Question 61512
Use calculus to determine the relative extrema of f(x)=x-lnx

Find the equation of the tangent line at x=1
To find the relative extrema take the first derivative and find the numbers that make it 0 or undefined as long as the original function is defined there.
f'(x)={{{1-1/x}}}
{{{0=1-1/x}}}
{{{-1=-1/x}}}
{{{1=1/x}}}
{{{x=1}}}
It's undefined at 0, but so is the original function, so we can't have a relative extrema there.
f(1)=1-ln(1)
f(1)=1-0
f(1)=1
There is a relative extrema at (1,1)
f''(x)={{{1/x^2}}}
f''(1)={{{1/1^2}}}
f''(1)=1  The second derivative is positive, so it's a local min.
:
f'(1)=1-1/1
f'(1)=1-1
f'(1)=0  The slope of the tangent line is 0.
:
We already found that f(1)=1
The equation of line with the slope of 0 going through the point (1,1) is {{{highlight(y=1)}}}
Happy Calculating!!!