# Calculus

## Applications of the Derivative # Global Extrema of Functions

### Definition of Global Maximum and Global Minimum

Consider a function $$y = f\left( x \right),$$ which is supposed to be continuous on a closed interval $$\left[ {a,b} \right].$$ If there exists a point $${x_0} \in \left[ {a,b} \right]$$ such that $$f\left( x \right) \le f\left( {{x_0}} \right)$$ for all $$x \in \left[ {a,b} \right],$$ then we say that the function $$f\left( x \right)$$ attains at $${x_0}$$ the maximum (greatest) value over the interval $$\left[ {a,b} \right].$$

The greatest value of the function $$f\left( x \right)$$ on the interval $$\left[ {a,b} \right]$$ is simultaneously the least upper bound of the range of the function on this interval and is denoted as

${f\left( {{x_0}} \right) = \max\limits_{x \in \left[ {a,b} \right]} f\left( x \right) } = {\sup\limits_{x \in \left[ {a,b} \right]} f\left( x \right).}$

Similarly, if there exists a point $${x_0} \in \left[ {a,b} \right]$$ such that $$f\left( x \right) \ge f\left( {{x_0}} \right)$$ for all $$x \in \left[ {a,b} \right],$$ then we say that the function $$f\left( x \right)$$ attains at $${x_0}$$ the minimum (least) value over the interval $$\left[ {a,b} \right].$$

The least value of the function $$f\left( x \right)$$ on the interval $$\left[ {a,b} \right]$$ is simultaneously the greatest lower bound of the range of the function on this interval and can be written as

${f\left( {{x_0}} \right) = \min\limits_{x \in \left[ {a,b} \right]} f\left( x \right) } = {\inf\limits_{x \in \left[ {a,b} \right]} f\left( x \right).}$

These concepts characterize the behavior of a function on a finite interval, in contrast to the local extremum, which describes the properties of the function in a small neighborhood of a point. Therefore, the maximum and minimum values of a function on an interval are often referred as the global (absolute) maximum or, respectively, the global minimum.

### The Weierstrass Extreme Value Theorem

According to the Weierstrass extreme value theorem for continuous functions, if a function $$f\left( x \right)$$ is continuous on a closed interval $$\left[ {a,b} \right],$$ then it attains the least upper and greatest lower bounds on this interval.

The proof of this theorem is based on the Weierstrass boundedness theorem, which is stated as follows:

If a function $$f\left( x \right)$$ is continuous on the interval $$\left[ {a,b} \right],$$ then it is bounded on it, i.e. there exists a real number $$M$$ such that $$\left| {f\left( x \right)} \right| \le M$$ for all $$x \in \left[ {a,b} \right].$$

Returning to the extreme value theorem, we denote by $$M$$ the least upper bound of the range of the function (or the maximum value of the function) on the interval $$\left[ {a,b} \right].$$ Assume the contrary, that the least upper bound is not attained, i.e. assume that

$f\left( x \right) \lt M\;\forall \;x \in \left[ {a,b} \right].$

Consider the auxiliary function:

$\varphi \left( x \right) = \frac{1}{{M – f\left( x \right)}}.$

Since the denominator is not zero, then the function $$\varphi \left( x \right)$$ is also continuous on $$\left[ {a,b} \right]$$ and hence, by the boundedness theorem, it is bounded on this interval: $$\varphi \left( x \right) \le L,$$ where $$L \gt 0.$$ It follows from here that

${\varphi \left( x \right) \le L,\;\;}\Rightarrow {\frac{1}{{M – f\left( x \right)}} \le L,\;\;}\Rightarrow {M – f\left( x \right) \ge \frac{1}{L},\;\;}\Rightarrow {f\left( x \right) \le M – \frac{1}{L}\;\forall \;x \in \left[ {a,b} \right].}$

In other words, the number $$M – {\large\frac{1}{L}\normalsize}$$ will be the least upper bound of the function $$f\left( x \right)$$, which contradicts the condition. (By the condition, the least upper bound of the function is equal to $$\left.{M}\right).$$

In the same way we can prove that a function $$f\left( x \right)$$ that is continuous in a closed interval $$\left[ {a,b} \right]$$ attains in this interval its greatest lower bound (or the minimum value).

Thus, according to the extreme value theorem, a function that is continuous on an interval always attains its maximum and minimum values on this interval.

### Finding the Maximum and Minimum Values of a Function

Let a function $$y = f\left( x \right)$$ be continuous on an interval $$\left[ {a,b} \right].$$

If the function on this interval has local maxima at the points $${x_1},{x_2}, \ldots ,{x_n},$$ then the maximum value of the function $$f\left( x \right)$$ on the interval $$\left[ {a,b} \right]$$ is equal to the greatest of the numbers

${f\left( a \right),f\left( {{x_1}} \right),f\left( {{x_2}} \right), \ldots ,} \kern0pt{f\left( {{x_n}} \right),f\left( b \right).}$

Similarly, if the function on this interval has local minima at the points $${{\bar x}_1},{{\bar x}_2}, \ldots ,{{\bar x}_k},$$ then the minimum value of the function $$f\left( x \right)$$ on $$\left[ {a,b} \right]$$ is equal to the least of the numbers

${f\left( a \right),f\left( {{{\bar x}_1}} \right),f\left( {{{\bar x}_2}} \right), \ldots ,} \kern0pt{f\left( {{{\bar x}_k}} \right),f\left( b \right).}$

Thus, the global maximum (minimum) values of a function are attained either on the boundary of the interval (Figure $$2$$), or at the points of local extrema inside the interval (Figure $$3$$).

### Special Case $$1$$

If there exists a unique extremum point $${x_1}$$ inside the interval $$\left[ {a,b} \right]$$ and this point is the local maximum (minimum), then the function attains the global maximum (minimum) at this point (Figure $$4$$).

### Special Case $$2$$

If the function $$y = f\left( x \right)$$ has no critical points on the interval $$\left[ {a,b} \right],$$ then the function has the global minimum at one endpoint of the interval and the global maximum on the other endpoint (Figure $$5\text{).}$$

### Special Case $$3$$

In practice, there are often cases when a differentiable and positively defined function $$f\left( x \right) \gt 0$$ is given on an interval $$\left[ {a,b} \right]$$ and is equal to zero at the endpoints of the interval: $$f\left( a \right) = f\left( b \right) = 0.$$ If this function has a unique stationary point $${x_1}$$ (where $$f’\left( {{x_1}} \right) = 0$$), then this point is not only the local maximum of the function, but also its global maximum on the interval $$\left[ {a,b} \right]$$ (see Figure $$6$$).

In the examples below, find the global maximum and minimum of the function on the given interval.

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Find the global maximum and minimum of the function on the given interval: ${f\left( x \right) = {x^2} – 2x + 5,\;\;\;}\kern-0.3pt{x \in \left[ { – 1,4} \right].}$

### 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 $$\left[ { – 2,4} \right].$$

### Example 3

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

### Example 4

Find the global maximum and minimum of the function on the given interval: ${f\left( x \right) = x + \frac{2}{x},\;\;\;}\kern-0.3pt{x \in \left[ {0.5,2} \right].}$

### Example 5

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

### Example 6

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

### Example 7

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

### Example 8

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

### Example 9

Find the global maximum and minimum of the function on the given interval: ${f\left( x \right) = x\left| {x – 2} \right|,\;\;\;}\kern-0.3pt{x \in \left[ {0,3} \right].}$

### Example 10

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

### Example 11

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

### Example 12

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

### Example 13

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

### Example 14

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

### Example 15

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

### Example 16

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

### Example 17

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

### Example 18

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

### Example 19

Find the global maximum and minimum of the function on the given interval: ${f\left( x \right) = {\cos ^2}x – 2\sin x,\;\;\;}\kern-0.3pt{x \in \left[ {0,2\pi } \right].}$

### Example 20

Find the global maximum and minimum of the function on the given interval: ${f\left( x \right) = \arctan \frac{{1 – x}}{{1 + x}},\;\;\;}\kern-0.3pt{x \in \left[ {0,1} \right].}$

### Example 21

Find the global maximum and minimum of the function on the given interval: ${f\left( x \right) = \left| {\sin x – \frac{{\sqrt 3 }}{2}} \right|,\;\;\;}\kern-0.3pt{x \in \left[ {0,\pi } \right].}$

### Example 1.

Find the global maximum and minimum of the function on the given interval: ${f\left( x \right) = {x^2} – 2x + 5,\;\;\;}\kern-0.3pt{x \in \left[ { – 1,4} \right].}$

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:

${f\left( 1 \right) = {1^2} – 2 \cdot 1 + 5 = 4,}\;\;\;\kern-0.3pt {f\left( { – 1} \right) = {\left( { – 1} \right)^2} – 2 \cdot \left( { – 1} \right) + 5 = 8,}\;\;\;\kern-0.3pt {f\left( 4 \right) = {4^2} – 2 \cdot 4 + 5 = 13.}$

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 $$f\left( x \right) = {x^3} – 6{x^2} – 15x + 100$$ on the interval $$\left[ { – 3,6} \right].$$

Solution.

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 $$f^\prime\left( c \right) = 0:$$

${3{c^2} – 12c – 15 = 0,}\;\; \Rightarrow {3\left( {{c^2} – 4c – 5} \right) = 0,}\;\; \Rightarrow {3\left( {c + 1} \right)\left( {c – 5} \right) = 0,}\;\; \Rightarrow {{c_1} = – 1,\;}\kern0pt{{c_2} = 5.}$

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

${f\left( { – 3} \right) }={ {\left( { – 3} \right)^3} – 6 \cdot {\left( { – 3} \right)^2} -}\kern0pt{ 15 \cdot \left( { – 3} \right) + 100 }={ 64,}$

${f\left( { – 1} \right) }={ {\left( { – 1} \right)^3} – 6 \cdot {\left( { – 1} \right)^2} -}\kern0pt{ 15 \cdot \left( { – 1} \right) + 100 }={ 108,}$

${f\left( {5} \right) }={ {{5}^3} – 6 \cdot {{5} ^2} -}\kern0pt{ 15 \cdot {5} + 100 }={ 0,}$

${f\left( {6} \right) }={ {{6}^3} – 6 \cdot {{6} ^2} -}\kern0pt{ 15 \cdot {6} + 100 }={ 10.}$

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: ${f\left( x \right) = x + \frac{2}{x},\;\;\;}\kern-0.3pt{x \in \left[ {0.5,2} \right].}$

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 can be seen, 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:

${f\left( {\sqrt 2 } \right) = \sqrt 2 + \frac{2}{{\sqrt 2 }} = 2\sqrt 2 \approx 2.83;\;\;\;}\kern-0.3pt {f\left( {0.5} \right) = 0.5 + \frac{2}{{0.5}} = 4.5;\;\;\;}\kern-0.3pt {f\left( 2 \right) = 2 + \frac{2}{2} = 3.}$

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 $$f\left( x \right) = {x^2} – \large{\frac{{16}}{x}}\normalsize$$ on 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 $$f\left( x \right) = 3{x^4} – 6{x^2} + 2$$ on 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:

${f\left( 0 \right) = 3 \cdot {0^4} – 6 \cdot {0^2} + 2 = 2;\;\;\;}\kern-0.3pt {f\left( { – 1} \right) = 3 \cdot {\left( { – 1} \right)^4} – 6 \cdot {\left( { – 1} \right)^2} + 2 = – 1;\;\;\;}\kern-0.3pt {f\left( { – 2} \right) = 3 \cdot {\left( { – 2} \right)^4} – 6 \cdot {\left( { – 2} \right)^2} + 2 = 26.}$

Since the function is even, we can write:

${f\left( 1 \right) = f\left( { – 1} \right) = – 1,}\;\;\;\kern-0.3pt {f\left( 2 \right) = f\left( { – 2} \right) = 26.}$

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 $$8.$$