# Calculus

## Applications of the Derivative # Global Extrema of Functions

## Definition of Global Maximum and Global Minimum

Consider a function y = f (x), which is supposed to be continuous on a closed interval [a, b]. If there exists a point x0 [a, b] such that f (x) ≤ f (x0) for all x [a, b], then we say that the function f (x) attains at x0 the maximum (greatest) value over the interval [a, b].

The greatest value of the function f (x) on the interval [a, b] 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 x0 [a, b] such that f (x) ≥ f (x0) for all x [a, b], then we say that the function f (x) attains at x0 the minimum (least) value over the interval [a, b].

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 $$M - {\frac{1}{L}}$$ 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

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

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

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

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 $$5\text{).}$$

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 $$6$$).