# `f(x) = 2x + 1/x` 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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### Textbook Question

Chapter 3, 3.3 - Problem 33 - 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)=2x+(1/x)=2x+1x^-1`

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

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

`2=1/x^2`

`2x^2=1`

`x^2=1/2`

`x=+-.707`

The critical value is at x=.707, x=-.707 and x=0. The original function f(x) is not defined at 0.

If f'(x)>0, then the function is increasing on that interval.

If f'(x)<0, then the function is decreasing on that interval.

Choose any value for x that is less than -.707.

f'(-1)=1 Since f'(-1)>0, the function is increasing on the interval (-oo, -.707).

Choose any value for x that is in the interval (-.707,0).

f'(-.5)=-2 Since f'(-.5)<0, the function is decreasing on the interval (-.707,0).

Choose any value for x that is in the interval (0, .707).

f'(.5)=-2 Since f'(.5)<0, the function is decreasing on the interval (0, .707).

Choose any value for x that is greater that .707.

f'(1)=1 Since f'(1)>0, the function is increasing on the interval (.707, `oo).`

Since the sign of the derivative changed from positive to negative, there will be a relative maximum at x=-.707. The relative maximum is the point (-.707, -2.828).

Since the sign of the derivative changed from negative to positive, there will be a relative minimum at x=.707. The relative minimum is the point (.707, 2.828).

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