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
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 :)
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
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.