`f(x) = x^(-2) ln(x)` Find the critical numbers of the function

Textbook Question

Chapter 4, 4.1 - Problem 44 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to find the critical points of the function, hence, you need to evaluate the solutions to the equation f'(x) = 0.

You need to evaluate the first derivative, using the product rule:

`f'(x)= -2x^(-3)*ln x + x^(-2)*1/x`

You need to solve for theta f'(x) = 0, such that:

`(2ln x)/(x^3) + 1/(x^3) = 0 => -2ln x + 1 = 0 => ln x = 1/2 => x = sqrt e`

Hence, evaluating the critical numbers of the function forf'(x) = 0, yields` x = sqrt e.`

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scisser | (Level 3) Honors

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Find the derivative and set the numerator equal to 0.

`f(x)=x^-2lnx`

`f’(x)=(1/x^3)(1-2lnx)`

Set numerator equal to 0.

`1-2lnx=0`

`-2lnx=-1`

`lnx=1/2`

`x=sqrte`

Therefore, the critical value is `x=sqrte`

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