Question 56086: . Find the ABSOULUTE EXTREMA of f(x)= x ^4/3 on the interval
> [-1,8].
> Use the EVT (Extreme Value Theorem)
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Find the ABSOULUTE EXTREMA of f(x)= x ^4/3 on the interval
> [-1,8].
Find the max on that interval:
f'(x)=(4/3)x^(1/3)
Let (4/3)x^(1/3)=0, thenx=0
So the function has a max at (0,0) on that interval.
Now look at the end points.
Evaluate f(x) at x=-1 and x=8
f(-1)=(-1)^(4/3)=1
f(8)=(8)^(4/3)=16
The absolute Extrema on that closed interval is 16.
Cheers,
Stan H.
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