Product Rule: Find where tangent to y=2(x-29)(x+1) horizonta

[K_Swiss]

Find the point(s) where the tangent to the curve is horizontal.

a) y = 2(x - 29)(x + 1)

y' = 0(x-29)(x+1) + 1(2)(x+1) + 1(2)(x-29)
y' = 0 + 2x + 2 + 2x - 58
y' = 4x - 56

4x - 56 = 0
4x = 56
x = 14

f'(14) = 4(14) - 56 = 0

Textbook Answer: (14, -450)

Where did I go wrong? Why did I get 0 for the y value? How do I get -450?

 

[skeeter]

Re: Product Rule

you found f'(14) rather than the y-value on the curve itself ... f(14).

the point on the curve where x = 14 is y = 2(14 - 29)(14 + 1) = -450

f'(14) = 0 is the slope at the point (14, -450)