Question

What is the relative extrema for the function
f(x) = 1/4 (x^4) -8x^2?

Answer 1

Find f'

Answer 2

set f'=0 and find where f' dne exist and use these values as critical numbers since that is the way they are defined.

Answer 3

what is F^1 of the problem?

Answer 4

Finf f'(x) and determine where it is zero or undefined. Then use First Derivative Test.

f ' (x) =x^3 - 16x
= x(x^2 - 16)
= x(x-4)(x+4)

So f ' (x) = 0 when x = 0, 4, and -4

Now use the First Derivative Test.

Answer 5

You can use the first derivative test to find the relative extrema.
Or you can use the second derivative.

f prime

Answer 6

what is the first derivate test and how would I apply it?

Answer 7

It says that where f' switches from negative to positive, there will exist a rel min; from positive to negative, will reult in a rel max. You want to test the intervals:
x< - 4
-4 < x < 0
0 < x < 4
x > 4

Check these 4 intervals and see if the derivative is positive or negative in that interval.

Then see if it switches from + to -, or - to +.

If it switches from + to + or - to -, then no rel max/min exists there.

Answer 8

Your welcome.

Answer 9

Got it. thank you