tangent line at a point

[rubing]

I am trying to find the line tangent to x - lnx at point (e, e - 1) <br><br>I have computed this 3 times using f'(x) = 1 - 1/x and get the following tangent line:<br><br>y = x + e/x - 2<br><br>,yet the book i am working from says the answer is x - x/e

 

[srmichael]

I am trying to find the line tangent to x - lnx at point (e, e - 1) <br><br>I have computed this 3 times using f'(x) = 1 - 1/x and get the following tangent line:<br><br>y = x + e/x - 2<br><br>,yet the book i am working from says the answer is x - x/e

\(\displaystyle \displaystyle f'(e)=1-\frac{1}{e}\)

Using y = mx + b and the point (e, e-1) to solve for b you get...

\(\displaystyle \displaystyle e-1=(1-\frac{1}{e})e+b\)

\(\displaystyle \displaystyle e-1=e-1+b\) after distributing the "e", so then b = 0.

Thus...

\(\displaystyle \displaystyle y=(1-\frac{1}{e})x+0\) or

\(\displaystyle \displaystyle y=x-\frac{x}{e}\) if you distribute the x.

 

[pka]

I am trying to find the line tangent to x - lnx at point (e, e - 1) <br><br>I have computed this 3 times using f'(x) = 1 - 1/x and get the following tangent line:<br><br>y = x + e/x - 2<br><br>,yet the book i am working from says the answer is x - x/e

The tangent line is \(\displaystyle y-y_0=f'(x_0)(x-x_0)\)

So \(\displaystyle y-[e-1]=\left[1-\frac{1}{e}\right](x-e)\).

See what that works out to be.