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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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- Thread starter frenkie
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- #1

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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.

- #2

arildno

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Need: Show your own work first.

- #3

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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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Are you absolutely sure you were asked to do this by aid of a calculator?

- #6

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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?

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?

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- #11

Hootenanny

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

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- #13

arildno

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So, can you find some GENERAL formula for the zeroes out of this?frenkie said:

(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? :-(

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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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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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- #18

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frenkie said:

First and second derivative tests maybe...

- #19

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- #20

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frenkie said:

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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- #25

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frenkie said:

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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