What are the interesting points on the graph of sin(x/2)?

  • Thread starter frenkie
  • Start date
  • Tags
    Derivative
In summary, the conversation is about finding the derivative of cos(x/2)/2 and setting it equal to 0 to solve for x. The correct derivative is cos(x/2) and the zeros of this function can be found by setting x/2 equal to values where cos(t) is zero. The conversation also discusses finding local extrema and points of inflection using the first and second derivatives. The zeros of the second derivative of sin(x/2)/4 can be found by setting sin(t) equal to zero, which leads to x values of 0, pi, and 2pi. The conversation also mentions that the graph of sin(x/2) has interesting points at 0, pi, and 2pi
  • #1
frenkie
37
0
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 what's on the bottom, but I don't know how he did it or if its right...help?
cos x=sqrt2/2
 
Physics news on Phys.org
  • #2
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 what's 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.
 
Last edited:
  • #3
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?
 
  • #4
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?
 
  • #5
x=Pi? I'm making this way too hard then it actually is.
 
  • #6
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...
 
  • #7
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?
 
  • #8
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
 
  • #9
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
 
  • #10
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?
 
  • #11
pi and 2pi which means that x=pi?
 
  • #12
or pi/2? getting confused all over again.
 
  • #13
You forgot 0 (and negative values). But you are otherwise correct.
 
  • #14
oh nvm my PIO is o and positive and negative 2pi? correct?
 
  • #15
is there any interesting points in the graph of sin(x/2)?
 

1. What is the meaning of "cos(x/2)/2 is my derivative"?

The statement "cos(x/2)/2 is my derivative" means that the function cos(x/2)/2 is the derivative of another function. In other words, when you take the derivative of the function cos(x/2)/2, you get another function as the result.

2. How do you know that cos(x/2)/2 is the derivative of another function?

This can be determined by using the rules of differentiation. The derivative of a function is the rate of change of that function, and there are specific rules and formulas for finding the derivative of different types of functions. In this case, cos(x/2)/2 follows the rule for finding the derivative of a composite function.

3. What is the process for finding the derivative of cos(x/2)/2?

The process for finding the derivative of cos(x/2)/2 involves using the chain rule. First, you would take the derivative of the outer function, which is 1/2. Then, you would take the derivative of the inner function, which is cos(x/2) using the chain rule. Finally, you would multiply these two derivatives together to get the overall derivative of cos(x/2)/2.

4. Can you provide an example of a function where cos(x/2)/2 is the derivative?

Yes, an example of a function where cos(x/2)/2 is the derivative is sin(x/2). If you take the derivative of sin(x/2), you will get cos(x/2)/2 as the result.

5. What are some real-world applications of understanding "cos(x/2)/2 is my derivative"?

Understanding the concept of "cos(x/2)/2 is my derivative" has many real-world applications, particularly in fields such as physics, engineering, and economics. For example, in physics, this concept is used to calculate the velocity of an object in motion, and in economics, it is used to analyze the rate of change of variables such as price and demand.

Similar threads

I Questions about these Trigonometry Graphs involving sin() and cos()
    • Calculus
Replies
3
Views
1K
B Question about change of variables
    • Calculus
Replies
29
Views
708
B Having trouble deriving the volume of an elliptical ring toroid
    • Calculus
Replies
4
Views
341
I Did Math DF's integral calculator make a glaring mistake?
    • Calculus
Replies
8
Views
166
I Integration of ##e^{-x^2}## with respect to ##x##
    • Calculus
Replies
4
Views
1K
I Why is the integral of ##\arcsin(\sin(x))## so divisive?
    • Calculus
Replies
3
Views
1K
I Is this Dirac delta function integral correct?
    • Calculus
Replies
4
Views
1K
B Deriving formula for sin(x-y)
    • General Math
Replies
5
Views
438
I How to evaluate the enclosed area of this implicit curve?
    • Calculus
Replies
2
Views
2K
B Why is Big-O about how rapidly the Taylor graph approaches that of f(x)?
    • Calculus
Replies
19
Views
2K

Hot Threads

Recent Insights

Back
Top