SOLUTION: Use calculus to determine the relative extrema of f(x)=x-lnx Find the equation of the tangent line at x=1 please help, i'm so confused!

Question 61512: Use calculus to determine the relative extrema of f(x)=x-lnx
Find the equation of the tangent line at x=1
please help, i'm so confused!
Found 2 solutions by venugopalramana, funmath:
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
Use calculus to determine the relative extrema of
f(x)=x-lnx
DF/DX =1-1/X = 0 AT EXTREMUM
X =1
D^F/DX^2 = 1/X^2 = 1/1=+VE
HENCE X=1 IS A MINIMUM
SLOPE OF TANGENT AT X=1 DY/DX AT X=1........THAT IS M = 0
AT X=1 ..Y = 1-LN1 =1
HENCE EQN.OF TANGENT AT X=1,Y=1 IS
Y-1=M(X-1)=0
Y = 1 IS THE EQN. OF THE TANGENT AT X=1
Find the equation of the tangent line at x=1
please help, i'm so confused!

Answer by funmath(2933) (Show Source): You can put this solution on YOUR website!
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)=

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)=
f''(1)=
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
Happy Calculating!!!
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