### 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 a problem to see the 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 \(x = \sqrt 2.\)