1. ## Finding stationary points

I can't figure this out at all, I've got the derivative

$g'(x)=\frac{1}{2}(4cos^{2}x+4cosx-3)e^{\frac{1}{2}x+cosx}$

and the domain of the function $g$ is $[0,2\pi]$

I've factorised to get

$(2cosx-1)(2cosx+3)=0$

But when I put the second bracket as zero to find $x$ for that bracket I get

$x=arccos(\frac{-3}{2})$

which doesn't get me anywhere on my calculator.

Have I made a mistake somewhere?

Many thanks.

2. ## Re: Finding stationary points

Why on Earth would you try to find the arc-cosine of -3/2?! The Domain of arccos(x) is [-1,1]. Never do that again.

3. ## Re: Finding stationary points

Sorry but I am still learning.

I'm pretty sure my factorising is correct, and making the first bracket equal to zero is giving me

$2cosx-1=0$

$2cosx=1$

$cosx=\frac{1}{2}$

$x=arccos(\frac{1}{2})\approx1.047$

which gives me one stationary point, but for the second bracket, like you say, I'm outside of the domain for arccos.

Any help?

4. ## Re: Finding stationary points

What kind of help do you need? You have correctly determined that the derivative is 0 only when 2 cos(x)- 1= 0 or when 2cos(x)+ 3= 0. Since the range of cosine is -1 to 1, there is NO value of x that makes 2cos(x)+ 3= 0 so the only possible critcal point is where 2 cos(x)- 1= 0.

5. ## Re: Finding stationary points

$cos{x} = \frac{1}{2}$ occurs at two angles on the unit circle ...

6. ## Re: Finding stationary points

Originally Posted by HallsofIvy
What kind of help do you need? You have correctly determined that the derivative is 0 only when 2 cos(x)- 1= 0 or when 2cos(x)+ 3= 0. Since the range of cosine is -1 to 1, there is NO value of x that makes 2cos(x)+ 3= 0 so the only possible critcal point is where 2 cos(x)- 1= 0.
Thank you so much,

All night last night I spent on that thinking I had to find two stationary points.

Thanks again.

7. ## Re: Finding stationary points

Originally Posted by Srengam
Thank you so much,

All night last night I spent on that thinking I had to find two stationary points.

Thanks again.
Look at skeeter's post again. There are two stationary points in $[0, 2\pi]$.