Main content

## AP®︎/College Calculus AB

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

Lesson 4: Using the first derivative test to find 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

# Relative minima & maxima review

Review how we use differential calculus to find relative extremum (minimum and maximum) points.

## How do I find relative minimum & maximum points with differential calculus?

A relative

**maximum**point is a point where the function changes direction from**increasing**to**decreasing**(making that point a "peak" in the graph).Similarly, a relative

**minimum**point is a point where the function changes direction from**decreasing**to**increasing**(making that point a "bottom" in the graph).Supposing you already know how to find increasing & decreasing intervals of a function, finding relative extremum points involves one more step: finding the points where the function changes direction.

*Want to learn more about relative extrema and differential calculus? Check out this video.*

## Example

Let's find the relative extremum points of f, left parenthesis, x, right parenthesis, equals, x, cubed, plus, 3, x, squared, minus, 9, x, plus, 7. First, we differentiate f:

f, prime, left parenthesis, x, right parenthesis, equals, 3, left parenthesis, x, plus, 3, right parenthesis, left parenthesis, x, minus, 1, right parenthesis

Our critical points are x, equals, minus, 3 and x, equals, 1.

Let's evaluate f, prime at each interval to see if it's positive or negative on that interval.

Interval | x-value | f, prime, left parenthesis, x, right parenthesis | Verdict |
---|---|---|---|

x, is less than, minus, 3 | x, equals, minus, 4 | f, prime, left parenthesis, minus, 4, right parenthesis, equals, 15, is greater than, 0 | f is increasing. \nearrow |

minus, 3, is less than, x, is less than, 1 | x, equals, 0 | f, prime, left parenthesis, 0, right parenthesis, equals, minus, 9, is less than, 0 | f is decreasing. \searrow |

x, is greater than, 1 | x, equals, 2 | f, prime, left parenthesis, 2, right parenthesis, equals, 15, is greater than, 0 | f is increasing. \nearrow |

Now let's look at the critical points:

x | Before | After | Verdict |
---|---|---|---|

minus, 3 | \nearrow | \searrow | Maximum |

1 | \searrow | \nearrow | Minimum |

**In conclusion, the function has a maximum point at x, equals, minus, 3 and a minimum point at x, equals, 1.**

## Want to join the conversation?

- (x−3)^2(1)+(x−2)[2(x−3)]

= (x-3)^2+((x-2)2)(x-3)

= (x-3)^2+(2x-4)(x-3)

.... then I am lost. how do I get from here to ...

=(x−3)(x−3+2x−4) ??

Thank you(13 votes)- Think of this equation: x^2 + 2x. You can split the x^2 term into the individual x's and reverse the distributive property. You end up with x(x+2). The same can be done with your equation.

(x-3)^2 = (x-3)(x-3)

(x-3)(x-3) + (2x-4)(x-3)

(x-3)((x-3) + (2x-4))(16 votes)

- I don understand how to do it without a graph.(1 vote)
- If you don't want to hear me take a long winded explanation of why it works then just skip.

If you think of maximum, it's like a hill. Say you can only climb the hill form the left to the right. If you're beginning to climb it, it's sloping up. But once you reach the top, it will start sloping down. And Since it must go from sloping up to sloping down in a continuous fashion the top point must have a slope of 0. (vice versa for minimum)

First you take the derivative of an arbitrary function f(x). So now you have f'(x). Find all the x values for which f'(x) = 0 and list them down. So say the function f'(x) is 0 at the points x1,x2 and x3. Now test the points in between the points and if it goes from + to 0 to - then its a maximum and if it goes from - to 0 to + its a minimum(32 votes)

- what if there are 2 variables F(x,y) = x^3 +y^2 -xy +x(3 votes)
- Then you're doing multivariable calculus, which is covered in a different section of Khan Academy. You can find it in the Courses dropdown.(7 votes)

- so if f'(x) goes from + to - at x=c then f(c) is an maximum but how do i know if it is a relative or absolute max(3 votes)
- You find all of the relative maxima, nondifferentiable points, and endpoints of the intervals, and see which one gives the highest function value.(3 votes)

- How do you solve for x^2/(x^2 -1)(3 votes)
- you can add and subtract 1 from numerator, so we will get:

x^2 + 1 - 1/(x^2-1) => x^2-1/x^2-1 + 1/x^2-1 => 1 + 1/x^2-1

you can solve it further given what the equation is equated to, whether it is 0 or other(3 votes)

- Values less than one are increasing (becoming less negative) and not decreasing, therefore there should only be one point with that is a retaliative minimum, that point being when x=1.(3 votes)
- Remember that the derivative is the slope at any given point. If the slope is less than one (negative) the graph is decreasing at that point.(2 votes)

- When we're doing the number line, what if it is negative before and after the point? Does that mean it is just continuously decreasing?(2 votes)
- I'm pretty sure that means there are no extreme values(4 votes)

- How would we calculate the local minimum and maximum of a function defined as min{abs(x-1), x}?(2 votes)
- Let 𝑓(𝑥) = min(|𝑥 − 1|, 𝑥)

What I did was to first write |𝑥 − 1| as a piecewise function:

1 − 𝑥 for 𝑥 < 1

𝑥 − 1 for 𝑥 ≥ 1

Then I found out when this function is less than 𝑥.

1 − 𝑥 < 𝑥 ⇒ 𝑥 > 1∕2

𝑥 − 1 < 𝑥 ⇒ −1 < 0, which is true for all 𝑥

That way, I could write 𝑓(𝑥) as a piecewise function:

𝑥 for 𝑥 ≤ 1∕2

1 − 𝑥 for 𝑥 ∈ (1∕2, 1)

𝑥 − 1 for 𝑥 ≥ 1

Both 𝑥 and |𝑥 − 1| are continuous and thereby 𝑓(𝑥) is also continuous.

Finally, I realized that

𝑥 has a positive slope,

1 − 𝑥 has a negative slope

and 𝑥 − 1 has a positive slope

which means that 𝑓(𝑥) has a relative maximum at 𝑥 = 1∕2

and a relative minimum at 𝑥 = 1(3 votes)

- ) f(x) = 12x5 – 45x4 + 40x3 + 5. Find the value of x for which the curve shows relative maxima & relative minima(2 votes)
- This is really simple if you watched videos. Find the first derivative of a function f(x) and find the critical numbers. Then, find the second derivative of a function f(x) and put the critical numbers. If the value is negative, the function has relative maxima at that point, if the value is positive, the function has relative maxima at that point. This is the Second Derivative Test. However, if you get 0, you have to use the First Derivative Test. Just find the first derivative of a function f(x) and critical numbers. Then, divide the domain (all real numbers) by the critical numbers. For example, if the critical numbers are -1, 4, 5, you should get 4 different domains which are (-∞, -1), (-1, 4), (4, 5), (5, ∞). Then, input any number inside each domain. If the value is negative and the value of next domain is positive, it has relative minima at that point and vice versa. For example, if you got a negative value in the interval (-1, 4) and positive value in the interval (4, 5), the function has relative minima at point 4.

Hope this helps! If you have any questions or need help, please ask! :)(3 votes)

- given the function f(x) = x^3+ax^2+bx+12, when x=3 there is a relative minimum value of -15. solve for a and b(2 votes)
- You know that plugging in x=3 will give a value of -15, so that gives you an equation in a and b.

If you take the derivative of f and plug in x=3, you know it will give 0, since you have a relative minimum. This gives you a second equation in a and b. Solve the system for a and b.(3 votes)