Q: Let f(x)= 3sin2x + 4cos3x. Determine where the tangent line is horizontal.
A: To have a horizontal (flat) tangent line, it means your slope is 0. Therefore, we must take the derivative (slope) and set it equal to 0.
f(x)= 3sin2x + 4cos3x
When taking the derivative, we must invoke the chain rule, which I like to describe as: the derivative of the outside, times the derivative of the inside.
So, the derivative of (something)2 = 2(something unchanged) * (derivative of that something)
f'(x)= 3sin2x + 4cos3x = 2(3sinx)*cosx + 3(4cos2x)*(-sinx)
f'(x) = 6sinx(cosx) – 12cos2x(sinx)
Now, we want to set the derivative = o
0 = 6sinx(cosx) – 12cos2x(sinx)
[factor the right side like so]:
0 = 6sinx(cosx) [1 - 2cosx]
So, either (1) 0 = 6sinx(cosx) OR (2) 0 = [1 - 2cosx]
(1) 0 = 6sinx(cosx) means that either sinx = 0 or cosx = 0
sinx = 0 when x = 2Πk , where k is any integer
cosx = 0 when x = Π/2 + Πk , where k in any integer
(2) 0 = [1 - 2cosx]
2cosx = 1
cosx = 1/2
This happens when x = Π/3 + 2Πk OR 5Π/3 + 2Πk
So, putting all of our answers together… We know that the tangent line is horizontal when:
x = 2Πk, Π/2 + Πk, Π/3 + 2Πk OR 5Π/3 + 2Πk
Wow! Long problem? Confused? Let me know.