Find the absolute max and min if they exist

Question

For

f(x) = 2 + 3x - x³ over (-1,2).

I found the first point is x = 1

Then took the limit as x goes towards 2 from the left and got 0

Then limit as x goes towards -1 from the right of the equation and got 4 but that was incorrect.. I'm not sure which part I did right and wrong

Answer 1

f '(x) = 3 - 3x² = 0

1 - x² = 0

x = 1, -1

f(-1) = 2 + 3(-1) - (-1)³ = 2 - 3 - (-1) = 2 - 3 + 1 = 0 ; I am not sure how you got 4 here...maybe 2 + 3 - 1???

f(1) = 2 + 3(1) - (1)³ = 2 + 3 - 1 = 4

f(2) = 2 + 3(2) - 2³ = 2 + 6 - 8 = 0

Therefore, the absolute max is at 1 and equals 4. The absolute min is at -1 and 2 and equals 0.