# Extreme Value Theorem

The **extreme value theorem** gives the existence of the extrema of a continuous function defined on a closed and bounded interval. Depending on the setting, it might be needed to decide the existence of, and if they exist then compute, the largest and smallest (extreme) values of a given function. For instance, a weather-related model based on collected data is to be analyzed with the means to explain wind speed of a storm. If only height and pressure are measured, by Bernoulli's principle, then wind behavior is a consequence of these measures. Understanding height and minimum pressure will lead to approximately measure maximum wind speed. In calculus, utilizing the derivative (and lack thereof) of a function at a point gives us a tool to find such extrema.

Theoretical consequences of EVT include results that are of fundamental importance to the study of differentiable functions. The most notable are Rolle's theorem and the mean value theorem.

The Extreme Value TheoremIf real numbers $a$ and $b$ are satisfies $a<b,$ and a function $f$ is continuous on $[a,b],$ then $f$ attains its maximum and minimum values on $[a,b].$

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## Applications

As mentioned in the introduction, knowing existence and computing extrema of a function will solve problems in many settings. Existence is given by the extreme value theorem, so the focus is now on how to compute these values.

Question: "How do we find the maxima and minima of a continuous function $f$ on a closed and bounded interval $[a,b]?$"

- Find the derivative $f'(x)$ of $f$.
- Find all critical values of $f$, let's say $c_1, c_2, …, c_k.$
- Compute $f(a)$ and $f(b)$, together with $f(c_1), f(c_2), …, f(c_k).$
- Compare (order) the values in step 3.
- Then the minimum is the lesser of the values in step 4 and the maximum is the greater.

## Examples

Given the function $f(x)=x^3-3x^2+3x+1$, find the maximum and minimum value of this function on the interval $[0,2]$.

This function is differentiable, hence continuous, and the interval in question is closed and bounded. Therefore we can apply the EVT.

We take the first derivative of this function:

$\begin{aligned} f'(x)&=3x^2-6x+3\\ &=3(x^2-2x+1). \end{aligned}$

Since $f$ is differentiable on its entire domain, we need only find the relative extrema and evaluate the endpoints of the interval. Setting $f'(x)=0$ and solving, we find that the only critical number is $x_1=1,$ which gives the value $y_1=2.$

Caution:It is a theorem of calculus that relative extrema only happen at critical numbers. The problem is to determine which critical numbers give us relative extrema. This is done in two ways:the first derivative testorthe second derivative test. By the first derivative test, $f'$ does not change sign before and after the critical number $x_1=1$. Therefore there are no relative extrema associated with this number.Computing the endpoints, we have $y_2=f(0)=1$ and $y_3=f(2)=3$. Therefore, the minimum value of $f$ on $[0,2]$ is 1 and its maximum on the same interval is 3. $_\square$

Given the function $f(x)=x^3-\frac{9}{2}x^2-12x+20$, find the maximum and minimum value of this function on the interval $[-2,6]$.

This function is differentiable, hence continuous, and the interval in question is closed and bounded. Therefore we can apply the EVT to this setting.

We take the first derivative of this function:

$\begin{aligned} f'(x)&=3x^2-9x-12\\ &=3(x^2-3x-4). \end{aligned}$

Since $f$ is differentiable on its entire domain, we need only find the relative extrema and evaluate the endpoints of the interval. Setting $f'(x)=0$ and solving, we find that the critical numbers are $x_1=-1$ and $x_2=4,$ which by the first derivative test give the values of the relative extrema $y_1=26.5$ and $y_2=-36$. Computing the endpoints, we have $y_3=f(-2)=18$ and $y_4=f(6)=2$. Therefore, the minimum value of $f$ on $[-2,6]$ is -36 and its maximum on the same interval is 26.5. $_\square$

Given the function $f(x)=\frac{3x}{x^2-1}$, find the critical numbers of $f$.

We take the first derivative of this function:

$\begin{aligned} f'(x)=\frac{3(x^2+1)}{({1-x^2})^2}. \end{aligned}$

Observe that $f'(x)=0$ has no solution on the real numbers and that $f'(x)$ is undefined for $x=-1$ and $x=1$. But note that the numbers $x=-1$ and $x=1$ are not critical numbers since they are not in the domain of $f$.

Therefore this function has no critical numbers. $_\square$

## See Also

**Cite as:**Extreme Value Theorem.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/extreme-value-theorem/