Maximum and minimum values
Many questions ask for the largest or smallest value of something: the most profit, the least cost, the highest point of a ball. In calculus, these are maximum and minimum values. Derivatives help us find them.
Absolute and local extreme values
Let $c$ be a number in the domain of $f$.
- $f(c)$ is the absolute maximum of $f$ if $f(c)\ge f(x)$ for every $x$ in the domain. It is the highest point of the whole graph.
- $f(c)$ is the absolute minimum of $f$ if $f(c)\le f(x)$ for every $x$ in the domain. It is the lowest point of the whole graph.
- $f(c)$ is a local maximum if $f(c)\ge f(x)$ for every $x$ near $c$. It is a peak: higher than the points around it.
- $f(c)$ is a local minimum if $f(c)\le f(x)$ for every $x$ near $c$. It is a valley: lower than the points around it.
Maximum and minimum values together are called extreme values, or extrema. "Absolute" is also called "global," and "local" is also called "relative."
A function can have more than one local maximum, but only one absolute maximum value. An absolute maximum inside the interval is also a local maximum. At an endpoint, we do not use the word "local."
The Extreme Value Theorem
Does a function always have a largest and a smallest value? Not always. But it does under two conditions.
The Extreme Value Theorem: If $f$ is continuous on a closed interval $[a,b]$, then $f$ has an absolute maximum and an absolute minimum on $[a,b]$.
A closed interval $[a,b]$ includes both endpoints. (See continuity.) Both conditions matter:
The theorem tells us that the extreme values exist. It does not tell us where they are. For that, we use derivatives.
Critical numbers
At a smooth peak or valley, the tangent line is horizontal. This gives the following theorem.
Fermat's Theorem: If $f$ has a local maximum or minimum at $c$, and $f'(c)$ exists, then
$f'(c)=0$.
Why: Suppose $f$ has a local maximum at $c$. Just to the left of $c$, the graph goes up to $f(c)$, so the slope is not negative. Just to the right, the graph comes down, so the slope is not positive. The only slope that is both is $0$.
Be careful. This does not work the other way:
- $f'(c)=0$ does not always give an extreme value. For $f(x)=x^3$, the derivative is $0$ at $x=0$, but the graph keeps rising through $0$.
- An extreme value can happen where $f'(c)$ does not exist. For $f(x)=|x|$, there is a minimum at the corner, $x=0$.
So we look at both kinds of points. A critical number of $f$ is a number $c$ in the domain of $f$ where
- $f'(c)=0$, or
- $f'(c)$ does not exist.
Every local maximum or minimum happens at a critical number. But not every critical number gives a maximum or minimum.
Example 1: Find the critical numbers of
$f(x)=x^3-3x^2$.
Solution:
$\begin{align*}&f'(x)\\&=3x^2-6x\\&=3x(x-2)\end{align*}$
The derivative exists everywhere. It is $0$ when $x=0$ or $x=2$. So the critical numbers are $0$ and $2$.
Example 2: Find the critical numbers of
$f(x)=x^{2/3}(x-5)$.
Solution:
First, multiply out:
$f(x)=x^{5/3}-5x^{2/3}$
Use the power rule:
$f'(x)=\dfrac{5}{3}x^{2/3}-\dfrac{10}{3}x^{-1/3}$
Write it as one fraction. Multiply the first term by $\dfrac{x^{1/3}}{x^{1/3}}$:
$\begin{align*}&f'(x)\\&=\dfrac{5x}{3x^{1/3}}-\dfrac{10}{3x^{1/3}}\\&=\dfrac{5x-10}{3x^{1/3}}\end{align*}$
- $f'(x)=0$ when the numerator is $0$: $5x-10=0$, so $x=2$.
- $f'(x)$ does not exist when the denominator is $0$: $x=0$. And $0$ is in the domain of $f$.
So the critical numbers are $0$ and $2$.
The closed interval method
On a closed interval, the absolute extreme values are at critical numbers or at the endpoints. So we only need to check a few points.
To find the absolute maximum and minimum of a continuous function $f$ on $[a,b]$:
- Find the critical numbers of $f$ in $(a,b)$.
- Find the value of $f$ at each critical number.
- Find the values of $f$ at the endpoints, $f(a)$ and $f(b)$.
- The largest value from steps 2 and 3 is the absolute maximum. The smallest is the absolute minimum.
Example 3: Find the absolute maximum and minimum of
$f(x)=x^3-3x^2+1$
on the interval $\left[-\tfrac{1}{2},4\right]$.
Solution:
Step 1: $f'(x)=3x^2-6x=3x(x-2)$. The critical numbers are $0$ and $2$. Both are in the interval.
Steps 2 and 3: Find the values.
| $x$ | $f(x)$ |
|---|---|
| $-\tfrac{1}{2}$ (endpoint) | $\tfrac{1}{8}$ |
| $0$ (critical) | $1$ |
| $2$ (critical) | $-3$ |
| $4$ (endpoint) | $17$ |
Step 4: The largest value is $17$, and the smallest is $-3$.
So the absolute maximum is $f(4)=17$, and the absolute minimum is $f(2)=-3$.
Notice that the absolute maximum is at an endpoint. If we had checked only the critical numbers, we would have missed it.
Example 4: Find the absolute maximum and minimum of
$f(x)=x-2\sin x$
on the interval $[0,2\pi]$.
Solution:
Step 1: $f'(x)=1-2\cos x$. Set it equal to $0$:
$\begin{align*}1-2\cos x&=0\\\cos x&=\dfrac{1}{2}\end{align*}$
In $(0,2\pi)$, this happens at two angles:
$x=\dfrac{\pi}{3}$ and $x=\dfrac{5\pi}{3}$.
Steps 2 and 3: Find the values. Use these sine values:
$\sin\dfrac{\pi}{3}=\dfrac{\sqrt{3}}{2}$ and $\sin\dfrac{5\pi}{3}=-\dfrac{\sqrt{3}}{2}$
With a calculator in radian mode:
| $x$ | $f(x)$ |
|---|---|
| $0$ | $0$ |
| $\dfrac{\pi}{3}$ | $\dfrac{\pi}{3}-\sqrt{3}\approx -0.685$ |
| $\dfrac{5\pi}{3}$ | $\dfrac{5\pi}{3}+\sqrt{3}\approx 6.968$ |
| $2\pi$ | $2\pi\approx 6.283$ |
Step 4: Compare the values.
The absolute maximum is at $x=\dfrac{5\pi}{3}$:
$f\left(\dfrac{5\pi}{3}\right)=\dfrac{5\pi}{3}+\sqrt{3}\approx 6.968$
The absolute minimum is at $x=\dfrac{\pi}{3}$:
$f\left(\dfrac{\pi}{3}\right)=\dfrac{\pi}{3}-\sqrt{3}\approx -0.685$
Example 5: A ball is thrown straight up. Its height after $t$ seconds is
$h(t)=2+19.6t-4.9t^2$ meters,
for $0\le t\le 4$. Find its greatest and least heights.
Solution:
Step 1: $h'(t)=19.6-9.8t$. It is $0$ when $t=2$.
Steps 2 and 3:
$\begin{align*}h(0)&=2\\h(2)&=2+39.2-19.6=21.6\\h(4)&=2+78.4-78.4=2\end{align*}$
Step 4: Compare the values.
The greatest height is $21.6$ m, at $t=2$ s.
The least height is $2$ m, at both $t=0$ and $t=4$.
At $t=2$, the velocity $h'(2)$ is $0$. The ball stops for an instant at the top.
Summary
- An absolute maximum or minimum is the highest or lowest value of the whole function. A local one is highest or lowest only near a point.
- A continuous function on a closed interval always has an absolute maximum and minimum.
- A critical number is where $f'(c)=0$ or $f'(c)$ does not exist. Local extremes happen only at critical numbers.
- On $[a,b]$, compare $f$ at the critical numbers and at the endpoints.
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