# Function + tangent line = 0

1. Aug 22, 2014

### 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!

2. Aug 22, 2014

### 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.)

3. Aug 22, 2014