To prove that these two functions meet only once

[willy0625]

Homework Statement

How would you go about proving that the functions

F(x)=\frac{2}{\pi}\arctan\Big(\frac{x}{a}\Big), x \geq 0

with a > 0

and

G(x)=1-\exp(-\lambda x),x\geq0

with \lambda>0meet only at one point for some x > 0

The Attempt at a Solution

At x=0, F and G takes the values 0s and both of the functions tend to 1 when x gets really large I think they must meet at least once. But still not sure about finding the actual point(s) since i do not think i can solve the equation
F=G analytically.

 

[Sweet_GirL]

Homework Statement

How would you go about proving that the functions

F(x)=\frac{1}{2}+\frac{1}{\pi}\arctan\Big(\frac{x}{a}\Big), x\in\mathbb{R}

with a > 0

and

G(x)=1-\exp(-\lambda x),x\geq0

with \lambda>0


meet only at one point for some x > 0

The Attempt at a Solution

At x=0, F and G takes the values 0.5 a 0, respectively and both of the functions tend to 1 when x gets really large I think they must meet a least once. But still not sure about finding the actual point(s) since i do not think i can solve the equation
F=G analytically.

as x becomes large, the two function tend to 1.
And clearly they are increasing functions.

so what do you suggest?

 

[Tinyboss]

I don't believe the problem is asking you to find x, rather to show it exists and is unique.

 

[willy0625]

Yeah showing the uniqueness of that particular point is enough.

 

[Tinyboss]

The fact that your functions have the same limit does not mean they ever meet. Take for instance 1/x and 1/(x+1), for x>0. Clearly they both tend to zero, and clearly they are never equal.

I don't see right away how to do existence or uniqueness. By continuity if you can find a point where F<G then you have existence, and then maybe you could look at the derivative of F-G to show uniqueness. Or maybe there's something prettier. What topics has your classed covered recently?

 

[willy0625]

This question poped up somewhere in my research.
I can use any results in calculus and analysis.

Note I have made some correction to the F function.