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## AP®︎/College Calculus AB

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

Lesson 4: Using the first derivative test to find relative (local) extrema

# Finding relative extrema (first derivative test)

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 $f(x)=\dfrac{x^2}{x-1}$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?

### 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?

### 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?

### 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 ?

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 ?

Want more practice? Try this exercise.

## Want to join the conversation?

• 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." • How do you get the "Test Value" #s? • 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. • My math teacher said something about first derivative proves increase/decrease, while first derivative test proves critical points, and that we should mention either first derivative or first derivative test while writing statements. Is this right, or have I mixed them up? Same with second derivative- plain derivative proves concavity, while the second derivative test proves inflection points. Thank you!
(1 vote) • You seem to have gotten the second derivative stuff mixed up, so I'll just correct them (You've also missed some key terms in the first derivative)

The first derivative proves the function's increase/decrease (if the first derivative is positive, the function is increasing and vice versa). You're right on the test though. The first derivative test is indeed used to prove the existence of critical points.

The second derivative itself doesn't prove concavity. Like the first derivative, the second derivative proves the first derivative's increase/decrease (if the second derivative is positive, the first derivative is increasing and vice versa). The second derivative test is used to find potential points of change in concavity (inflection points). To prove whether or not the point is actually an inflection point, you can do two things:

1. Check if the second derivative changes signs before and after the inflection point

2. See the third derivative (which isn't really required here. You can stick with the first method)
• In problem 3, would x = -1 and x = -1 be the relative minimum points?
(1 vote) • 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) • "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) • 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! 