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# Calculus

Applications of the Derivative

# Global Extrema of Functions

Page 1
Problems 1-2
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Problems 3-10

### 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 $$M$$.)

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.

Fig.1 Karl Weierstrass
(1815-1897)

### 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 $$3$$).

Figure 2.

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 $$5\text{).}$$

Figure 4.

Figure 5.

Figure 6.

### 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 on problem description to see solution.

### ✓Example 1

${f\left( x \right) = {x^2} – 2x + 5,\;\;\;}\kern-0.3pt{x \in \left[ { – 1,4} \right].}$

### ✓Example 2

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

### ✓Example 3

${f\left( x \right) = 3{x^4} – 6{x^2} + 2,\;\;\;}\kern-0.3pt{x \in \left[ { – 2,2} \right].}$

### ✓Example 4

${f\left( x \right) = x\left| {x – 2} \right|,\;\;\;}\kern-0.3pt{x \in \left[ {0,3} \right].}$

### ✓Example 5

${f\left( x \right) = \sqrt {3 – 2x} ,\;\;\;}\kern-0.3pt{x \in \left[ { – 3,1} \right].}$

### ✓Example 6

$f\left( x \right) = \frac{{{x^2}}}{{{2^x}}},\;\;x \in \left[ { – 1,3} \right].$

### ✓Example 7

${f\left( x \right) = \sqrt[\large x\normalsize]{x},\;\;\;}\kern-0.3pt{x \in \left[ {2,3} \right].}$

### ✓Example 8

${f\left( x \right) = {\cos ^2}x – 2\sin x,\;\;\;}\kern-0.3pt{x \in \left[ {0,2\pi } \right].}$

### ✓Example 9

${f\left( x \right) = \arctan \frac{{1 – x}}{{1 + x}},\;\;\;}\kern-0.3pt{x \in \left[ {0,1} \right].}$

### ✓Example 10

${f\left( x \right) = \left| {\sin x – \frac{{\sqrt 3 }}{2}} \right|,\;\;\;}\kern-0.3pt{x \in \left[ {0,\pi } \right].}$

### Example 1.

${f\left( x \right) = {x^2} – 2x + 5,\;\;\;}\kern-0.3pt{x \in \left[ { – 1,4} \right].}$

#### 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.

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

#### 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 $$x = \sqrt 2.$$

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Problems 1-2
Page 2
Problems 3-10