Finding locations of absolute extrema

[Mathamateur]

Trying to review for my test next week and this is an example problem that I have a feeling will be on the test yet I am somewhat lost on how to solve. Thanks in advance for the help!


Find the locations of all absolute extrema for the function....

f(x) = 2x^3 + 3x^2 -36x +12 with the domain [ -2, 4 ]

 

[skeeter]

use the first derivative test to locate any relative extrema, then compare the extrema to determine which are absolute extrema.

you do know what the first derivative test is, correct?

 

[Mathamateur]

Alright, I used the first derivative test and here is what I have so far.

f'(x) = 6x^2 + 6x - 36
0 = 6x^2 + 6x - 36
6(x^2 + x - 6) = 0
x^2 + x - 6 = 0
(x+3) (x-2) = 0

x= 2 or x= -3


Now how do I solve from here? Could someone please show me how?

 

[galactus]

You're working with the domain [-2,4].

The Extreme-Value Theorem tells us that if the function is continuous(yours is because it's a polynomial) on a closed interval [a,b], then it has a max and min on the interval.

1. Find your critical points in [-2,4].

2. Evaluate the function at the critical points and the endpoints.

3. The largest of the values you found in 2 is the max and the smallest is the min.

The point x=-3 is outside your domain, but 2 is inside the domain.

We find that x=2 gives y=-32, a minimum.

x=-2 gives y=80, an endpoint. A max.

 

[tkhunny]

You're working with the domain [-2,4].
2. Evaluate the function at ... the endpoints.

In my view, this is the most important part of this problem, because it is so often overlooked. Derivatives do not exist at endpoints. You have to check the endpoints yourself.