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

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.

arildno
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Need: Show your own work first.

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?

Hootenanny
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There will be an infinite number of roots. What is the interval you are required to solve for?

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?

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.;-(

arildno
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Well, let's take the roots first:
Letting y=x/2, when is sin(y)=0?

when x=0, is that the only root? or is there more?

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?

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?

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 $\pi$, for example, $\pi , 2\pi , 3\pi$ etc.

Now you need to think about your function $f(x) = \sin\left( \frac{x}{2} \right)$, where will the crossing points be?

i believe they will be at 0, negative pie and 2pie. since one cycle i pie. and there are 2 complete cycles.

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).

plug in Pi for x in the original equation? :-(

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.

arildno
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Well:
What do you think the following expressions equals:
$$\sin(-3\pi), \sin(4\pi), \sin(7\pi)$$

What is the common feature with these expressions?

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.

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...

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.

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.

derivative of sin(x/2) is cos(x/2)? and the second derivative is -sin(x/2)?

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.

what is the derivative of (x/2)?

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?

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..