the quartic function y=x^4-2x^2-1.
Figure 10.

Take the derivative:

\[f^\prime\left( x \right) = \left( {{x^4} - 2{x^2} - 1} \right)^\prime = 4{x^3} - 4x.\]

Determine the critical points:

\[f^\prime\left( c \right) = 0,\;\; \Rightarrow 4{c^3} - 4c = 0,\;\; \Rightarrow 4c\left( {{c^2} - 1} \right) = 0,\;\; \Rightarrow 4c\left( {c - 1} \right)\left( {c + 1} \right) = 0,\;\; \Rightarrow {c_1} = - 1,\;{c_2} = 0,\;{c_3} = 1.\]

Only \(c = 0\) and \(c = 1\) belong to the given interval. The function has no other critical points, so we must calculate the following function values:

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

The last point \(\left( {1, - 2} \right)\) is a global minimum. The function has no global maximum since \(\lim\limits_{x \to \infty } f\left( x \right) = + \infty .\)

Example 9.

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

Solution.

By expanding the absolute value, we can re-write the function in the form:

\[ f(x) = \begin{cases} - x\left( {x - 2} \right), & x \lt 2 \\ x\left( {x - 2} \right), & x \ge 2 \end{cases},\;\;\Rightarrow f(x) = \begin{cases} - {x^2} + 2x, & x \lt 2 \\ {x^2} - 2x, & x \ge 2 \end{cases}.\]

As it can be seen, the function consists of two quadratic functions \({f_1}\left( x \right) = - {x^2} + 2x\) and \({f_2}\left( x \right) = {x^2} - 2x.\) The function has a kink at the point \(x = 2,\) where the two branches are connected (Figure \(11\)), i.e. the derivative does not exist at this point.

Global maximum and minimum points of the piecewise function y=x|x-2|.
Figure 11.

Determine the other critical points of both branches of the function:

\[{f'_1}\left( x \right) = 0,\;\; \Rightarrow \left( { - {x^2} + 2x} \right)^\prime = 0,\;\; \Rightarrow - 2x + 2 = 0,\;\; \Rightarrow x = 1;\]
\[{f'_2}\left( x \right) = 0,\;\; \Rightarrow \left( {{x^2} - 2x} \right)^\prime = 0,\;\; \Rightarrow 2x - 2 = 0,\;\; \Rightarrow x = 1.\]

Here the root \(x = 1\) makes sense only for the first branch of solutions, which exists for \(x \lt 2.\) Next, we calculate the values of the function at the critical points \(x = 1,\) \(x = 2\) and at the endpoints of the interval at \(x = 0\) and \(x = 3:\)

\[f\left( 0 \right) = 0 \cdot \left| {0 - 2} \right| = 0,\;\;f\left( 1 \right) = 1 \cdot \left| {1 - 2} \right| = 1,\;\;f\left( 2 \right) = 2 \cdot \left| {2 - 2} \right| = 0,\;\;f\left( 3 \right) = 3 \cdot \left| {3 - 2} \right| = 3.\]

This shows that the function has the minimum value \(y = 0\) at two points: at \(x = 0\) and \(x = 2.\) The maximum value is respectively \(f\left( 3 \right) = 3.\)

Example 10.

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

Solution.

The function is defined under the condition

\[3 - 2x \ge 0,\;\; \Rightarrow 2x \le 3,\;\; \Rightarrow x \le \frac{3}{2}.\]

The interval specified in the problem falls into the domain of the function. The function is differentiable on this interval. Therefore its extrema (if they exist) can be determined from the equation \(f'\left( x \right) = 0.\) Find the derivative:

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

It follows that the equation \(f'\left( x \right) = 0\) has no solutions, i.e. the function has no local extrema. Since the derivative is negative, the function \(f\left( x \right)\) is monotonically decreasing on the interval \(\left[ { - 3,1} \right].\) Calculating the value of the function at the boundary points:

\[f\left( { - 3} \right) = \sqrt {3 - 2 \cdot \left( { - 3} \right)} = 3,\;\;f\left( 1 \right) = \sqrt {3 - 2 \cdot 1} = 1,\]

we find that the maximum value is equal to \(3\) at \(x = -3\) and the minimum value is \(1\) at \(x = 1.\)

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