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