Question 1001594: Please check my answers:
Q: Let f(x) = 2x^3 - 3x^2 + 6. Find the max and min values of f on [-1,1]
A: f'(x) = 6x^2 - 6x
= 6x(x - 1) =0
x=0, x=+1 critical values. However, +1 is already there so 2nd der test is not needed.
f"(x) = 12x-6
f"(0) = 12(0)-6=-6<0, therefore a MIN occurs at 0.
plugging in the x-values given.
f(1) = 2(1)^3-3(1)^2+6 = 5
f(-1) = 2(-1)^3-3(-1)^2+6 = 1
After graphing it looks like (-1,7) is the MIN and (0,6) is the MAX.
Is this correct?
Thank you
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! 2nd Derivative Test: if f '' (x) is negative on some interval, then the region on f(x) is concave DOWN (not up). So we have a max on this interval. That means a local max is at x = 0.
I think you meant to say (-1,1) instead of (-1,7)
Summary:
Local Max = (0,6)
Local Min = none
Absolute Max = (0,6)
Absolute Min = (-1,1)
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