Draw the graph of `y = e^(x) sin 2pix`.

Explanation

We have `y = f(x) = e^(x) sin 2pix`
First draw the graph of `y = +-e^(x).`
Now `sin 2pix = 0` when `2pix = npi` or x = n/2, where `n in Z`
Also `sin 2pix = 1` for `2 pix = (4n + 1) (pi)/(2)` or `x = n + (1)/(4), n in Z`
and `sin 2pix = - 1` for `2 pix = (4n - 1) (pi)/(2)` or `x = n - (1)/(4), n in Z`
For `f(n + (1)/(4)) = e^(n + (1)/(4))` and `f(n - (1)/(4)) = -e^(n + (1)/(4))`
All points `f(n + (1)/(4), e^(n + (1)/(4)))` lie on the graph of `y = e^(x)`
and all points `f(n - (1)/(4), -e^(n - (1)/(4)))` lie on the graph of `y = -e^(x)`
Thus, the graph of the function is as shown in the following figure.