Recall that sin(2x) = 2sin(x)cos(x)
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??
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?
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.
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?
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?
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