calc
posted by APpreciate student on .
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) = lnx + (1/x)
0 = lnx + (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