# Finding a parameter that makes a function tangentially touch another

I'm tasked with finding the value of parameter $m$ so that $f(x)=\sqrt{mx}$ tangentially touches $g(x)=e^x$. The solution is $m=2e$ however I don't know how to obtain this. In order for them to tangentially touch one another, they have to be: 1) $f(x_1)=g(x_1)$ and 2) $f'(x_1)=g'(x_1)$ for some $x_1$. Another idea was to set $f(x)=g(x)$ and then express $x$ in terms of $m$ so we could transform $g(x)=e^x$ into some function of $m$ and then by deriving both sides with respect to $m$ and solving we could obtain the solution, however the equations that pop out during this are weird and I don't know how to solve them.

The second part of the task is that I'm supposed to find the area bounded by those two curves and the y-axis, however even if I just take $m=2e$ I can't find the value of $x$ where they touch so I don't know the upper bound of integration. Perhaps a switch to polar coordinates could be the solution or some type of parametrisation ?

You can exploit the fact that, for the curve $y=e^x$ the gradient is always equal to the $y$ value. So if the curves are going to touch, the same will have to be true, at the point of contact, for the curve $f(x)=\sqrt{m}x^{\frac 12}$

Accordingly we solve $f'(x)=f(x)$ and get $$\frac 12\sqrt{m}x^{-\frac 12}=\sqrt{m}x^{\frac 12}\implies x=\frac 12$$

Then we have $$f(\frac 12)=\sqrt{\frac m2}=g(\frac 12)=e^{\frac 12}\implies m=2e$$

Now you have the $x$ value for the integral you wanted...

Moving forward by your methods :-

We have, $f(x) = a\sqrt{x}$ (let $a= \sqrt{m}$)

Also, $g(x)=e^x$

So by your method we have, $$a\sqrt{x}=e^x.....................(1)$$ $$\frac{a}{2\sqrt{x}}=e^x.....................(2)$$ Equating (1) and (2) $$\frac{a}{2\sqrt{x}}=a\sqrt{x}$$ $$\implies x=\frac{1}{2}$$ You just got the intersection point, feed it in (1) $$a\sqrt{\frac{1}{2}}= e^{\frac{1}{2}}$$ $$a=\frac{1}{2}$$ $$\implies \sqrt{m}=\sqrt{2e}$$ I presume you have the answer.