If f(x)=(x^3/3)-x^2 on [-2, 3], f(x) has an absolute minimum at

a. x=-2
b. x=0
c. x=2
d. x=3
e. x=0 and x=2

I chose A. Because when I plug -2 into the function, it gives me the smallest value. Am I right? If not, tell me what to do with steps. Thanks.

Question

If f(x)=(x^3/3)-x^2 on [-2, 3], f(x) has an absolute minimum at 

a. x=-2 
b. x=0
c. x=2 
d. x=3 
e. x=0 and x=2 

I chose A. Because when I plug -2 into the function, it gives me the smallest value. Am I right? If not, tell me what to do with steps. Thanks.

inkyvoyd · Answer

Instead of plugging stuff in, you want to find the first derivative.

anonymous · Answer

in order to find an absolute minimum at a closed interval, first find the first derivative of an equation. first derivative is x^2 - 2x.

anonymous · Answer

set first derivative equal to zero. ie, 0=x^2 - 2x => x=0,2

anonymous · Answer

now pick any number between -2 and 0 and plug into first derivative. if positive function is increasing, if negative function is decreasing. do the same for the interval (0, 2) and (2,3).

anonymous · Answer

either x=2 or x=-2,3 (endpoints) is your answer, depending on which value is smallest (ie, smallest out of f(2), f(-2), f(3))

anonymous · Answer

going over first derivative  would be helpful, thanks