1. Tangency

Another problem I'm stuck on.

"If k >/= 1 (greater than or equal to 1), the graphs of $y=sin(x)$ and y=ke^-x intersect for x >/= 0. Find the smallest value of k for which the graphs are tangent. What are the coordinates of the point of tangency?"

2. Hello, Rumor!

I think I have a solution.
Please check my reasoning and my work.

If $k \geq 1$, the graphs of: $y\:=\:\sin x$ and $y\:=\:ke^{-x}$ intersect for $x \geq 0.$
Find the smallest value of $k$ for which the graphs are tangent.
What are the coordinates of the point of tangency?
We have: . $\begin{array}{ccc} f(x) &=& \sin x \\ g(x) &=& ke^{-x} \end{array}$

The graphs intersect at the point of tangency.
. . Hence: . $f(x) \:=\:g(x) \quad\Rightarrow\quad \sin x \:=\:ke^{-x} \quad\Rightarrow\quad e^x\sin x \:=\:k$ [1]

Their slopes are equal at the point of tangency.
. . Hence: . $f'(x) = g'(x) \quad\Rightarrow\quad \cos x \:=\:-ke^{-x} \quad\Rightarrow\quad e^x\cos x \:=\:-k$ [2]

Divide [1] by [2]: . $\frac{e^x\sin x}{e^x\cos x} \:=\:\frac{k}{\text{-}k} \quad\Rightarrow\quad \tan x \:=\:-1 \quad\Rightarrow \quad x \:=\:\frac{3\pi}{4}$

We have: . $\begin{array}{ccccccc}f\left(\frac{3\pi}{4}\right) &=& \sin\frac{3\pi}{4} &=& \frac{1}{\sqrt{2}} \\ \\[-3mm]
g\left(\frac{3\pi}{4}\right) &=& ke^{-\frac{3\pi}{4}}\end{array}$

Since $g\left(\tfrac{3\pi}{4}\right) \,=\,f\left(\tfrac{3\pi}
{4}\right)$
, we have: . $ke^{-\frac{3\pi}{4}} \:=\:\frac{1}{\sqrt{2}}$

Therefore: . $\boxed{k \:=\:\frac{e^{\frac{3\pi}{4}}}{\sqrt{2}}}$

And the point of tangency is: . $\left(\frac{3\pi}{4},\:\frac{1}{\sqrt{2}}\right)$

3. Ah, of course!

Your work looks sound to me. I don't know why I couldn't remember to set them equal. Thank you very much!