2x^3+12x^2-72x+4

Question

find the points on the curve y=2x^3+15x^2+36x+(-2) at which the tangent line is horizontal

Answer 1

2x^3 + 12x^2 - 72x + 4

Tangent line horizontal means slope = 0.

Slope is first derivative, so:

f(x) = 2x^3 + 12x^2 - 72x + 4

f ' (x) = 6x^2 + 24x - 72

6x^2 + 24x - 72 = 0 solve for x

6(x^2 + 4x - 72) = 0

x^2 + 4x - 72 = 0

Doesn't look like it factors easy, so let's try the quadratic formula.

x = (-4 +- sqrt(16 - 4(1)(-72) ) / 2

x = (-4 +- sqrt(16 + 288) ) / 2

x = (-4 +- sqrt(304) ) / 2

x = (-4 +- sqrt(16*19) ) / 2

x = (-4 +- 4sqrt(19) ) / 2

x = -2 +- 2sqrt(19)

The tangent line will be horizontal at x = -2 + 2sqrt(19) and x = -2 - 2sqrt(19).