Question

Ok, so I had to make a function and show examples of the first and second derivative tests to find relative extrema.

f(x) = x^3 – 3x^2

Answer 1

First Derivative Test:

f’(x) = 3x2 – 6x = 3x(x-2) = 0 x = 0, 2

Test x = 0: f(-1) = (-1^)2 – 3(-1)^2 = -1 – 3 = -4
f(1/2) = (1/2)^3 – 3(1/2)^2 = 1/8 – 3/4 = 7/8

Test x = 2: f(1) = (1)^3 – 3(1)^2 = 1 – 3 = -2
f(4) = (4)^3 – 3(4)^2 = 64 – 48 = 16

f(0) is a relative minimum because f’(x) changes from negative to positive at x=0.
f(2) is a relative minimum because f’(x) changes from negative to positive at x=2.

Is that right?

Answer 2

I got this for the second derivative test:
Second Derivative Test:

f”(x) = 6x – 6

f”(0) = 6(0) – 6 = 0 -6 = -6
f”(2) = 6(2) -6 = 12 – 6 = 6

But that means one of my previously stated rel. min. is now a max. Can someone go through the process with me and clear it up?

Answer 3

check that you are actually calculating f'(1) and the others above -- there's a small problem there

Answer 4

Am I supposed to test with f'(x) instead of f(x) in the first derivative test?

Answer 5

Yes, calculate f'(-1), f'(1/2), f'(1) and f'(4). This will tell you the slope of f(x) before and after the extrema you found

Answer 6

Ok, thank you! I'll fix that and come back if I run into any more problems. :)

Answer 7

ok, just let me know if need me to check later -- you are definitely on the right track!