The smallest slope of f(x)=6x^2-x^3 for 0 is less than or equal to x is less than or equal to 6 occurs at x=

Question

A. 0
B. 2
C. 3
D. 4
E. 6

I got E

Answer 1

You're asking for
minimum value of f'(x) for 0≤x≤6 where f(x)=6x²-x³.

First we need to know if f'(x) is increasing, decreasing, or there's a minimum within [0,6].
We can find out by first finding
f'(x)=12x-3x²

Find critical points for f'(x):
f"(x)=12-6x
so f"(x)=0 at x=2.
So check f'(x) for x=0,2,6 and locate the minimum of f'(x):
f'(0)=0
f'(2)=12
f'(6)=-36
so f'(6) is least, so choice E is correct.