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### Course: AP®︎/College Calculus AB>Unit 5

Lesson 5: Using the candidates test to find absolute (global) extrema

# 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 $h\left(x\right)=2{x}^{3}+3{x}^{2}-12x$ over the interval $-3\le x\le 3$.
${h}^{\prime }\left(x\right)=6\left(x+2\right)\left(x-1\right)$, so our critical points are $x=-2$ and $x=1$. They divide the closed interval $-3\le x\le 3$ into three parts:
Interval$x$-value${h}^{\prime }\left(x\right)$Verdict
$-3$x=-\frac{5}{2}$${h}^{\prime }\left(-\frac{5}{2}\right)=\frac{21}{2}>0$$h$ is increasing $↗$
$-2$x=0$${h}^{\prime }\left(0\right)=-12<0$$h$ is decreasing $↘$
$1$x=2$${h}^{\prime }\left(2\right)=24>0$$h$ is increasing $↗$
Now we look at the critical points and the endpoints of the interval:
$x$$h\left(x\right)$BeforeAfterVerdict
$-3$$9$$-$$↗$Minimum
$-2$$20$$↗$$↘$Maximum
$1$$-7$$↘$$↗$Minimum
$3$$45$$↗$$-$Maximum
On the closed interval $-3\le x\le 3$, the points $\left(-3,9\right)$ and $\left(1,-7\right)$ are relative minima and the points $\left(-2,20\right)$ and $\left(3,45\right)$ are relative maxima.
$\left(1,-7\right)$ is the lowest relative minimum, so it's the absolute minimum point, and $\left(3,45\right)$ 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
$f\left(x\right)={x}^{3}-3{x}^{2}+12$
What is the absolute maximum value of $f$ over the closed interval $\left[-2,4\right]$?

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 $f\left(x\right)=x$ 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 $g\left(x\right)=x{e}^{3x}$.
${g}^{\prime }\left(x\right)={e}^{3x}\left(1+3x\right)$, so our only critical point is $x=-\frac{1}{3}$.
Interval$x$-value${g}^{\prime }\left(x\right)$Verdict
$\left(-\mathrm{\infty },-\frac{1}{3}\right)$$x=-1$${g}^{\prime }\left(-1\right)=-\frac{2}{{e}^{3}}<0$$g$ is decreasing $↘$
$\left(-\frac{1}{3},\mathrm{\infty }\right)$$x=0$${g}^{\prime }\left(0\right)=1>0$$g$ is increasing $↗$
Let's imagine ourselves walking on the graph of $g$, starting all the way to the left (from $-\mathrm{\infty }$) and going all the way to the right (until $+\mathrm{\infty }$).
We will start by going down and down until we reach $x=-\frac{1}{3}$. Then, we will be forever going up. So $g$ has an absolute minimum point at $x=-\frac{1}{3}$. The function doesn't have an absolute maximum value.
Problem 1
$g\left(x\right)=\frac{\mathrm{ln}\left(x\right)}{x}$
What is the absolute maximum value of $g$ ?

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

## Want to join the conversation?

• Please, tell me What is the meaning of before and after in the above table and also whats the meaning of that table... Please Help me!
• The "before" simply refers to whether the function is increasing or decreasing BEFORE the x-value in question (e.g. -3). The "after" refers to whether the function is increasing or decreasing AFTER the x-value in question. The places where you see a dash (or what looks like a minus sign) are where it doesn't matter because the x-value is an endpoint (e.g. for -3 and +3, since these are the endpoints, what happens before -3 and what happens after +3 are irrelevant).
• Does x^3 - 3x^2 + 12 have a global maximum? For my sign chart, it shows that the function continues to increase as x approaches positive and negative infinity. How do I determine if there are global extrema based on the sign chart?
(1 vote)
• If the function continues to increase as x → +∞ (or -∞) then there can never be a maximum value and therefore no global maximum.

This is definitely true for x³ - 3x² + 12 and I believe will be true for any single variable polynomial with an odd degree ... in fact I don't think any such polynomial could ever have any global extrema (maximum or minimum) ...
• Why is this function defined for x=0?
• When we are looking at the critical points and the endpoints to compare them the table has a before and after columns. What is the before and after referring to? Before and after what?
• We need to determine whether the slope is positive or negative before and after the critical points and the end points. Hence, the comparison table helps to determine this.
• At the Finding absolute extrema on entire domain section in the first solution of the exercise, how is it possible that for g(x) = ln(x)/x, g'(x) is defined for all real numbers of g(x) when g'(x) = (1-ln(x))/x^2) is undefined for x=0 (i.e. dividing by 0^2)?.
• If the absolute value of a graph falls on a removable discontinuity or a hole does it exist? The function value does not exist as it is an open dot, but can it still be considered the absolute max/min?
(1 vote)
• No. In that case, there would be no extremum on that particular interval containing the discontinuity.

However, a special case can be made for something like f(x) = x^2 if x ≠ 0, -1 if x = 0, where a relative minimum does exist. So in general, if a function is undefined somewhere, you should still check for extrema.
• If we already have to calculate f(x) by substituting critical and end points, why do we need to find f'(x)?
• You need f’(x) to find the critical points in the first place. Critical points occur wherever f’(x) = 0 or is undefined.

Once you know the critical points, you can use them to determine the maxima and minima.
(1 vote)
• In the 1st example, how do you know the point (-3,9) is a minimum point?
(1 vote)
• It is given that -3 is the least x value that the function can take (in other words, for our purpose, the function starts at x = -3). Now, it has been shown that on the interval (-3,-2), the function is increasing. So, this implies that whatever value the function took at x = -3 was the minimum value in that interval (because after x = -3, the function increases and takes higher values). Hence, the point would be a minimum