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

We first take the derivative to see that

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

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

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

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

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

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

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