`f(x) = (x^5 - 5x)/(5)` 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 25 - 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^5-5x)/5=(1/5)x^5-x`

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

`f'(x)=x^4-1=0`

`(x^2+1)(x^2-1)=0`

`x=1, x=-1`

The critical values for x are x=1 and x=-1. 

If f'(x)>0 the function is increasing over an interval.

If f'(x)<0 the function is increasing over an interval.

Choose an x value less than -1.

f'(-2)=15 Since f'(-2)>0 the function is increasing in the interval (-`oo, -1).`

Choose an x value between -1 and 1.

f'(0)=-1 Since f'(0)<0 the function is decreasing in the interval (-1, 1).

Choose an x value greater than 1.

f'(2)=15 Since f'(2)>0 the function is increasing in the interval (1, `oo).`

Because the function changed direction from increasing to decreasing a relative maximum will exist at x=-1. The relative maximum occurs at the point

(-1, 4/5).

Because the function changed direction from decreasing to increasing a relative minimum will exist at x=1. The relative minimum occurs at the point (1, -4/5).

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