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How many Tangent lines go to a certain point.

  • Thread starter Hypnos_16
  • Start date
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1. Homework Statement

How many tangent lines to the curve y = x / (x + 1) pass through the point (1,2)? At which points do these tangent lines touch the curve?

2. Homework Equations

n/a

3. The Attempt at a Solution

I'm completely drawing a blank on this one. Help.
 

Mentallic

Homework Helper
3,797
94
Let's first simply imagine a tangent line to the curve passing through the point (1,2).
What would the equation of this tangent be?

Well we know a point, and we don't know what the gradient is but we do know it is tangent to the curve so this would point towards the derivative equation, yes?

So we have, [tex]y-2=m(x-1)[/tex] where [tex]m=\frac{dy}{dx}=\frac{d}{dx}\left(\frac{x}{x+1}\right)[/tex]

And this line equation touches the curve, so what can we do here?
 
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So we have, [tex]y-2=m(x-1)[/tex] where [tex]m=\frac{dy}{dx}=\frac{d}{dx}\left(\frac{x}{x+1}\right)[/tex]

And this line equation touches the curve, so what can we do here?
Find the slope of the line, and then follow it outward to see at what other points it would touch the graph?
 

Mentallic

Homework Helper
3,797
94
Find the slope of the line, and then follow it outward to see at what other points it would touch the graph?
Trial and error is one way, but a very bad way. Let's do this more rigorously :smile:

You already have the equation of the tangent line(s), [tex]y=m(x-1)+2[/tex] and the equation of the curve [tex]y=\frac{x}{x+1}[/tex], and we know that the line touches the curve. How can we find the point where they touch by mathematical means?

If you can't quite figure it out, it's probably that you're just being overwhelmed by the problem at hand. As a simpler example, it's equivalent to finding the points of intersection between the equations [tex]y=x+1[/tex] and [tex]y=2x[/tex]
 
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you could make them equal one another. Since they both have y = (something) then solve for the variable. Or that's what i'd try in your example there. But i don't know how to ramp that up to make it work on a bigger scale, like the example i have.
 

Mentallic

Homework Helper
3,797
94
That's it.

So we just have to solve [tex]m(x-1)+2=\frac{x}{x+1}[/tex] for x, where [tex]m=\frac{dy}{dx}[/tex]. This will give you the points where the line and curve intersect, which means you'll have the x values of where the tangent(s) to the curve passing through (1,2) touch the curve.
 
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Alright, so then i would have to first get the slope of the line though wouldn't i? how can i do that if i only have one point?
 

Mentallic

Homework Helper
3,797
94
Alright, so then i would have to first get the slope of the line though wouldn't i? how can i do that if i only have one point?
Well we know the gradient is going to be the derivative at the point x that we solve for, so we can just find the derivative and replace the variable m for it.
 
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So i got a value of x to be -2 - 1√3 and -2 + 1√3. I'm thinking something has gone wrong, cause that would be impossible to try and work out.
 

Mentallic

Homework Helper
3,797
94
What do you mean it would be impossible to try and work out?

So the x values are [tex]x=-2\pm \sqrt{3}[/tex], now find the corresponding y values to complete the answer. If you want to check to be sure you have the right answer, find the derivative at those points, then test to see if the line going through that point with such a gradient passes through (1,2).

But I'll save you the effort and tell you now that it's right.
 
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haha thanks. I just had a moment where i forgot these can also be decimal values. As i read that too i thought, this problem doesn't seem worth all this effort. But thanks for checking them for me. I appreciate it man. Thanks for all your help again.
 

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