Originally Posted by

**smartcar29** hi

have function $\displaystyle g(x)=cosxexp(\frac{2}{3}sinx)$

has derivative $\displaystyle g'(x)=(2-3sinx-2sin^2x)exp(\frac{2}{3}sinx)$

i factor out 1/3 and ignored $\displaystyle exp(\frac{2}{3}sinx)$ as it always positive and got $\displaystyle (1-2sinx)(2+sinx)$ then set it to zero and got x= $\displaystyle \frac{\pi}{6}$ and $\displaystyle \frac{5\pi}{6}$ used these to get coordinates, which i am happy with but when i try to find the max and min locals of g(x) i get the oposite result to the graph of g(x). no matter how i try to look at it i get $\displaystyle \frac{\pi}{6}$,$\displaystyle \frac{\sqrt{3}}{2}$ as minimum point and $\displaystyle \frac{5\pi}{6}$,$\displaystyle \frac{-\sqrt{3}}{2}$ as a maximum point. have been using first derivitive test as i am unable to work out how to get g''(x)

any help would be great