Question 56092: > 3. The function f(x)= (x^2+2x)^2/3: (USE CALCULUS to answer a-e)
>
> a. Increasing on what interval(s)
> b. Decreasing on what interval(s)
> c. Has critical numbers?
> d. Has Relative Maximum?
> e. Has Relative Minimum?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! f(x)= (x^2+2x)^2/3: (USE CALCULUS to answer a-e)
f'(x)=(2/3)(2x+2)/(x^2+2x)^(1/3)
f'(x)=(4/3)(x+1)/(x^2+2x)^(1/3)
f'(x)=(4/3)(x+1)/[x(x+2)]^(1/3)
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Critical points are where numerator is zero or
denominator is zero.
Critical points are x=-2, x=-1,x=0
These break the number line into four intervals.
Int I: Test x=-10, f'(-10)= neg/pos<0 so f(x)decreasing on (-inf,-2)
Int II: Test x=-(3/2), f'(-3/2)=neg/neg>0 so f(x) increasing on (-2,-1)
Int III: Test x=-1/2, f'(-1/2)=pos/neg<0 so f(x) decreasing on (-1,0)
Int IV: Test x=10, f'(10)=pos/pos>0 so f(x) increasing on (0,inf)
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> a. Increasing on what interval(s)
See above
> b. Decreasing on what interval(s)
See above
> c. Has critical numbers?
See above
> d. Has Relative Maximum?
At x=-1 because f(x) increasing to the left and decreasing to the right.
> e. Has Relative Minimum
At x=-2 and x=0 because f(x) decreasing to the left and increasing to the right
Cheers,
Stan H.
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