Can You Solve f(x) + g(x,y) = 0 Where g Is Tangent to f?

[ekkilop]

Say we have two functions with the following properties:

f(x) is negative and monotonically approaches zero as x increases.
g(x,y) is a linear function in x and is, for any given y, tangent to f(x) at some point x_0(y) that depends on the choice of y in a known way.
Additionally, for any given y, f(x) \leq g(x,y) for all x, with equality only at x = x_0.

It is then true that for each y, f(x) + g(x,y) = 0 at precisely one value of x. I'm trying to find this value.

Writing g as a tangent line;

g(x,y) = f(x_0(y)) + f'(x_0(y))(x - x_0(y))

seems to be the obvious place to start, but trying to solve the above equaiton I find myself forced to try to invert f(x) at some point, which unfortunately cannot be done in terms of standard functions for the cases in which I'm interested.

My question is thus; can this be done in a nice way? And if not generally, are there specific circumstances under which this may be done?

Thank you!

 

[HallsofIvy]

It is not necessary to solve that equation. If you show that, for some x, f(x)+ g(x,y) is negative, that for another x, f(x)+ g(x, y) is positive, and that f(x)+ g(x, y) is monotone, then there must exist a unique x such that f(x)+ g(x,y)= 0.
(That's true if f(x) is continuous. You don't give that as a hypothesis but it is implied.)

 

[ekkilop]

Thank you for your reply!

Yes, f(x) is continuous. And indeed f(x) + g(x,y) is monotone.
What I meant to ask was if there is a way to explicitly find that value for x at which f+g=0 other than the "brute force" way of inverting the expression? Or perhaps, more generally - does an equation of the form "function + tangent line = 0" have a known solution under some assumptions/circumstances?