relative extrema

[twinke]

Find the values of any relative extrema.

f(x) = x^3 - 3x^2 + 1

I am stuck on this problem. I am thinking it is a relative maximum of 1 at 0; relative minimum of -3 at 2. But I am still a bit confused and I want to make sur eI am doing this right. This problem in on a worksheet and I can't find any formulas in my text. If anyone can give me some direction I would appreciate it.

Thanks

 

[skeeter]

f(x) = x^3 - 3x^2 + 1

f'(x) = 3x^2 - 6x

set f'(x) = 0 ...

3x^2 - 6x = 0

3x(x - 2) = 0

critical values of x at x = 0 and x = 2

f"(x) = 6x - 6

f"(0) = -6 ... f is concave down at x = 0, therefore f has a relative max at x = 0

f"(2) = 6 ... f is concave up at x = 2, therefore f has a relative min at x = 2.

 

[twinke]

Skeeter, see where I am confused is that this is a mulitple choice problem and these are the options I am given. But I am not sure which one meets this.

a. No relative extrema.
b. Relative maximum of 1 at 0; Relative minimum of -3 at 2.
c. Relative maximum of 1 at 0.
d. Relative maximum of 0 at 1; Relative minimum of -3 at -2.

 

[skeeter]

relative max at x = 0 ... f(0) = 1

relative min at x = 2 ... f(2) = -3

extrema are y-values ... x values only give their location.