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## Homework Statement

x/√(x

^{2}+y

_{1}

^{2})-(l-x)/√((l-x)

^{2}+y

_{2}

^{2})=0

How do I prove that the above equation has a solution for x in ℝ and that the solution is unique?

(y

_{1}, y

_{2}, and l are constants.)

## Homework Equations

x√((l-x)

^{2}+y

_{2}

^{2})-(l-x)√(x

^{2}+y

_{1}

^{2})=0

x√((l-x)

^{2}+y

_{2}

^{2})+x√(x

^{2}+y

_{1}

^{2})=l√(x

^{2}+y

_{1}

^{2})

x[1+√((l-x)

^{2}+y

_{2}

^{2})/√(x

^{2}+y

_{1}

^{2})]=l=f(x)

I actually don't think it's plausible to isolate x, since it'll result in a quartic equation, which is messy.

## The Attempt at a Solution

Existence

Observe that the square root of a real number is always positive. Then lim(x→-∞)f(x)=-∞ and lim(x→∞)f(x)=∞. Since l is a continuous function, there must be at least one value of x that equals the constant l.

Uniqueness

I was told that in general, existence proofs start with the assumption that two different values x

_{1}and x

_{2}satisfy the equation, and then show that the two values are actually the same, but I really don't feel like solving a quartic equation.

So I'm going to make an observation. Observe that f(x) is a purely increasing function. Hence, df/dx is always positive.

df/dx=[1+√((l-x)

^{2}+y

_{2}

^{2})/√(x

^{2}+y

_{1}

^{2})]+x*(d/dx)[√((l-x)

^{2}+y

_{2}

^{2})/√(x

^{2}+y

_{1}

^{2})]

(d/dx)[√((l-x)

^{2}+y

_{2}

^{2})/√(x

^{2}+y

_{1}

^{2})]=-(l-x)/[√((l-x)

^{2}+y

_{2}

^{2})√(x

^{2}+y

_{1}

^{2})]-x√((l-x)

^{2}+y

_{2}

^{2})/(x

^{2}+y

_{1}

^{2})

^{3/2}

df/dx=[1+√((l-x)

^{2}+y

_{2}

^{2})/√(x

^{2}+y

_{1}

^{2})]+x*[-(l-x)/[√((l-x)

^{2}+y

_{2}

^{2})√(x

^{2}+y

_{1}

^{2})]-x√((l-x)

^{2}+y

_{2}

^{2})/(x

^{2}+y

_{1}

^{2})

^{3/2}]

So messy...

I'm not even sure if df/dx is always positive. It just appears to be that way when I graph it.

-----

By the way, I know this "proof" isn't rigorous at all, but I've never taken an analysis course. I also don't know how large the difference is between "sufficient" and "complete", but I think the assumption of continuity and limits is sufficient here.

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