Find the points on the curve $y = 2 x ^ { 3 } + 3 x ^ { 2 } - 36 x + 19$ where the tangent is horizontal.
A) $( - 3,100 ) , ( 2 , - 25 )$
B) $( - 3,88 ) , ( 4,39 )$
C) $( - 4,71 ) , ( 4,39 )$
D) $( - 4,71 ) , ( 2 , - 37 )$
E) $( - 3,37 ) , ( 2 , - 37 )$

Question

Quizplus · Accepted Answer

To find the points where the tangent to the curve is horizontal, we need to find where the derivative of the given function equals zero. The derivative of $y = 2x^3 + 3x^2 - 36x + 19$ is $y' = 6x^2 + 6x - 36$. Setting this equal to zero gives $6x^2 + 6x - 36 = 0$, which simplifies to $x^2 + x - 6 = 0$. Factoring this equation gives $(x + 3)(x - 2) = 0$, so $x = -3$ or $x = 2$. Substituting these x-values back into the original equation gives the points $(-3, 100)$ and $(2, -25)$.

Find the Points on the Curve y=2x3+3x2−36x+19y = 2 x ^ { 3 } + 3 x ^ { 2 } - 36 x + 19y=2x3+3x2−36x+19

Find the Points on the Curve $y = 2 x ^ { 3 } + 3 x ^ { 2 } - 36 x + 19$