Main content

## Differential Calculus

### Course: Differential Calculus > Unit 5

Lesson 4: Relative (local) extrema- Introduction to minimum and maximum points
- Finding relative extrema (first derivative test)
- Worked example: finding relative extrema
- Analyzing mistakes when finding extrema (example 1)
- Analyzing mistakes when finding extrema (example 2)
- Finding relative extrema (first derivative test)
- Relative minima & maxima
- Relative minima & maxima review

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# 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:

**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:

Interval | Test x-value | f, prime, left parenthesis, x, right parenthesis | Conclusion |
---|---|---|---|

left parenthesis, minus, infinity, comma, 0, right parenthesis | x, equals, minus, 1 | f, prime, left parenthesis, minus, 1, right parenthesis, equals, 0, point, 75, is greater than, 0 | f is increasing \nearrow |

left parenthesis, 0, comma, 1, right parenthesis | x, equals, 0, point, 5 | f, prime, left parenthesis, 0, point, 5, right parenthesis, equals, minus, 3, is less than, 0 | f is decreasing \searrow |

left parenthesis, 1, comma, 2, right parenthesis | x, equals, 1, point, 5 | f, prime, left parenthesis, 1, point, 5, right parenthesis, equals, minus, 3, is less than, 0 | f is decreasing \searrow |

left parenthesis, 2, comma, infinity, right parenthesis | x, equals, 3 | f, prime, left parenthesis, 3, right parenthesis, equals, 0, point, 75, is greater than, 0 | f 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

*f, prime changes signs.***and**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.

### 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.

### 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.

### 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

*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."(11 votes)
- 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.(1 vote)

- How do you get the "Test Value" #s?(2 votes)
- 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.(4 votes)

- 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)- 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)

- 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)(2 votes)

- In problem 3, would x = -1 and x = -1 be the relative minimum points?(1 vote)
- 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)

- 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)
- 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)

- 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!(0 votes)- 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)(0 votes)