Answer
The curves $~~y = e^{-x}~~$ and $~~y = -e^{-x}~~$ touch the curve $~~y = e^{-x}~sin~x~~$ at its inflection points.
Work Step by Step
$y = e^{-x}~sin~x$
$y' = -e^{-x}~sin~x+ e^{-x}~cos~x$
$y'' = e^{-x}~sin~x- e^{-x}~cos~x -e^{-x}~cos~x- e^{-x}~sin~x$
$y'' = -2e^{-x}~cos~x$
We can find the values of $x$ such that $y''=0$:
$y'' = -2e^{-x}~cos~x = 0$
$cos~x = 0$
$x = \frac{\pi}{2}+\pi~n$, where $n$ is an integer
The graph of $y$ has inflection points at these values of $x$.
Suppose the curve $y = e^{-x}$ touches $y = e^{-x}~sin~x$:
Then $sin~x = 1$
Then $x = \frac{\pi}{2}+2\pi~n$, where $n$ is an integer
These values of $x$ are inflection points of $y = e^{-x}~sin~x$
Suppose the curve $y = -e^{-x}$ touches $y = e^{-x}~sin~x$:
Then $sin~x = -1$
Then $x = \frac{3\pi}{2}+2\pi~n$, where $n$ is an integer
These values of $x$ are inflection points of $y = e^{-x}~sin~x$
Therefore, the curves $~~y = e^{-x}~~$ and $~~y = -e^{-x}~~$ touch the curve $~~y = e^{-x}~sin~x~~$ at its inflection points.
