`f(x) = x^(1/3) + 1` 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 27 - 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^(1/3)+1`

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

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

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

`1=0`

1=0 is not a true statement. A critical value cannot be found using the first derivative. 

If f'(x)>0 the function increases on the interval.

If f'(x)<0 the function decreases on the interval.

The domain for the function is all real values for x. 

Notice that f'(0)=undefined. This means the slope of the function at x=0 does not exist.

Choose an x value less than 0.

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

Choose an x value greater than 0.

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

Since the function does not change direction there will not be a relative extrema.

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