Solve f(x)=sin(x/2)

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  • #1
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original equation: f(x)=sin(x/2)
need: show work on how to find roots, POI, min, max.
intervals of increase/descrease.
intervals of concavity, end behavior.
please help..any help is greatly appreciated.
 

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  • #2
arildno
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Need: Show your own work first.
 
  • #3
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to find a root i plugged the equation into the calculator in the y= and i graphed it and pushed 2nd trace and pushed #2, and then it asked me to pick a number less then o and greater then o so I picked -2 and 2 and it gave me a root at x=0.

for the POI, i graphed it and looked at where the concavity changes. (not sure if its correct)

for limits i just looked at the graph and saw that as x goes to infinity y goes to 1 and as x goes to negative infinity y goes to -1.

and for the mins and max's i also looked at the graph and everytime thre was a concave up i put a min and everytimet here was a concave down i put a max. right?
 
  • #4
Hootenanny
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There will be an infinite number of roots. What is the interval you are required to solve for?
 
  • #5
arildno
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You are using TEXAS, right?

Are you absolutely sure you were asked to do this by aid of a calculator?
 
  • #6
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need to do it by hand...i know that to find the root i need to set the original equation to 0 and solve, but i'm stuck because i never did it with trig functions. i also konw that to find mins and maxs you are suppose to set the second derivative = to 0, but again, stuck.

and the intervals are -infinity to infinity.;-(
 
  • #7
arildno
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Well, let's take the roots first:
Letting y=x/2, when is sin(y)=0?
 
  • #8
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when x=0, is that the only root? or is there more?
 
  • #9
Hootenanny
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frenkie said:
when x=0, is that the only root? or is there more?

Think about a sin function. Where does it cross the x-axis?
 
  • #10
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from the interval of -10 to 10 sin function crosses the x-axis at 3, 6, 9 same for the negative. and sin(x/2) crosses at -6,0,6...so these are the 3 roots of the equation from -10 to 10?
 
  • #11
Hootenanny
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I assume your working in radians.Your answers are correct, however it is more usual to give them in terms of [itex]\pi[/itex], for example, [itex]\pi , 2\pi , 3\pi[/itex] etc.

Now you need to think about your function [itex]f(x) = \sin\left( \frac{x}{2} \right)[/itex], where will the crossing points be?
 
  • #12
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i believe they will be at 0, negative pie and 2pie. since one cycle i pie. and there are 2 complete cycles.
 
  • #13
arildno
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frenkie said:
i believe they will be at 0, negative pie and 2pie. since one cycle i pie. and there are 2 complete cycles.
So, can you find some GENERAL formula for the zeroes out of this?
(Hint: It has something to do with multiples of a famous number).
 
  • #14
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plug in Pi for x in the original equation? :-(
 
  • #15
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frenkie said:
plug in Pi for x in the original equation? :-(

But if the original equation is y = sin(x/2) then letting x = pi you get

y = sin(pi/2) = 1 So that certainly isn't a zero.
 
  • #16
arildno
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Well:
What do you think the following expressions equals:
[tex]\sin(-3\pi), \sin(4\pi), \sin(7\pi)[/tex]

What is the common feature with these expressions?
 
  • #17
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yeah true..I don't konw what the equation is...anybody know how to find POI of the equation and min/max? i know how to find it on the graph but I don't know how to do it and show work.
 
  • #18
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frenkie said:
yeah true..I don't konw what the equation is...anybody know how to find POI of the equation and min/max? i know how to find it on the graph but I don't know how to do it and show work.

First and second derivative tests maybe...
 
  • #19
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you mean set the first derivative equal to 0? and then the numbers you get you plug into the original equation? because when i set the first derivative equal to 0 i get x=0 as my only answer.
 
  • #20
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frenkie said:
you mean set the first derivative equal to 0? and then the numbers you get you plug into the original equation? because when i set the first derivative equal to 0 i get x=0 as my only answer.

The first derivative of that function certainly has more than 1 zero, and x=0 is definitely not one of them.
 
  • #21
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derivative of sin(x/2) is cos(x/2)? and the second derivative is -sin(x/2)?
 
  • #22
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frenkie said:
derivative of sin(x/2) is cos(x/2)? and the second derivative is -sin(x/2)?

Close, but you need to remember the chain rule.
 
  • #23
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what is the derivative of (x/2)?
 
  • #24
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1st derivative of y=sin(x/2) i found to be y=cos(x/2)/4 and second derivative y=-1sin(x/2)/4...but I am not sure how i got it. any ideas?
 
  • #25
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frenkie said:
1st derivative of y=sin(x/2) i found to be y=cos(x/2)/4 and second derivative y=-1sin(x/2)/4...but I am not sure how i got it. any ideas?

How are you not sure how you got it? And oddly enough your second derivative is correct but the first derivative you found is wrong..
 

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