# Equal tangent lines

1. Feb 4, 2008

### TOONCES

If the tangent line to y = f (x) at x = a is the same as the tangent line to y = g(x) at x = b, find two equations that must be satisfied by a and b.

I don't understand how to go about this problem. I tried putting the equations of the tangent lines together using the variables in the problem but i don't know what to do with my results. This is all the info i have. Any help is appreciated.

2. Feb 4, 2008

### belliott4488

Why don't you try showing us what you have so far? That usually gets more of a response.

3. Feb 4, 2008

### TOONCES

Something I tried

-tangent line equations
y-f(x) = f '(x)(x-a) y-g(x) = g '(x)(x-b)
y = f(x)+f '(x)(x-a) y = g(x)+g '(x)(x-b)

-set them equal to each other

f(x)+f '(x)(x-a) = g(x)+g '(x)(x-b)

-distributed and simplified

f '(x)*x -g '(x)*x = g(x)-g '(x)*b-f(x)+f '(x)*a

-solved for x

x = [g(x)-g '(x)*b-f(x)+f '(x)*a]/[f '(x)-g '(x)]

- I'm really not sure if i can do anything with this but i did it anyway.
- i also tried trial and error with random functions as a last ditch kinda effort but that gave me pretty much the same results.

4. Feb 5, 2008

### HallsofIvy

Staff Emeritus
You haven't "solved for x" because you still have "x" inside each function! I have no idea why you want to "solve for x" because x is a variable, not a particular number. The problem asked you to "find two equations that must be satisfied by a and b."

No, those are not correct. In order that they be lines, the only "x" must be the x in (x-a) and (x- b). You evaluate the function and its derivative at a and b:
y- f(a)= f'(a)(x- a) and y- g(a)= g'(a)(x- a) so y= f(a)+ f'(a)(x-a) and y= g(b)+ g'(b)(x- b).

Saying that the two lines are the same means f(a)+ f'(a)(x- a)= g(b)+ g'(b)(x- b) for all x. That, in turn, means that the coefficient of x and the constant term must be the same on both sides. Those are your "two equations".

5. Feb 5, 2008

### TOONCES

thanks I'll try that and see what my professor says