Determine the concavity and find the points of inflection.
y=2cosx + sin2x 0≤x≤2pi
y'=-2sinx + 2cos2x
y"=-2cosx-4sinx
How do I find the IP(s)?
Your second derivative's last term should be -4sin2x
set y'' = 0 for points of inflection
-2cosx - 4sin2x = 0
cosx + 2sin2x = 0
cosx + 4sinxcosx = 0
cosx(1 + 4cosx) = 0
cosx = 0 or cosx = -1/4
You seem to know what you are doing, and can probably take it from here.
Concavity in short:
If for some f(x), the second derivative exists,
the curve is concave upwards if f''(x) > 0
the curve is concave downwards if f''(x) < 0
if f''(x) = 0 it is neither and you have a point of inflection as seen above.
I got-4sin2x for the second term, I made a typo.
How do you translate sinx=-1/4 to decimal numbers?
To find the points of inflection (IPs) of a function, you need to find the values of x at which the concavity changes. In other words, you are looking for the x-values where the second derivative, y'', equals zero or is undefined.
In this case, we have the function y = 2cos(x) + sin(2x) with the second derivative y" = -2cos(x) - 4sin(x).
To find the points of inflection, we need to solve the equation y" = 0.
-2cos(x) - 4sin(x) = 0
Now, let's solve this equation. First, rearrange it:
-2cos(x) = 4sin(x)
Next, divide both sides by -2:
cos(x) = -2sin(x)
Now, we can use the trigonometric identity: tan(x) = sin(x) / cos(x). Rearrange it to:
cos(x) = sin(x) / tan(x)
Now substitute tan(x) = sin(x) / cos(x):
cos(x) = sin(x) / [sin(x) / cos(x)]
cos(x) = cos(x)
That means the equation is satisfied for all values of x between 0 and 2π.
So, there are no points of inflection in the given interval (0 ≤ x ≤ 2π) for the function y = 2cos(x) + sin(2x).
from cosx + 4sinxcosx = 0
it should have been
cosx(1 + 4sinx) = 0
cosx = 0 or sinx = -1/4