Finding stationary points of a function

T

Tiaan

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Aug 15, 2009
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Hello,

I have had a very long question involving derivatives and using different rules to find them.

I have found that the function g(x) = cosexp(2/3sinx) has the derivative g'(x) = 1/3(2-3sinx-2sin^2x)exp(2/3sinx). Now I have to find any stationary points of the function and use the first derivative test to classify each stationry point as a local maximum or minimum of g(x). The domain of the function is (0<x<2pi)
(0,2pi) and sinx <1 for all values of x.

I can find the stationary points of functions usually but because of the trinonometric identities here (my week spot) I am really stugiling to get started with this one.

Thank you Hope that is all clear.

g(x) is the function. and g"(x) is its derivative .

The way it is wriiten in my book is :

cos x exp(2/3 sinx) (exp (x) = e^x)

That is excatly how it is written
Is that any clearer???
So i need to solve for g'(x) = 0

Another Edit ;

I need to find the 2 values of x for
2-3sinx-2sin^2x=0 these will be the stationary points
 
It appears to be \(\displaystyle g(x)=cos(e^{\frac{2}{3}sin(x)}) \;\ or \;\ cos(e^{\frac{2}{3sin(x)}})\), but I could be wrong. No solid replies until it is verified because the wrong assumption may be made.
 
The way it is written in the book is not what you originally presented to us. Is that any clearer?

What is wrong with you? Are you all right?

Perhaps you should ponder getting a mental evaluation.

\(\displaystyle g(x) \ = \ cosx \ exp(2/3 \ sinx), \ (the \ way \ it \ is \ written \ in \ the \ book) \ = \ cos(x)e^{\frac{2sin(x)}{3}}\)

\(\displaystyle Hence, \ g \ ' \ (x) \ = \ e^{\frac{2sin(x)}{3}}[-sin(x)+2/3-(2/3)sin^{2}(x)]\)

\(\displaystyle Setting \ g \ ' \ (x) \ to \ zero, \ gives \sin(x) \ = \ 1/2, \ therefore \ x \ = \ \frac{\pi}{6}, \ \frac{5\pi}{6}.\)

\(\displaystyle Ergo, \ max \ = \ (\frac{\pi}{6},\frac{\sqrt 3}{2}e^{1/3}) \ and \ min \ = \ (\frac{5\pi}{6},\frac{-\sqrt 3}{2}e^{1/3})\)
 
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