Horizontal Tangent Line

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.

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