Question

Find and classify as local maxima, local minima, or neither, all critical points of f(x)=x^(3)+x^(2)-4x-4
Do I just find the derivative and set it equal to zero?

Answer 1

Yes Yes you do :)

Answer 2

Find the derivative and equate it to zero. Get all the values..

Then find the second derivative and plug in those values.

Answer 3

What does the second derivative tell me?

Answer 4

If after plugging in those values, the second derivative turns out >0, you have your local minima. If it is <0, local maxima.

Answer 5

Ok cool. so my first derivative = \[3x^2+2x-4\] and when i set it equal to zero I get positive and negative 1.535 right?

Answer 6

No.

Answer 7

-1.535 and 0.868

Answer 8

Oh yea that's right. Sorry! So what do those mean? Where it's increasing and decreasing?

Answer 9

Both of these are the critical points. Now you need to evaluate it further to identify whether they are local maxima or minima.

Answer 10

I do that by taking the second derivative, correct?

Answer 11

The second derivative comes as positive for both of this values; however, not both of them can be the local minimas.

Answer 12

Yep.

Answer 13

So how do I tell which one to go with?

Answer 14

Plug in the values in the original function. The solution with the smaller answer is the global minima and the value that makes the function attain the global minima is the local minima.

The other value will just be a critical point.

Answer 15

ok So for this question, I have critical points at -1.535 and 0.868. Then I have a local max @ .879 and a local min @ -6.06 Are these correct?

Answer 16

f(x)=x^3 + x^2 - 4x - 4
f'(x) = 3x^2 + 2x - 4 = 0. Use quadratic formula to find the roots.
You have critical points at x = -1.535 and at x = 0.869
f''(x) = 6x + 2
f''(-1.535) = 6(-1.535) + 2 = -7.21. Therefore, x = -1.535 is a local maxima.
f''(0.869) = 6(.869) + 2 which is positive. Therefore, x = 0.869 is a local minima.

Answer 17

Thank you guys so much!

Answer 18

You are welcome.