**The extreme value theorem states that a function has both a maximum and a minimum value in a closed interval $[a,b]$ if it is continuous in $[a,b]$.**

We are interested in finding the maxima and the minima of a function in many applications. For example, a function describes the oscillation behavior of an object; it will be natural for us to be interested in the highest and lowest point of the oscillating wave.

In this topic, **we will discuss in detail about extreme value theorem**, its proof, and how to calculate the minima and maxima of a continuous function.

## What Is Extreme Value Theorem?

The extreme value theorem is a theorem that **determines the maxima and the minima of a continuous function defined in a closed interval**. We would find these extreme values either on the endpoints of the closed interval or on the critical points.

On critical points, **the derivative of the function is zero**. For any continuous closed interval function, the first step is to find all the critical points of a function and then determine the values on these critical points.

Also, evaluate the function on the endpoints of the interval. **The highest value** of the function would be **the maxima**, and** the lowest value** of the function would be **the minima**.

## How To Use Extreme Value Theorem

The procedure of using the extreme value theorem is given i*n the following steps:*

- Make sure the function is continuous on a closed interval.
- Find all the critical points of the function.
- Calculate the value of the function at those critical points.
- Calculate the value of the function on the endpoints of the interval.
- The highest value among all the calculated values is the maxima, and the least value is the minima.

**Note:** If you have confusion regarding a continuous function and a closed interval, see the definitions at the end of this article.

### Proof of Extreme Value TheoremÂ

If $f(x)$ is a continuous function in $[a,b]$, then it must have a least upper bound in $[a,b]$ (by the Boundedness theorem). Let $M$ is **the least upper bound**. We have to show that for a certain point $x_o$ in the closed interval $[a,b]$, $f(x_o)=M$.Â

We will prove this by using the contradictory method.

Suppose there is no such $x_o$ in $[a,b]$ where $f$ **has a maximum value** $M$.

*Consider a function:*

$g(x) =Â \dfrac{1}{M\hspace{1mm} – \hspace{1mm}f(x)}$

As we have assumed there is no M for the function f(x), hence g(x) > 0Â for all values of x and as M â€“ f(x) is continuous, **so the function** $g(x)$ **will also be a continuous function**.

So function g is bounded in the closed interval $[a,b]$ (again by Boundedness theorem), and hence there must be a $C > 0$ such that $g(x) \leq C$ for every value of $x$ in $[a,b]$.

$g(x) \leq C$

$\dfrac{1}{M\hspace{1mm} – \hspace{1mm}f(x)} \leq C$

$M â€“ f(x) \leq \dfrac{1}{C}$

$M – \dfrac{1}{c}\geq f(x)$Â Â Â Â Â Â Â Â Â Â Â Â Â (1)

So according to equation (1),Â $M – \dfrac{1}{C}$ **is the upper bound of function** $f(x)$, but it is smaller than $M$, so it contradicts the definition of M being the least upper bound of $f$. As we have derived a contradiction, our original assumption must be false and hence it is proved there is a point $x_o$ in the closed interval $[a,b]$ where $f(x_o) = M$.Â

We can obtain the proof for minima by** applying the above arguments on** $-f$.

### Example 1:

Find the extreme values for the function $f(x) = x^{2} – 6x + 10$ on the closed interval $[0,4]$.

__Solution:__

This is a quadratic function; the given function is continuous and is bounded by the closed interval $[0,4]$. The first step is to** find the critical values of the given function**. To find the critical values, we have to differentiate the function and put it equal to zero.

$f(x) = x^{2} – 6x + 10$

$f'(x) = 2x â€“ 6$

Now by putting $f'(x) = 0$, we get

$2x â€“ 6 = 0$

$2x = 6$

$x = \dfrac{6}{2}$

$x = 3$

So $x = 3$ is the only critical value of the given function. Moreover, **the calculated critical value lies in the given interval** $[0,4]$.

The absolute extremes of a function must occur at endpoints on the bounded interval (in this case, $0$ or $4$) or at the calculated critical values, so in this case, **the points where the absolute extreme will occur are** $0$, $4$, or $3$; hence we have to calculate the value of the given function at these points.

The value of $f(x)$ at $x = 0$

$f (0) = (0)^{2} â€“ 6 (0) + 10 = 10$

The value of $f(x)$ at $x = 4$

$f (4) = (4)^{2} â€“ 6 (4) + 8 = 16 â€“ 24 + 10 = 2$Â Â

The value of $f(x)$ at $x = 3$

$f (3) = (3)^{2} â€“ 6 (3) + 10 = 1$Â Â

The highest or maximum value is $10$ at $x = 0$ and the lowest or minimum value is $1$ at $x = 3$. With this, we can conclude that **the maximum value of the given function is** $10$, which occurs on the left end point at $x = 0$ while** the minimum value occurs at the critical point** $x = 3$.Â

Â Â

### Example 2:Â

Find the extreme values for the function $f(x) = 2x^{3} – 6x^{2} + 8$ on the closed interval $[-2,5]$.

__Solution:__

Â Â Â Â Â Â Â Â Â Â Â $f(x) = 2x^{3} – 6x^{2} + 8$

Â Â Â Â Â Â Â Â Â Â Â $f'(x) = 6x^{2} â€“ 12x$

Â Â Â Â Â Â Â Â Â Â Â $6x^{2} â€“ 12x = 0$

Â Â Â Â Â Â Â Â Â Â Â $6x (x â€“ 2) = 0$

So $x = 0$Â and $x = 2$ are **the critical values of the given function**. Hence the maxima and minima of the given function will either be at the endpoints of the interval $[-2, 5]$ or at the critical points $0$ or $2$. Calculate the value of the function on all four points.

The value of $f(x)$ at $x = 0$

$f (0) = 2(0)^{3} â€“ 6(0)^{2} + 8 = 8$Â

The value of $f(x)$ at $x = 2$

$f (2) = 2(2)^{3} â€“ 6(2)^{2} + 8 = 16 â€“ 24 + 8 = 0$

The value of $f(x)$ at $x = -2$

$f (-2) = 2(-2)^{3} â€“ 6(-2)^{2} + 8 = -16 – 24 + 8 = -32$

The value of $f(x)$ at $x = 5$

$f (5) = 2(5)^{3} â€“ 6(5)^{2} + 8 = 250-150+8 = 108$

The highest or **maximum value is** $108$ at $x = 5$ and the lowest or **minimum value is** $-32$ at $x = -2$.

Â Â

### Example 3:Â

Find the extreme values for the function $f(x) = 8x^{3} – 12x^{2}$ on the closed interval $[0, 4]$.

__Solution:__

Â Â Â Â Â Â Â Â Â Â Â $f(x) = 8x^{3} – 12x^{2}$

Â Â Â Â Â Â Â Â Â Â Â $f'(x) = 24x^{2} â€“ 24x$

Â Â Â Â Â Â Â Â Â Â Â $24x^{2} â€“ 24x = 0$

Â Â Â Â Â Â Â Â Â Â Â $24x (x â€“ 1) = 0$

So $x = 0$Â and $x = 1$ are **the critical values of the given function**. Hence the maxima and minima of the given function will either be at the $0$, $2$, or $4$. Calculate the value of the function on all three points.

The value of $f(x)$ at $x = 0$

$f (0) = 8(0)^{3} â€“ 12(0)^{2} = 0$Â

The value of $f(x)$ at $x = 1$

$f (1) = 8(1)^{3} â€“ 12(1)^{2} = 8 â€“ 12 Â = -4$

The value of $f(x)$ at $x = 4$

$f (4) = 8(4)^{3} â€“ 12(4)^{2}Â = 512 – 192 = 320$

The highest or **maximum value is** $320$ at $x = 4$ and the lowest or **minimum value is** $-4$ at $x = 1$.

### Example 4:Â

Find the extreme values for the function $f(x) = sinx^{2}$ on the closed interval $[-3,3]$.

__Solution:__

$f(x) = sinx^{2}$

$f'(x) = 2x cosx^{2}$

$2x cosx^{2} = 0$

$2x = 0$ and $cosx^{2} = 0$

$f'(x) = 0$ at $x = 0$, so one of **the critical point is** $x = 0$ while the rest of critical points where value $x^{2}$ is such that it makes $cosx^{2} = 0$. We know that $cos(x) = 0$ at $x = \pm\dfrac{\pi}{2}, \pm\dfrac{3\pi}{2}, \pm\dfrac{5\pi}{2}$â€¦

So, $cosx^{2} = 0$ when $x = \pm\sqrt{\dfrac{\pi}{2}}, \pm\sqrt{\dfrac{3\pi}{2}}, \pm\sqrt{\dfrac{5\pi}{2}}$â€¦

Hence the maxima and minima of the given function **will either be at the endpoints of the interval** $[-3, 3]$ **or at the critical points** $0$,$\pm\sqrt {\dfrac{\pi}{2}}$, $\pm\sqrt {\dfrac{3\pi}{2}}$ and $\pm\sqrt {\dfrac{5\pi}{2}}$.

**Calculate the value of the function** on all these points.

The value of $f(x)$ at $x = 0$

$f (0) = sin(0)^{2}Â = 0$Â

The value of $f(x)$ at $x = \sqrt{\dfrac{\pi}{2}}$

$f (\sqrt{\pi}) = sin(\sqrt{\dfrac{\pi}{2}})^{2} = 1$

The value of $f(x)$ at $x = -\sqrt{\dfrac{\pi}{2}}$

$f (-\sqrt{\pi}) = sin(-\sqrt{\pi})^{2} Â = 1$Â Â

The value of $f(x)$ at $x = \sqrt{\dfrac{3\pi}{2}}$

$f (\sqrt{\dfrac{3\pi}{2}}) = sin(\sqrt{\dfrac{3\pi}{2}})^{2} = -1$

The value of $f(x)$ at $x = -\sqrt{\dfrac{3\pi}{2}}$

$f (-\sqrt{\dfrac{3\pi}{2}}) = sin(-\sqrt{\dfrac{3\pi}{2}})^{2}Â = -1$

The value of $f(x)$ at $x = \sqrt{\dfrac{5\pi}{2}}$

$f (\sqrt{\dfrac{5\pi}{2}}) = sin(\sqrt{\dfrac{5\pi}{2}})^{2} = 1$

The value of $f(x)$ at $x = -\sqrt{\dfrac{5\pi}{2}}$

$f (-\sqrt{\dfrac{5\pi}{2}}) = sin(-\sqrt{\dfrac{5\pi}{2}})^{2}Â = 1$

The value of f(x) at $x = 3$

$f (0) = sin(3)^{2}Â = 0.412$Â

The value of $f(x)$ at $x = -3$

$f (0) = sin(-3)^{2}Â = 0.412$

## Important Definitions

### Continuous Function

A function is known as a continuous function if** the graph of the said function is continuous without any breakpoints**. The function will be continuous on all the points of the given interval. For example, $x^{2}$, $x^{4}$, $\sqrt{x}$ are all continuous functions. Mathematically,Â a function $f(x)$ is continuous in $[a,b]$ if $\lim x \to c f(x) = f(c)$ for all $c$ in $[a,b]$.

The differentiation of a function can only be carried out if the function is continuous; **the critical points of a function are found using differentiation**. So to find the extreme values of a function, it is essential that the function must be continuous.

### Closed Interval

A closed interval is an interval that **includes all the points within the given limit, and square brackets denote it**, i.e., [ ]. For example, the interval $[3, 6]$ includes all the greater and equal points to $3$ and less than or equal to $6$.Â

*Practice Questions:*

- Find the extreme values for the function $f(x) = 6x^{2} -3x +12$ on the closed interval $[0, 3]$.
- Find the extreme values for the function $f(x) = xe^{6x}$ on the closed interval $[-2, 0]$.Â

*Answer Key:*

1.

$f(x) = 6x^{2} -3x +12$

$f^{â€˜}(x) = 12x -3 $

$= 12x -3 = 0$

$x Â = \dfrac{1}{4}$

So $x = \dfrac{1}{4}$ is **the critical value of the given function**. Hence, the maxima and minima of the given function will either be at the $\dfrac{1}{4}$, $0$, or $3$.

*Calculating the value of the function on all three points:*

The value of $f(x)$ at $x = 0$

$f (0) = 6(0)^{2} â€“ 3(0) +12 = 12$Â

The value of $f(x)$ at $x = 3$

$f (3) = 6(3)^{2} â€“ 3(6) +12 = 54 â€“ 9 + 12 = 57$

The value of $f(x)$ at $x = \dfrac{1}{4}$

$f (4) = 6(\dfrac{1}{4})^{2} â€“ 3(\dfrac{1}{4}) +12Â = \dfrac{3}{8}+\dfrac{3}{4}+ 12 = 13.125$

The highest or **maximum value is** $48$ at $x = 3$ and the lowest or **minimum value is** $12$ at $x = 0$.

2.

$f(x) = xe^{6x}$

*Applying chain rule to differentiate the above function*:

$ f^{â€˜}(x) = 1. e^{6x} + 6x. e^{6x} = e^{6x}(1+6x)$

Now putting $f^{â€˜}(x) = 0$

$e^{6x}(1+6x) = 0$

$ 1+6x = 0$

$ x = – \dfrac{1}{6}$

So $x = -\dfrac{1}{6}$ is **the critical value of the given function**. Hence, the maxima and minima of the given function will either be at the $-\dfrac{1}{6}$, $-2$ or $0$.

*Calculating the value of the function on all three points:*

The value of $f(x)$ at $x = 0$

$f (0) = 0. e^{0} = 0$Â

The value of $f(x)$ at $x = -2$

$f (3) = -2. e^{-12} = -1.22 \times 10^{-5}$

The value of $f(x)$ at $x = -\dfrac{1}{6}$

$f (3) = -\dfrac{1}{6}. e^{-1} = 0.06131$Â Â Â Â