`f(x) = x^(2/3) - 4` Find the critical numbers, open intervals on which the function is increasing or decreasing, apply first derivative test to identify all relative extrema.

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Chapter 3, 3.3 - Problem 28 - Calculus of a Single Variable (10th Edition, Ron Larson).
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mathace | (Level 3) Assistant Educator

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Given: `f(x)=x^(2/3)-4`

Find the critical value(s) by setting the first derivative equal to zero and solving for the x value(s).

`f'(x)=(2/3)x^(-1/3)=0`

`2/(3x^(1/3))=0`

`2=0`

Because 2=0 is not a true statement the first derivative will not produce a critical value.

If f'(x)>0 the function will increase in the interval.

If f'(x)<0 the function will decrease in the interval.

Notice f'(x)=undefined. This means the function is not differentiable at x=0.

Choose a value for x that is less than zero.

f'(-1)=-2/3 Since f'(-1)<0 the function is decreasing on the interval (-`oo,0).`

Choose a value for x that is greater than zero.

f'(1)=2/3  Since f'(1)>0 the function is increasing on the interval (0, `oo).`

Because the function changes direction from decreasing to increasing a relative minimum exists. The relative minimum exists at the point (0, -4) even though the derivative does not exist at that point. 

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