Single Variable Calculus, Chapter 3, 3.3, Section 3.3, Problem 71

Find the points that has a curve $y = 2x^3+3x^2-12x+1$ where the tangent is horizontal.
Given: $y = 2x^3 + 3x^2 - 12x + 1$
$y'=m_T = 0$ slope is 0 when tangent is horizontal.


$
\begin{equation}
\begin{aligned}
y &= 2x^3 + 3x^2 - 12x +1\\
\\
y'&= m_T = 2 \frac{d}{dx}(x^3) + 3 \frac{d}{dx}(x^2) - 12 \frac{d}{dx}(x) + \frac{d}{dx} (1)
&& \text{Derive each term}\\
\\
y'&= (2)(3x^2)+(3)(2x)-(12)(1)+0
&& \text{Simplify the equation}\\
\\
m_T &= 6x^2 + 6x - 12
&& \text{Substitute the value of slope } (m_T) \text{ which is 0}\\
\\
0 &= 6x^2+6x-12
&& \text{Get the factor of the equation.}\\
\end{aligned}
\end{equation}
$


$
\begin{equation}
\begin{aligned}
\text{Equation 1:}& &&& \text{Equation 2:}&\\
(6x & -7)
&&& (x & +2) &&& \text{Equate both Equation to 0}\\
6x-6 &= 0
&&& x+2 &= 0\\
\frac{\cancel{6}x}{\cancel{6}} &= \frac{\cancel{6}}{\cancel{6}}
&&& x &= -2 \\
x &= 1
\end{aligned}
\end{equation}
$


In Equation 1, add 6 to each sides and in Equation 2, add -2 to each sides.

@ $ x = 1$

$
\begin{equation}
\begin{aligned}
y &= 2x^3 + 3x^2 - 12x + 1 && \text{Substitute the computed values of }x \text{ on the curve } y = 2x^3 + 3x^2 - 12x +1\\
\\
y &= 2(1)^3 + 3(1)^2 - 12 (1) + 1\\
\\
y &= 2 + 3 - 12 + 1\\
\\
y &= -6
\end{aligned}
\end{equation}
$


@ $ x= -2$

$
\begin{equation}
\begin{aligned}
y &= 2 (-2)^3 + 3(-2)^{-2} - 12 (-2) + 1\\
\\
y &= -16+12+24+1\\
\\
y &= 21
\end{aligned}
\end{equation}
$


The points on the curve $y = 2x^3 + 3x^2 - 12x + 1$ are $(1,-6), (-2,21)$