# Cos(x/2)/2 is my derivative

cos(x/2)/2 is my derivative. and when I set it equal to 0, i try to solve for x.

so cos(x/2)/2=0
cos(x/2)=0/2 this is the part where i'm stuck...a friend of mine did the rest and he came up with whats on the bottom, but I don't know how he did it or if its right...help?
cos x=sqrt2/2

## Answers and Replies

frenkie said:
cos(x/2)/2 is my derivative. and when I set it equal to 0, i try to solve for x.

so cos(x/2)/2=0
cos(x/2)=0/2 this is the part where i'm stuck...a friend of mine did the rest and he came up with whats on the bottom, but I don't know how he did it or if its right...help?
cos x=sqrt2/2

You should have put this in your other thread, and yes that would be a correct derivative, but I don't know what the heck your friend did to find the zeros of that function.. This should really not be that difficult.. What do you KNOW about cos(x), think about that for a second, maybe replace x/2 with t and figure out where cos(t) is zero, if you know anything about trigonometry this should be relatively easy and by know I really hope you know that there are an infinite number of them unless you are being restricted to a specific domain. Once you have the values of t where cos(t) is zero, you know that t = x/2 so from that you can find the values of x that make cos(x/2) zero.

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cos x is zero at Pi/2 and 3pi/2 on the unit circle so when i plug these two numbers into x in my derivative equation i get o for the answer

so x is Pi/2 and 3pi/2?

frenkie said:
cos x is zero at Pi/2 and 3pi/2 on the unit circle so when i plug these two numbers into x in my derivative equation i get o for the answer

so x is Pi/2 and 3pi/2?

NO!!! Firstly there are more zeros then that, a lot more(read my last post). Second those value are correct for a function of the form y = cos(t), but you have something of the form y = cos(x/2) so then what would the relation between x and t be?

It looks to me like it should be that

x/2 = t.

Do you think you can handle it from there?

x=Pi? i'm making this way too hard then it actually is.

frenkie said:
x=Pi? i'm making this way too hard then it actually is.

That is ONE of them, but only one. If you would post the original problem it would be much easier to help you, because I would think that you would be given some bounds within which to find the max/min/points of inflection for a function like this if not then well...

the teacher said that it would be alright if the domain iz [-2Pi, 2Pi] so i think that -Pi and Pi are the x values...which means that -Pi is a min and Pi is a max and 0 is a PIO.? do u agree?

frenkie said:
the teacher said that it would be alright if the domain iz [-2Pi, 2Pi] so i think that -Pi and Pi are the x values...which means that -Pi is a min and Pi is a max and 0 is a PIO.? do u agree?

What are you basing this on? You need to show proof to justify your claims using the first and second derivative tests not just random points where funtions are zero. The first derivative in that domain has only 2 zeros, which means only 2 places that could be extrema, you then need to find the seond derivative to find possible points of inflection.

The zeros of the first derivative you have are correct though. Pi and negative Pi

I'm not sure how to solve for x when i set my second derivative equal to 0.

-1sin(x/2)/4=0
if i do PEMDAS in reverse i get
sqrt2/1 multiplied by 4 but i don't think its right

frenkie said:
I'm not sure how to solve for x when i set my second derivative equal to 0.

-1sin(x/2)/4=0
if i do PEMDAS in reverse i get
sqrt2/1 multiplied by 4 but i don't think its right

Where is sqrt2 coming from?

This is the same as the first problem cos(x/2)=0, but with a sin instead of a cos. Where is sin t = 0?

pi and 2pi which means that x=pi?

or pi/2? getting confused all over again.

You forgot 0 (and negative values). But you are otherwise correct.

oh nvm my PIO is o and positive and negative 2pi? correct?

is there any interesting points in the graph of sin(x/2)?