Science:Math Exam Resources/Courses/MATH100/December 2010/Question 01 (l)/Solution 1

We first take the derivative to see that

$\displaystyle f'(x)=e^{x}(\sin x-\cos x)+e^{x}(\cos x+\sin x)=2e^{x}\sin x$

Setting this to zero, we see that the function is zero whenever $\displaystyle \sin x=0$ (the exponential function is always bigger than 0) and on our interval $[{\tfrac {\pi }{2}},2\pi ]$, this occurs at $\displaystyle \pi$ and at $\displaystyle 2\pi$.

Checking the endpoints and values at the critical points, we have

$\displaystyle f({\tfrac {\pi }{2}})=e^{\tfrac {\pi }{2}}(\sin({\tfrac {\pi }{2}})-\cos({\tfrac {\pi }{2}}))=e^{\tfrac {\pi }{2}}$

$\displaystyle f(\pi )=e^{\pi }(\sin(\pi )-\cos(\pi ))=e^{\pi }$

$\displaystyle f(2\pi )=e^{2\pi }(\sin(2\pi )-\cos(2\pi ))=-e^{2\pi }$

Hence, our function obtains its absolute maximum at $\displaystyle \pi$ and the value is $\displaystyle e^{\pi }$.