# Absolute minima & maxima review

Review how we use differential calculus to find absolute extremum (minimum and maximum) points.

## How do I find absolute minimum & maximum points with differential calculus?

An absolute maximum point is a point where the function obtains its greatest possible value. Similarly, an absolute minimum point is a point where the function obtains its least possible value.
Supposing you already know how to find relative minima & maxima, finding absolute extremum points involves one more step: considering the ends in both directions.

## Finding absolute extrema on a closed interval

Extreme value theorem tells us that a continuous function must obtain absolute minimum and maximum values on a closed interval. These extreme values are obtained, either on a relative extremum point within the interval, or on the endpoints of the interval.
Let's find, for example, the absolute extrema of $h(x)=2x^3+3x^2-12x$ over the interval $-3\leq x\leq 3$.
$h'(x)=6(x+2)(x-1)$, so our critical points are $x=-2$ and $x=1$. They divide the closed interval $-3\leq x\leq 3$ into three parts:
Interval$x$-value$h'(x)$Verdict
$-3$x=-\dfrac52$$h'\left(-\dfrac52\right)=\dfrac{21}{2}>0$$h$ is increasing $\nearrow$
$-2$x=0$$h'\left(0\right)=-12<0$$h$ is decreasing $\searrow$
$1$x=2$$h'\left(2\right)=24>0$$h$ is increasing $\nearrow$
Now we look at the critical points and the endpoints of the interval:
$x$$h(x)$BeforeAfterVerdict
$-3$$9$$-$$\nearrow$Minimum
$-2$$20$$\nearrow$$\searrow$Maximum
$1$$-7$$\searrow$$\nearrow$Minimum
$3$$45$$\nearrow$$-$Maximum
On the closed interval $-3 \leq x \leq 3$, the points $(-3,9)$ and $(1,-7)$ are relative minima and the points $(-2,20)$ and $(3,45)$ are relative maxima.
$(1,-7)$ is the lowest relative minimum, so it's the absolute minimum point, and $(3,45)$ is the largest relative maximum, so it's the absolute maximum point.
Notice that the absolute minimum value is obtained within the interval and the absolute maximum value is obtained on an endpoint.
Problem 1
$f(x)=x^3-3x^2+12$
What is the absolute maximum value of $f$ over the closed interval $[-2,4]$?

Want to try more problems like this? Check out this exercise.

## Finding absolute extrema on entire domain

Not all functions have an absolute maximum or minimum value on their entire domain. For example, the linear function $f(x)=x$ doesn't have an absolute minimum or maximum (it can be as low or as high as we want).
However, some functions do have an absolute extremum on their entire domain. Let's analyze, for example, the function $g(x)=xe^{3x}$.
$g'(x)=e^{3x}(1+3x)$, so our only critical point is $x=-\dfrac{1}{3}$.
Interval$x$-value$g'(x)$Verdict
$(-\infty,-\dfrac13)$$x=-1$$g'(-1)=-\dfrac{2}{e^3}<0$$g$ is decreasing $\searrow$
$(-\dfrac{1}{3},\infty)$$x=0$$g'(0)=1>0$$g$ is increasing $\nearrow$
Let's imagine ourselves walking on the graph of $g$, starting all the way to the left (from $-\infty$) and going all the way to the right (until $+\infty$).
We will start by going down and down until we reach $x=-\dfrac{1}{3}$. Then, we will be forever going up. So $g$ has an absolute minimum point at $x=-\dfrac{1}{3}$. The function doesn't have an absolute maximum value.
$g(x)=\dfrac{\ln(x)}{x}$
What is the absolute maximum value of $g$ ?