Find extrema of f(x)=1-(x to the 2/3)
on interval [-1,8]
there are two possibilities:
The extrema is between -1,8
or
the extrema is at the endpoints.
f'=-2/3 x ^(-1/3)=0 or x=0
and at f(o) f(x)=1
f"=2/9 x(-4/3) and at x=0, this is positive, so f(o) is a relative min.
Now examine the the endpoints:
f(-1)=1-(-1^2/3)=1+1=2
f(8)=1-(8^2/3)=1-4=-3
first, check my math. Then, if it is right, it appears that f(0) is a relative minimum, and f(8) is a maximum