I am not sure how to approach this...I have to find the value of x that makes the derivate of f(x) equal to zero...here's what I have so far (is it correct?) and how is the problem finished?

Thank you very much!

f(x) = (1/x) + ln(x)
f'(X) = ln|x| + (1/x)
0 = ln|x| + (1/x)
how do I solve for x??
Thanks again :)

your derivative is incorrect,

the derivative of 1/x is -1/x^2 , not lnx

so f'(x) = -1/x^2 + 1/x

-1/x^2 + 1/x = 0
multiply by -x^2
1 - x = 0
x =1

To solve for the value of x that makes the derivative of f(x) equal to zero, you need to find the x-values where the derivative equation equals zero. Let's review your work so far:

The function you provided is:
f(x) = (1/x) + ln(x)

To find the derivative, you correctly used the sum rule and the derivative of the natural logarithm:

f'(x) = d/dx(1/x) + d/dx(ln(x))
= -1/x^2 + 1/x

To find the x-values where the derivative is zero, you want to solve the equation:

0 = -1/x^2 + 1/x

Now, let's solve for x. To do this, you can start by getting rid of the fractions by multiplying the entire equation by x^2:

0 = -x + x^2

Rearrange the equation:

x^2 - x = 0

Factor out x:

x(x - 1) = 0

Now, set each factor equal to zero and solve for x:

x = 0 or x - 1 = 0

From these equations, you find that x can be either 0 or 1.

So, the values of x that make the derivative of f(x) equal to zero are x = 0 and x = 1. These are the critical points at which the slope of the function f(x) changes.