1. ## Calculus problem...

How would you show 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 $\lambda>0$

meet (only)once at some $x>0$?

2. Originally Posted by willy0625
How would you show 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 $\lambda>0$

meet (only)once at some $x>0$?
The will intersect where they are equal.

So $\frac{2}{\pi}\arctan{\frac{x}{a}} = 1 - e^{-\lambda x}$

or $\frac{2}{\pi}\arctan{\frac{x}{a}} - 1 - e^{-\lambda x} = 0$.

If there was more than one intersection, then the graph must turn around on itself (by Rolle's Theorem). So see if there are any turning points.

3. I thought about this for a while and my lecturer gave me the following result to prove the uniqueness of the point of the intersection.

$\noindent{\bf Claim}$ Let F and G be function defined as

$F(t):= \frac{1}{\pi}\arctan\Big(\dfrac{t}{a}\Big)$

$G(t):= 1-\exp(-\lambda t)$

for $t\in(0,\infty).$

Suppose there exist x,y $\in(0,\infty)$ with $0 such that

$F(y) > G(y) \quad\cdots$(1)
$F(x) < G(x) \quad\cdots$(2)
$F'(y) > G'(x) \quad\cdots$(3)

Then there exists a unique $t^{*}\in[x,y]$ such that
$F(t^{*})=G(t^{*}).$

${\bf Proof}$
Let $H0,\infty) \rightarrow [0,1]" alt=" H0,\infty) \rightarrow [0,1]" /> be a function defined by
$H(t):= F(t)-G(t)$
and let $x,y\in(0,\infty)$ WLOG $0.
It follows that H is continuous since it is the difference of continuous functions F and G.

Then by (1) and (2), we have
$H(y)=F(y)-G(y) > 0$
$H(x)=F(x)-G(x) < 0$

Then it follows from the Bolzano's Intermediate Value Theorem that there exists a $t^{*}\in[x,y]$ such that
$H(t^{*})=0$
which is equivalently,
$F(t^{*})=G(t^{*})$

$\hspace{\stretch{1}} \blacksquare$

I think this point is unique in the interval $(0,\infty)$.

I'm not so sure about this because although I have proved that there is a point of intersection, yet the rate of increase for each function changes over different values for $x^{*}. I do not think the assumption (3) guarantees that G never over takes F again since we can have for some value z that is greater than y such that $F(z) < G(z)$.