Originally Posted by
Chaim Well I have a problem like:
f(x) = 3x^{2}^{/3} - 2x, Interval: [-1, 1]
So this is what I did first:
Found the derivative: 2x^{-1/3}-2
0 = 2x^{-1/3}-2
x = 1
OR
f'(x) = 2/x^{1/3} - 2
x does not equal 0
Critical numbers: x = 1, x = 0
Used a chart:
x [-1, 0] 0 (0, 1) 1
f(x) negative undefined positive 0
Minimum occurs at x=0 since it goes from negative to positive
But I'm a bit confused on the maximum
I plugged in the points f(-1) = 3(-1)^{2/3} - 2(-3)
=5, so the point is (-1, 5)
Then I plugged in 1
f(1) = 3(1)^{2/3} - 2(1)
=1, so the point is (1, 1)
So I'm a bit confused, since at the back of my book,
The minimum is: (0, 0)
While the only maximum is (-1, 5)
So how come (1, 1) didn't get included as well