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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

# Relative minima & maxima review

AP.CALC:
FUN‑4 (EU)
,
FUN‑4.A (LO)
,
FUN‑4.A.2 (EK)
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.

## 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.
Intervalx-valuef, prime, left parenthesis, x, right parenthesisVerdict
x, is less than, minus, 3x, equals, minus, 4f, prime, left parenthesis, minus, 4, right parenthesis, equals, 15, is greater than, 0f is increasing. \nearrow
minus, 3, is less than, x, is less than, 1x, equals, 0f, prime, left parenthesis, 0, right parenthesis, equals, minus, 9, is less than, 0f is decreasing. \searrow
x, is greater than, 1x, equals, 2f, prime, left parenthesis, 2, right parenthesis, equals, 15, is greater than, 0f is increasing. \nearrow
Now let's look at the critical points:
xBeforeAfterVerdict
minus, 3\nearrow\searrowMaximum
1\searrow\nearrowMinimum
In conclusion, the function has a maximum point at x, equals, minus, 3 and a minimum point at x, equals, 1.

Problem 1
• Current
h, left parenthesis, x, right parenthesis, equals, minus, x, cubed, plus, 3, x, squared, minus, 4
For what value of x does h have a relative maximum ?

Want to try more problems like this? Check out this exercise.

## 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
• 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))
• 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
• what if there are 2 variables F(x,y) = x^3 +y^2 -xy +x
• Then you're doing multivariable calculus, which is covered in a different section of Khan Academy. You can find it in the Courses dropdown.
• 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
• You find all of the relative maxima, nondifferentiable points, and endpoints of the intervals, and see which one gives the highest function value.
• How do you solve for x^2/(x^2 -1)
• 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
• 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.
• 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.
(1 vote)
• ) f(x) = 12x5 – 45x4 + 40x3 + 5. Find the value of x for which the curve shows relative maxima & relative minima
(1 vote)
• 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! :)