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Finding relative extrema (first derivative test)

AP.CALC:
FUN‑4 (EU)
,
FUN‑4.A (LO)
,
FUN‑4.A.2 (EK)
The first derivative test is the process of analyzing functions using their first derivatives in order to find their extremum point. This involves multiple steps, so we need to unpack this process in a way that helps avoiding harmful omissions or mistakes.
What if we told you that given the equation of the function, you can find all of its maximum and minimum points? Well, it's true! This process is called the first derivative test. Let's unpack it in a way that helps avoiding harmful omissions or mistakes.

Example: finding the relative extremum points of f, left parenthesis, x, right parenthesis, equals, start fraction, x, squared, divided by, x, minus, 1, end fraction

Step 1: Finding f, prime, left parenthesis, x, right parenthesis
To find the relative extremum points of f, we must use f, prime. So we start with differentiating f:
f, prime, left parenthesis, x, right parenthesis, equals, start fraction, x, squared, minus, 2, x, divided by, left parenthesis, x, minus, 1, right parenthesis, squared, end fraction
Step 2: Finding all critical points and all points where f is undefined.
The critical points of a function f are the x-values, within the domain of f for which f, prime, left parenthesis, x, right parenthesis, equals, 0 or where f, prime is undefined. In addition to those, we should look for points where the function f itself is undefined.
The important thing about these points is that the sign of f, prime must stay the same between two consecutive points.
In our case, these points are x, equals, 0, x, equals, 1, and x, equals, 2.
Step 3: Analyzing intervals of increase or decrease
This can be done in many ways, but we like using a sign chart. In a sign chart, we pick a test value at each interval that is bounded by the points we found in Step 2 and check the derivative's sign on that value.
This is the sign chart for our function:
IntervalTest x-valuef, prime, left parenthesis, x, right parenthesisConclusion
left parenthesis, minus, infinity, comma, 0, right parenthesisx, equals, minus, 1f, prime, left parenthesis, minus, 1, right parenthesis, equals, 0, point, 75, is greater than, 0f is increasing \nearrow
left parenthesis, 0, comma, 1, right parenthesisx, equals, 0, point, 5f, prime, left parenthesis, 0, point, 5, right parenthesis, equals, minus, 3, is less than, 0f is decreasing \searrow
left parenthesis, 1, comma, 2, right parenthesisx, equals, 1, point, 5f, prime, left parenthesis, 1, point, 5, right parenthesis, equals, minus, 3, is less than, 0f is decreasing \searrow
left parenthesis, 2, comma, infinity, right parenthesisx, equals, 3f, prime, left parenthesis, 3, right parenthesis, equals, 0, point, 75, is greater than, 0f is increasing \nearrow
Step 4: Finding extremum points
Now that we know the intervals where f increases or decreases, we can find its extremum points. An extremum point would be a point where f is defined and f, prime changes signs.
In our case:
  • f increases before x, equals, 0, decreases after it, and is defined at x, equals, 0. So f has a relative maximum point at x, equals, 0.
  • f decreases before x, equals, 2, increases after it, and is defined at x, equals, 2. So f has a relative minimum point at x, equals, 2.
  • f is undefined at x, equals, 1, so it doesn't have an extremum point there.
Problem 1
Jason was asked to find where f, left parenthesis, x, right parenthesis, equals, 2, x, cubed, plus, 18, x, squared, plus, 54, x, plus, 50 has a relative extremum. This is his solution:
Step 1: f, prime, left parenthesis, x, right parenthesis, equals, 6, left parenthesis, x, plus, 3, right parenthesis, squared
Step 2: The solution of f, prime, left parenthesis, x, right parenthesis, equals, 0 is x, equals, minus, 3.
Step 3: f has a relative extremum at x, equals, minus, 3.
Is Jason's work correct? If not, what's his mistake?
Choose 1 answer:
Choose 1 answer:

Common mistake: not checking the critical points

Remember: We must not assume that any critical point is an extremum. Instead, we should check our critical points to see if the function is defined at those points and the derivative changes signs at those points.
Problem 2
Erin was asked to find if g, left parenthesis, x, right parenthesis, equals, left parenthesis, x, squared, minus, 1, right parenthesis, start superscript, 2, slash, 3, end superscript has a relative maximum. This is her solution:
Step 1: g, prime, left parenthesis, x, right parenthesis, equals, start fraction, 4, x, divided by, 3, cube root of, x, squared, minus, 1, end cube root, end fraction
Step 2: The critical point is x, equals, 0.
Step 3:
IntervalTest x-valueg, prime, left parenthesis, x, right parenthesisVerdict
left parenthesis, minus, infinity, comma, 0, right parenthesisx, equals, minus, 3g, prime, left parenthesis, minus, 3, right parenthesis, equals, minus, 2, is less than, 0g is decreasing \searrow
left parenthesis, 0, comma, infinity, right parenthesisx, equals, 3g, prime, left parenthesis, 3, right parenthesis, equals, 2, is greater than, 0g is increasing \nearrow
Step 4: g decreases before x, equals, 0 and increases after, so there is a relative minimum at x, equals, 0 and no relative maximum.
Is Erin's work correct? If not, what's her mistake?
Choose 1 answer:
Choose 1 answer:

Common mistake: not including points where the derivative is undefined

Remember: When we analyze increasing and decreasing intervals, we must look for all points where the derivative is equal to zero and all points where the function or its derivative are undefined. If you miss any of these points, you will probably end up with a wrong sign chart.
Problem 3
Jake was asked to find whether h, left parenthesis, x, right parenthesis, equals, x, squared, plus, start fraction, 1, divided by, x, squared, end fraction has a relative maximum. This is his solution:
Step 1: h, prime, left parenthesis, x, right parenthesis, equals, start fraction, 2, left parenthesis, x, start superscript, 4, end superscript, minus, 1, right parenthesis, divided by, x, cubed, end fraction
Step 2: The critical points are x, equals, minus, 1 and x, equals, 1, and h is undefined at x, equals, 0.
Step 3:
IntervalTest x-valueh, prime, left parenthesis, x, right parenthesisVerdict
left parenthesis, minus, infinity, comma, minus, 1, right parenthesisx, equals, minus, 2h, prime, left parenthesis, minus, 2, right parenthesis, equals, minus, 3, point, 75, is less than, 0h is decreasing \searrow
left parenthesis, minus, 1, comma, 0, right parenthesisx, equals, minus, 0, point, 5h, prime, left parenthesis, minus, 0, point, 5, right parenthesis, equals, 15, is greater than, 0h is increasing \nearrow
left parenthesis, 0, comma, 1, right parenthesisx, equals, 0, point, 5h, prime, left parenthesis, 0, point, 5, right parenthesis, equals, minus, 15, is less than, 0h is decreasing \searrow
left parenthesis, 1, comma, infinity, right parenthesisx, equals, 2h, prime, left parenthesis, 2, right parenthesis, equals, 3, point, 75, is greater than, 0h is increasing \nearrow
Step 4: h increases before x, equals, 0 and decreases after it, so h has a maximum point at x, equals, 0.
Is Jake's work correct? If not, what's his mistake?
Choose 1 answer:
Choose 1 answer:

Common mistake: forgetting to check the domain of the function

Remember: After we've found points where the function changes its direction, we must check whether the function is defined at those points. Otherwise, this isn't a relative extremum.

Practice applying the first derivative test

Problem 4
Let f, left parenthesis, x, right parenthesis, equals, x, cubed, plus, 6, x, squared, minus, 15, x, plus, 2.
For what value of x does f have a relative maximum ?
Choose 1 answer:
Choose 1 answer:

Problem 5
Let g be a polynomial function and let g, prime, its derivative, be defined as g, prime, left parenthesis, x, right parenthesis, equals, x, left parenthesis, x, plus, 2, right parenthesis, left parenthesis, x, plus, 4, right parenthesis, squared.
At how many points does the graph of g have a relative maximum ?
Choose 1 answer:
Choose 1 answer:

Want more practice? Try this exercise.

Want to join the conversation?

  • female robot grace style avatar for user RihamSoliman
    I see Matthew already made a point about "the sign of f' must stay the same between 2 consecutive critical points". But can someone confirm whether x=1 is a critical point in the example where f(x)=x^2 / (x−1)? From what I understand if a particular x is undefined in both f(x) and f'(x) then it is not considered a critical point for the function. But the example states explicitly "In our case, the critical points are x=0, x=1, and x=2."
    (8 votes)
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  • female robot ada style avatar for user Mr. K
    Is there a typo in problem 2? In the first line of the table, shouldn't there be g'(-3) = -2 < 0 instead of g'(3) = -2 < 0?
    (2 votes)
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  • blobby green style avatar for user Nikki Thomas
    How do you get the "Test Value" #s?
    (2 votes)
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    • male robot hal style avatar for user Miller, Kaleb
      You look at the interval say (0,2) then you pick a number that is within the interval, not 0 or 2 but between 0 and 2. In this case you should choose 1 because that is the easiest number to plug back into the equation and see if that interval is increasing or decreasing.
      (3 votes)
  • leaf green style avatar for user tdavidpearce
    Can it be a "critical" point without being defined in the original function?

    i.e.
    Does this requirement of being defined in the original function apply to extremum only or critical points as well?

    I don't recall seeing a Sal-signature intuition-explanation for this rule. Did I miss it? Video suggestion?

    I could see how an "extreme" needs to be part of the function itself, but a critical point could just be describing the function's behavior around that point.
    (2 votes)
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    • aqualine ultimate style avatar for user Martin
      A critical point is a point in a function where the derivative either equals zero or doesn't exist.
      So if a point isn't defined by the function it cannot be a critical point because for the function this point doesn't exist, so talking about the derivative wouldn't make much sense.
      (2 votes)
  • duskpin ultimate style avatar for user -PaperAngel-
    In problem 3, would x = -1 and x = -1 be the relative minimum points?
    (1 vote)
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    • piceratops ultimate style avatar for user Hecretary Bird
      Yes. Whenever your function changes from decreasing to increasing, or when your first derivative changes from negative to positive, you have a relative minimum (and vice versa for relative maximums). This is true for x = -1 and x = 1, so both of them are relative minimums.
      (1 vote)
  • blobby green style avatar for user Etienne
    This confuses me, I have a few questions:

    1) In the first part (step 2) what it means "The important thing about these points is that the sign of f' must stay the same between two consecutive points."

    2) Why, since we have points where the function f is undefined, we can still differentiate the function? (and how are we able to tell if it's decreasing or increasing after an undefined point of f).

    Thanks in advance for the help!
    (1 vote)
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    • hopper cool style avatar for user JPOgle ✝
      1) A function that is continuous on its domain cannot change sign unless it equals 0 or is undefined.(Critical points and points on the original function that are undefined)
      2) We cannot differentiate where a function is undefined. However, we can differentiate at the points around the undefined point. (Unless they are undefined)
      (1 vote)
  • blobby green style avatar for user Dima Escaroda
    What does this phrase mean? (step 2 of 1st example) "The important thing about these points is that the sign of f' must stay the same between two consecutive points". Aren't we talking about relative extrema where "f' changes signs"? I'm confused a bit. Am I missing something or it's line of text that is wrong?
    (1 vote)
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  • blobby green style avatar for user zhiyi li
    "x=1 is critical points" from "In our case, the critical points are x=0, x=1, and x=2." makes me confuse.
    (1 vote)
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    • blobby green style avatar for user Ajay Zutshi
      Yes, but a critical point must be in the domain of the function. Strictly speaking, x=1 is NOT a critical point for this function b/c x=1 is NOT in the domain of the function.

      f(c) must be defined for x=c to be a critical point. In this case f(1) is undefined so it should not be a critical point.
      (1 vote)