Question 15

To find the points of intersection, set the equations equal:

$e^{-x} = e^{-x}(\sin x + \cos x)$

Since $e^{-x} \neq 0$ for any $x$, we can divide by $e^{-x}$:

$1 = \sin x + \cos x$

Divide by $\sqrt{2}$ on both sides:

$\frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x = \frac{1}{\sqrt{2}}$

$\sin(x + \frac{\pi}{4}) = \sin(\frac{\pi}{4})$

The general solutions are:

1. $x + \frac{\pi}{4} = 2k\pi + \frac{\pi}{4} \implies x = 2k\pi$

2. $x + \frac{\pi}{4} = (2k+1)\pi - \frac{\pi}{4} \implies x = 2k\pi + \frac{\pi}{2}$

For $x \in [0, 10\pi]$:

From $x = 2k\pi$: $x = 0, 2\pi, 4\pi, 6\pi, 8\pi, 10\pi$ (6 points)

From $x = 2k\pi + \frac{\pi}{2}$: $x = \frac{\pi}{2}, \frac{5\pi}{2}, \frac{9\pi}{2}, \frac{13\pi}{2}, \frac{17\pi}{2}$ (5 points)

Total number of intersection points $n = 6 + 5 = 11$.