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

Expert Answers
sciencesolve eNotes educator| Certified Educator

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.`

scisser | Student

Find the derivative and set the numerator equal to 0.



Set numerator equal to 0.





Therefore, the critical value is `x=sqrte`