Question

How do I find all extrema of this function?

Answer 1

f(x) = 3x-3x^(2/3) on interval [-1, 3] ?????

Answer 2

find f'(x). do a sign chart.

Answer 3

\[f'(x) = 3-2x^{-1/3}\]
set equal to 0
\[2x^{-1/3} = 3\]
\[x^{-1/3} = \frac{3}{2}\]
\[x = (\frac{3}{2})^{-3} = (\frac{2}{3})^{3} = \frac{8}{27}\]
so there is a local maximum in interval [-1,3] at x=8/27

Answer 4

sorry mistake, its a minimum

Answer 5

great! so if I had 2-3x^(2/3) would the answer be local min on interval [-1, 3] at x =1 ?

Answer 6

Would the local min be x=-1 and then max x=3