Find stationary points of the following function...

Question

anonymous · Answer

$$g(x)=(1+ 2sinx) \exp ^{1/2x+cosx}   $$
Interval 0 to 2pi

.sam. · Answer

Differentiate g(x) then set it =0

cwrw238 · Answer

is ir e^( 1 / (2x+ cos x))  ?

anonymous · Answer

No. as it is.

.sam. · Answer

did you get
$$\frac{\left(e^x\right)^{\frac{1}{2 x}+\cos (x)} \left(2 x^2 (2 \sin (x)+3) \cos (x)-(2 \sin (x)+1) \left(2 x^2 \log \left(e^x\right) \sin (x)-x+\log \left(e^x\right)\right)\right)}{2 x^2}$$
?

cwrw238 · Answer

(1/2x ) + cos x

cwrw238 · Answer

thats a complicated one - not sure how .sam as done it - ill have to do this on paper...

anonymous · Answer

I have worked out the derivative but can't work out the stationary points. Any help?$$1/2(4\cos^2 x + 4\cos x -3)\exp ^{1/2x + cosx}$$

cwrw238 · Answer

ok - i got the same expression that wolfram alpha did and here it is http://www.wolframalpha.com/input/?i=differentiate+%281%2B2sinx%29%28e%5E%28%281%2F2x%29%2Bcos+x%29%29

anonymous · Answer

Did you manage to work out the stationary points?

cwrw238 · Answer

no - did you see how complicated the equation is?
how did you get that equation in cos x?

cwrw238 · Answer

that can be solved quite easily

anonymous · Answer

Which one? the one I derived?

cwrw238 · Answer

yes

anonymous · Answer

I can't seem to do it to get =0. Could you give me some guidance.

cwrw238 · Answer

(4cos^2x+4cosx−3) = 0   should be solvable

cwrw238 · Answer

but i don't know how the heck you got that one

anonymous · Answer

I used the trig identities

anonymous · Answer

(2cosx -1) (2cosx+3) =0
So x = 1 and x=-3 
These are the stationary points?