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) (d)/(dx) \sin x+9\sin x (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=(3\pi)/(4)`

`f\left((3\pi)/(4) \right)=9e^(3\pi/4)\sin \left((3\pi)/(4) \right)=(9√(2)e^(3\pi/4))/(2)`

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