Question

find the point on the graph of y=1+1/x at which the tangent line passes through the point (3,0)

Answer 1

Let \((x_0,y_0)\) be the intersection of the given curve and the tangent line at \(x=x_0\).
The tangent line equation is
\[y-y_0=y'(x_0)(x-x_0)\\
y-\left(1+\frac{1}{x_0}\right)=-\frac{1}{(x_0)^2}(x-x_0)\\
\color{red}{y=-\frac{1}{(x_0)^2}(x-x_0)+\left(1+\frac{1}{x_0}\right)}.\]
It passes through the point (3, 0), so you know it can also be written as
\[y-0=m(x-3)\\
y=m(x-3).\]
Since these two equations represent the same line, they must have the same slope, so the last equation can be written as
\[\color{blue}{y=-\frac{1}{(x_0)^2}(x-3)}.\]
Set the red equation equal to the blue, and solve for \(x_0\). This gives you the x-coordinate of the point you're looking for, so you'll also have to solve for \(y_0\) by finding \(1+\frac{1}{x_0}\).

Answer 2

Wow thanks so much. I was completely stumped on this one.

Answer 3

Actually, I may have made some computational mistakes at some point (in my own work on paper)... I'll go over this again. But the procedure is right.

Answer 4

Alright, thanks again.

Answer 5

You're welcome

Answer 6

And here's what you should get:
http://www.wolframalpha.com/input/?i=y%3D1%2B1%2Fx%2C+y%3D3-x
and
http://www.wolframalpha.com/input/?i=y%3D1%2B1%2Fx%2C+y%3D-1%2F9%28x-3%29

Answer 7

For some reason I can't get it to turn out right when I set them equal to each other. I keep getting weird values. Should I get an exact number or should it still have the x variable in it?

Answer 8

The x should cancel.
\[\color{red}{-\frac{1}{(x_0)^2}(x-x_0)+\left(1+\frac{1}{x_0}\right)}=\color{blue}{-\frac{1}{(x_0)^2}(x-3)}\]
Multiplying both sides by \(-(x_0)^2\), you have
\[x-x_0-(x_0)^2-x_0=x-3\\
-(x_0)^2-2x_0=-3\\
(x_0)^2+2x_0-3=0\]

Answer 9

Okay, thanks. I was making it harder than it had to be