Question

Find the absolute extreme of the function on the closed interval. y=9e^xsinx , [0,pi]

Answer 1

For convenience sake, I will let $y=f(x)=9e^{x}\sin x$

First, we evaluate the function at the endpoints of the interval.

$f(0)=9e^{0}\sin(0)=0\\\\f(\pi)=9e^{\pi}\sin(\pi)=0$

Then, we need to find the critical points.

We can start by taking the derivative using the power rule.

$f'(x)=9e^{x} \frac{d}{dx} \sin x+9\sin x \frac{d}{dx} e^{x}=9e^{x}(\cos x+\sin x)$

Setting this equal to 0,

$9e^x (\cos x+\sin x)=0$

Since $9e^x > 0$, we can divide both sides by $9e^x$.

$\cos x+\sin x=0\\\\\sin x=-\cos x\\\\\tan x=-1\\\\x=\frac{3\pi}{4}$

$f\left(\frac{3\pi}{4} \right)=9e^{3\pi/4}\sin \left(\frac{3\pi}{4} \right)=\frac{9\sqrt{2}e^{3\pi/4}}{2}$

So, the absolute minimum is $\boxed{\left(\frac{3\pi}{4}, \frac{9\sqrt{2}e^{3\pi/4}}{\sqrt{2}} \right)}$ and the absolute minima are $\boxed{(0,0), (\pi, 0)}$