For convenience sake, I will let [tex]y=f(x)=9e^{x}\sin x[/tex]
First, we evaluate the function at the endpoints of the interval.
[tex]f(0)=9e^{0}\sin(0)=0\\\\f(\pi)=9e^{\pi}\sin(\pi)=0[/tex]
Then, we need to find the critical points.
We can start by taking the derivative using the power rule.
[tex]f'(x)=9e^{x} \frac{d}{dx} \sin x+9\sin x \frac{d}{dx} e^{x}=9e^{x}(\cos x+\sin x)[/tex]
Setting this equal to 0,
[tex]9e^x (\cos x+\sin x)=0[/tex]
Since [tex]9e^x > 0[/tex], we can divide both sides by [tex]9e^x[/tex].
[tex]\cos x+\sin x=0\\\\\sin x=-\cos x\\\\\tan x=-1\\\\x=\frac{3\pi}{4}[/tex]
[tex]f\left(\frac{3\pi}{4} \right)=9e^{3\pi/4}\sin \left(\frac{3\pi}{4} \right)=\frac{9\sqrt{2}e^{3\pi/4}}{2}[/tex]
So, the absolute minimum is [tex]\boxed{\left(\frac{3\pi}{4}, \frac{9\sqrt{2}e^{3\pi/4}}{\sqrt{2}} \right)}[/tex] and the absolute minima are [tex]\boxed{(0,0), (\pi, 0)}[/tex]
