So I have this equation

f" (x) = -2sinx - 4sin2x

My question is what can I factor out of this equation? my guess is - 2sin or is it just -2?? I'm confused....

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- Nov 17th 2012, 02:07 PMasilvester635Simple Factoring
So I have this equation

f" (x) = -2sinx - 4sin2x

My question is what can I factor out of this equation? my guess is - 2sin or is it just -2?? I'm confused.... - Nov 17th 2012, 02:12 PMMacstersUndeadRe: Simple Factoring
Recall that sin(2x) = 2sin(x)cos(x)

- Nov 17th 2012, 02:21 PMasilvester635Re: Simple Factoring
So I'm asked to find the points of inflection and to discuss the concavity of the graph of the function

In order to do that I know i need to take the first and second derivative... so here's the work that I have so far

- Original Function

f(x) = 2sinx + sin2x

- First Derivative

f' (x) = 2cosx + 2cos2x

- Second Derivative

f" (x) = -2sinx - 4sin2x

I only needed to take up to the second derivative.... if I take the sin2x from 4sin2x and turn it into 2sin(x)cos(x) wouldn't I be taking the third derivative when i only needed to take the second derivative?? - Nov 17th 2012, 02:25 PMMacstersUndeadRe: Simple Factoring
In the case of substituting sin(2x) with 2sin(x)cos(x), all we would be using is a

*trigonometric identity*to simplify.

f''(x) = -2sin(x) - 4sin(2x) = -2sin(x) - 4*(2sin(x)cos(x) = -2sin(x) - 8sin(x)cos(x) = -2sin(x) [ 1 + 4cos(x) ] - Nov 17th 2012, 02:45 PMasilvester635Re: Simple Factoring
So to find the inflection points i need to set x equal to zero... heres what i got

-2sinx = 0

sinx = -2

and the other one is...

1 + 4cos(x) = 0

4cos(x) = 1

cos(x) = -1/4

So when i set both x equal to zero i got sinx = -2 and cos(x) = -1/4

NOW at this point I'm lost.... I'm not familiar with this type of situation... usually i'm used to seeing sinx = 1/2, and i know that equals to Pi/6 and 5Pi/6 in the unit circle and just plug it in a number line and check for concavity... but now i'm confused, what do i do now? - Nov 17th 2012, 02:56 PMMacstersUndeadRe: Simple Factoring
Simple arithmetic error

-2sin(x) = 0

sin(x) = 0 [divide by -2 on both sides]

which makes sense since the zeroes wouldn't change if you were to apply a vertical stretch by a factor of -2.

For cos(x) = -1/4, we know that a solution exists since -1/4 is in [-1,1] so apply cos^-1 to both sides to get your solution. - Nov 17th 2012, 03:17 PMasilvester635Re: Simple Factoring
okay so since sinx = 0, then that corresponds to Pi and 0 in the unit circle

and cosx is equal to -1 and 1? if it is then cos of -1 corresponds to Pi while cos of 1 corresponds to 0... am i right? - Nov 17th 2012, 03:19 PMasilvester635Re: Simple Factoring
okay so since sinx = 0, then that corresponds to Pi and 0 in the unit circle

and cosx is equal to -1 and 1? if it is then cos of -1 corresponds to Pi while cos of 1 corresponds to 0... am i right? - Nov 17th 2012, 03:33 PMMacstersUndeadRe: Simple Factoring
No. [-1,1] is the interval in which cos^-1 is formally defined. For example, if we had cos^-1 = -5, then the solution would not exist. but since -1/4 is in the interval [-1,1], then the solution does exist.

cos^-1(-1/4) is approx. 1.82 radians - Nov 17th 2012, 04:02 PMasilvester635Re: Simple Factoring
thank you so much.... but i am right with the "sinx = 0, then that corresponds to Pi and 0 in the unit circle" part right?

- Nov 17th 2012, 04:03 PMMacstersUndeadRe: Simple Factoring
Yes.