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AP®︎/College Calculus AB
Course: AP®︎/College Calculus AB > Unit 5
Lesson 5: Using the candidates test to find absolute (global) extremaAbsolute minima & maxima review
AP.CALC:
FUN‑4 (EU)
, FUN‑4.A (LO)
, FUN‑4.A.3 (EK)
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.
Want to learn more about absolute extrema and differential calculus? Check out this video.
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 parenthesis, x, right parenthesis, equals, 2, x, cubed, plus, 3, x, squared, minus, 12, x over the interval minus, 3, is less than or equal to, x, is less than or equal to, 3.
h, prime, left parenthesis, x, right parenthesis, equals, 6, left parenthesis, x, plus, 2, right parenthesis, left parenthesis, x, minus, 1, right parenthesis, so our critical points are x, equals, minus, 2 and x, equals, 1. They divide the closed interval minus, 3, is less than or equal to, x, is less than or equal to, 3 into three parts:
| Interval | x-value | h, prime, left parenthesis, x, right parenthesis | Verdict |
|---|---|---|---|
| minus, 3, is less than, x, is less than, minus, 2 | x, equals, minus, start fraction, 5, divided by, 2, end fraction | h, prime, left parenthesis, minus, start fraction, 5, divided by, 2, end fraction, right parenthesis, equals, start fraction, 21, divided by, 2, end fraction, is greater than, 0 | h is increasing \nearrow |
| minus, 2, is less than, x, is less than, 1 | x, equals, 0 | h, prime, left parenthesis, 0, right parenthesis, equals, minus, 12, is less than, 0 | h is decreasing \searrow |
| 1, is less than, x, is less than, 3 | x, equals, 2 | h, prime, left parenthesis, 2, right parenthesis, equals, 24, is greater than, 0 | h is increasing \nearrow |
Now we look at the critical points and the endpoints of the interval:
| x | h, left parenthesis, x, right parenthesis | Before | After | Verdict |
|---|---|---|---|---|
| minus, 3 | 9 | minus | \nearrow | Minimum |
| minus, 2 | 20 | \nearrow | \searrow | Maximum |
| 1 | minus, 7 | \searrow | \nearrow | Minimum |
| 3 | 45 | \nearrow | minus | Maximum |
On the closed interval minus, 3, is less than or equal to, x, is less than or equal to, 3, the points left parenthesis, minus, 3, comma, 9, right parenthesis and left parenthesis, 1, comma, minus, 7, right parenthesis are relative minima and the points left parenthesis, minus, 2, comma, 20, right parenthesis and left parenthesis, 3, comma, 45, right parenthesis are relative maxima.
left parenthesis, 1, comma, minus, 7, right parenthesis is the lowest relative minimum, so it's the absolute minimum point, and left parenthesis, 3, comma, 45, right parenthesis 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.
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 parenthesis, x, right parenthesis, equals, 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 parenthesis, x, right parenthesis, equals, x, e, start superscript, 3, x, end superscript.
g, prime, left parenthesis, x, right parenthesis, equals, e, start superscript, 3, x, end superscript, left parenthesis, 1, plus, 3, x, right parenthesis, so our only critical point is x, equals, minus, start fraction, 1, divided by, 3, end fraction.
| Interval | x-value | g, prime, left parenthesis, x, right parenthesis | Verdict |
|---|---|---|---|
| left parenthesis, minus, infinity, comma, minus, start fraction, 1, divided by, 3, end fraction, right parenthesis | x, equals, minus, 1 | g, prime, left parenthesis, minus, 1, right parenthesis, equals, minus, start fraction, 2, divided by, e, cubed, end fraction, is less than, 0 | g is decreasing \searrow |
| left parenthesis, minus, start fraction, 1, divided by, 3, end fraction, comma, infinity, right parenthesis | x, equals, 0 | g, prime, left parenthesis, 0, right parenthesis, equals, 1, is greater than, 0 | g is increasing \nearrow |
Let's imagine ourselves walking on the graph of g, starting all the way to the left (from minus, infinity) and going all the way to the right (until plus, infinity).
We will start by going down and down until we reach x, equals, minus, start fraction, 1, divided by, 3, end fraction. Then, we will be forever going up. So g has an absolute minimum point at x, equals, minus, start fraction, 1, divided by, 3, end fraction. The function doesn't have an absolute maximum value.
Want to learn more about absolute extrema over entire domain? Check out this video.
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!(1 vote)
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).(10 votes)
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.
ADDENDUM:
This is definitely true forx³ - 3x² + 12and 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) ...(6 votes)
- Why is this function defined for x=0?(3 votes)
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?(0 votes)
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.(7 votes)
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.(3 votes)
- Thank you for this helpful summary.
I noticed that when finding global extrema on a closed interval, once you know the critical points and endpoints that are candidates for being global extrema, you just have to plug those values into the original function along with the endpoints on the closed interval. You do not have to actually evaluate the derivative in between each critical point (although this would further inform your answer from an intuitive point of view).
My question is whether it is completely necessary to always evaluate the derivative in between critical and end points 'when finding global extrema' or whether you can just rely on the output of those values when plugged into the initial function. (This way, the critical/end point with the highest output for the original function is the global maximum... etc.) I think this only works on closed intervals with endpoints.
Thank you in advance(2 votes) 
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(3 votes)
what if the lowest value within the interval is a relative maximum, wouldn't that be the abs. minimum of the interval? because "absolute minimum" is defined to be the "lowest possible value" within something.(1 vote)- If a point is a relative maximum, that means there are nearby points with lower y-values, so the absolute minimum cannot be a relative maximum. Likewise, the absolute maximum cannot be a relative minimum.(2 votes)
So for the both the absolute and relative extrema, you would test x-values around the critical points of the FIRST DERIVATIVE to determine an increasing or decreasing value. But for absolute extrema, you must also plug in the endpoints into the ORIGINAL equation and that also determines if the function is increasing or decreasing?(0 votes)- You've almost got it. When testing the endpoints, you don't strictly need to test if the function is increasing or decreasing there. If it's a factor, your analysis of the critical points will catch those relative extrema. The concern with endpoints is that the function might be increasing towards some critical value out of the domain in question, and so could be larger than any of the local extrema in the domain without being a critical point.(4 votes)
- 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)?.(1 vote)
