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
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 :)
1 answer