How do you get the "Test Value" #s?

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.

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.

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.

Now I'm a bit confused about labeling and names are extrema, extremum, maxima, and maximum the same. If they're the same then why do we take all these names? The same goes for minimum and minima.

maximum: the point at which the function is the largest minimum: the point at which the function is the smallest extremum: a maximum or minimum maxima, minima, extrema: the plural of maximum, minimum, and extremum respectively

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!

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)

Main content

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)

Google Classroom

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 $f (x) = \frac{x^{2}}{x - 1}$ ‍

Step 1: Finding $f^{'} (x)$ ‍

To find the relative extremum points of

f

, we must use

f^{'}

. So we start with differentiating

f

f^{'} (x) = \frac{x^{2} - 2 x}{(x - 1)^{2}}

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^{'} (x) = 0

or where

f^{'}

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^{'}

must stay the same between two consecutive points.

In our case, these points are

x = 0

x = 1

, and

x = 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:

Interval	Test $x$ ‍-value	$f^{'} (x)$ ‍	Conclusion
$(- \infty, 0)$ ‍	$x = - 1$ ‍	$f^{'} (- 1) = 0.75 > 0$ ‍	$f$ ‍ is increasing $↗$ ‍
$(0, 1)$ ‍	$x = 0.5$ ‍	$f^{'} (0.5) = - 3 < 0$ ‍	$f$ ‍ is decreasing $↘$ ‍
$(1, 2)$ ‍	$x = 1.5$ ‍	$f^{'} (1.5) = - 3 < 0$ ‍	$f$ ‍ is decreasing $↘$ ‍
$(2, \infty)$ ‍	$x = 3$ ‍	$f^{'} (3) = 0.75 > 0$ ‍	$f$ ‍ is increasing $↗$ ‍

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^{'}

changes signs.

In our case:

$f$ ‍ increases before $x = 0$ ‍, decreases after it, and is defined at $x = 0$ ‍. So $f$ ‍ has a relative maximum point at $x = 0$ ‍.
$f$ ‍ decreases before $x = 2$ ‍, increases after it, and is defined at $x = 2$ ‍. So $f$ ‍ has a relative minimum point at $x = 2$ ‍.
$f$ ‍ is undefined at $x = 1$ ‍, so it doesn't have an extremum point there.

Problem 1

Jason was asked to find where

f (x) = 2 x^{3} + 18 x^{2} + 54 x + 50

has a relative extremum. This is his solution:

Step 1:

f^{'} (x) = 6 (x + 3)^{2}

Step 2: The solution of

f^{'} (x) = 0

x = - 3

Step 3:

f

has a relative extremum at

x = - 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 (x) = (x^{2} - 1)^{2 / 3}

has a relative maximum. This is her solution:

Step 1:

g^{'} (x) = \frac{4 x}{3 \sqrt[3]{x^{2} - 1}}

Step 2: The critical point is

x = 0

Step 3:

Interval	Test $x$ ‍-value	$g^{'} (x)$ ‍	Verdict
$(- \infty, 0)$ ‍	$x = - 3$ ‍	$g^{'} (- 3) = - 2 < 0$ ‍	$g$ ‍ is decreasing $↘$ ‍
$(0, \infty)$ ‍	$x = 3$ ‍	$g^{'} (3) = 2 > 0$ ‍	$g$ ‍ is increasing $↗$ ‍

Step 4:

g

decreases before

x = 0

and increases after, so there is a relative minimum at

x = 0

and no relative maximum.

Is Erin's work correct? If not, what's her mistake?

(Choice A)
Erin's work is correct.
(Choice B)
Step 2 is incorrect. Erin should have also looked for points where $g$ ‍ or $g^{'}$ ‍ are undefined.
(Choice C)
Step 3 is incorrect. $g^{'} (- 3)$ ‍ is positive so $g$ ‍ actually increases before $x = 0$ ‍.
(Choice D)
Step 3 is incorrect. Erin should have evaluated $g (3)$ ‍ instead of $g^{'} (3)$ ‍.

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 (x) = x^{2} + \frac{1}{x^{2}}

has a relative maximum. This is his solution:

Step 1:

h^{'} (x) = \frac{2 (x^{4} - 1)}{x^{3}}

Step 2: The critical points are

x = - 1

and

x = 1

, and

h

is undefined at

x = 0

Step 3:

Interval	Test $x$ ‍-value	$h^{'} (x)$ ‍	Verdict
$(- \infty, - 1)$ ‍	$x = - 2$ ‍	$h^{'} (- 2) = - 3.75 < 0$ ‍	$h$ ‍ is decreasing $↘$ ‍
$(- 1, 0)$ ‍	$x = - 0.5$ ‍	$h^{'} (- 0.5) = 15 > 0$ ‍	$h$ ‍ is increasing $↗$ ‍
$(0, 1)$ ‍	$x = 0.5$ ‍	$h^{'} (0.5) = - 15 < 0$ ‍	$h$ ‍ is decreasing $↘$ ‍
$(1, \infty)$ ‍	$x = 2$ ‍	$h^{'} (2) = 3.75 > 0$ ‍	$h$ ‍ is increasing $↗$ ‍

Step 4:

h

increases before

x = 0

and decreases after it, so

h

has a maximum point at

x = 0

Is Jake's work correct? If not, what's his mistake?

(Choice A)
Jake's work is correct.
(Choice B)
Step 1 is incorrect. Jake didn't differentiate $h$ ‍ correctly.
(Choice C)
Step 2 is incorrect. There are more critical points that he didn't find.
(Choice D)
Step 4 is incorrect. $x = 0$ ‍ isn't in the domain of $h$ ‍ so it cannot be an extremum.

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 (x) = x^{3} + 6 x^{2} - 15 x + 2

For what value of $x$ ‍ does $f$ ‍ have a relative maximum ?

Problem 5

Let

g

be a polynomial function and let

g^{'}

, its derivative, be defined as

g^{'} (x) = x (x + 2) (x + 4)^{2}

At how many points does the graph of $g$ ‍ have a relative maximum ?

Want more practice? Try this exercise.

Want to join the conversation?

Sort by:

RihamSoliman
Posted 6 years ago. Direct link to RihamSoliman's post “I see Matthew already mad...”
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."
Button navigates to signup pageComment on RihamSoliman's post “I see Matthew already mad...”
(12 votes)
Answer
- Gitanjali
  Posted a year ago. Direct link to Gitanjali's post “critical points are when ...”
  critical points are when the value of f'(x) is either 0 or does not exist.
  
  In the example above, x = 0 and x = 2 are cases when f'(x) = 0 whereas for x = 1, the value of f'(x) = -1/0 which does not exist.
  
  Hence 0, 1, and 2 are the critical points.
  Comment on Gitanjali's post “critical points are when ...”
  (0 votes)
Nikki Thomas
Posted 6 years ago. Direct link to Nikki Thomas's post “How do you get the "Test ...”
How do you get the "Test Value" #s?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Miller, Kaleb
  Posted 5 years ago. Direct link to Miller, Kaleb's post “You look at the interval ...”
  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.
  Button navigates to signup page
  (5 votes)
tdavidpearce
Posted 4 years ago. Direct link to tdavidpearce's post “Can it be a "critical" po...”
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.
Button navigates to signup pageComment on tdavidpearce's post “Can it be a "critical" po...”
(2 votes)
Answer
- Martin
  Posted 4 years ago. Direct link to Martin's post “A critical point is a poi...”
  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.
  Button navigates to signup page
  (2 votes)
Dima Escaroda
Posted 4 years ago. Direct link to Dima Escaroda's post “What does this phrase mea...”
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?
Button navigates to signup pageComment on Dima Escaroda's post “What does this phrase mea...”
(2 votes)
Answer
zainrajab5
Posted 6 months ago. Direct link to zainrajab5's post “Now I'm a bit confused ab...”
Now I'm a bit confused about labeling and names
are extrema, extremum, maxima, and maximum the same.
If they're the same then why do we take all these names?
The same goes for minimum and minima.
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- KLaudano
  Posted 6 months ago. Direct link to KLaudano's post “maximum: the point at whi...”
  maximum: the point at which the function is the largest
  
  minimum: the point at which the function is the smallest
  
  extremum: a maximum or minimum
  
  maxima, minima, extrema: the plural of maximum, minimum, and extremum respectively
  Button navigates to signup page
  (3 votes)
Jenna Kim
Posted a year ago. Direct link to Jenna Kim's post “My math teacher said some...”
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!
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Venkata
  Posted a year ago. Direct link to Venkata's post “You seem to have gotten t...”
  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)
  Button navigates to signup page
  (3 votes)
BijanS
Posted a month ago. Direct link to BijanS's post “Ok in the previous Lesson...”
Ok in the previous Lesson, in one of the examples it said "If the function is not defined at a particular point, then that point cannot be a critical point". Now in this example 1, he contradicts that by saying x=1 is a critical point...?
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(1 vote)
Answer
- Pirie
  Posted a month ago. Direct link to Pirie's post “The reason x=1 was put as...”
  The reason x=1 was put as a critical point is because f' is undefined at that point. However, it later notes how x=1 is not an extremum point because f is not defined there. You are right though that x=1 is not a critical point it just isn't proven when the critical points are listed.
  Button navigates to signup page
  (1 vote)
paperangel220
Posted 3 years ago. Direct link to paperangel220's post “In problem 3, would x = -...”
In problem 3, would x = -1 and x = -1 be the relative minimum points?
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Hecretary Bird
  Posted 3 years ago. Direct link to Hecretary Bird's post “Yes. Whenever your functi...”
  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.
  Button navigates to signup page
  (1 vote)
Avin
Posted a month ago. Direct link to Avin's post “I'm still a bit confused....”
I'm still a bit confused. I've read through other questions posted regarding the same problem, though I still don't understand what the statement, "the sign of f' must stay the same between 2 consecutive critical points" means.
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(1 vote)
Answer
zhiyi li
Posted 5 years ago. Direct link to zhiyi li's post “"x=1 is critical points" ...”
"x=1 is critical points" from "In our case, the critical points are x=0, x=1, and x=2." makes me confuse.
Button navigates to signup pageComment on zhiyi li's post “"x=1 is critical points" ...”
(1 vote)
Answer
- Ajay Zutshi
  Posted 5 years ago. Direct link to Ajay Zutshi's post “Yes, but a critical point...”
  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.
  Button navigates to signup page
  (1 vote)

AP®︎/College Calculus AB

Course: AP®︎/College Calculus AB > Unit 5

Example: finding the relative extremum points of f(x)=x2x−1‍

Common mistake: not checking the critical points

Common mistake: not including points where the derivative is undefined

Common mistake: forgetting to check the domain of the function

Practice applying the first derivative test

Want to join the conversation?

Example: finding the relative extremum points of $f (x) = \frac{x^{2}}{x - 1}$ ‍