Thread: Finding stationary points

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



and the domain of the function is

I've factorised to get



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



which doesn't get me anywhere on my calculator.

Have I made a mistake somewhere?

Many thanks.

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.

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









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

Any help?

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.

occurs at two angles on the unit circle ...

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.

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 .