Single Variable Calculus, Chapter 3, 3.3, Section 3.3, Problem 72

Find the values of $x$ that has a graph of $f(x) = xX^3 + 3x^2 + x + 3$, where tangent is horizontal.
Given: $f(x) = x^3 + 3x^2 + x + 3$
$f'(x) = m_T = 0$ slope is 0 when tangent is horizontal.


$
\begin{equation}
\begin{aligned}
f(x) & = x^3 + 3x^2 + x + 3\\
f'(x)& = m_T = \frac{d}{dx}(x^3) + 3 \frac{d}{dx} (x^2) + \frac{d}{dx} (x) + \frac{d}{dx} (3)
&& \text{Derive each terms}\\
\\
f'(x)& = 3x^2 + (3)(2x) + 1 + 0
&& \text{Simplify}\\
\\
m_T & = 3x^2 + 6x + 1
&& \text{Substitute the value of the slope } (m_T)\\
\\
0 & = 3x^2 + 6x + 1
&& \text{Multiply both sides by } \frac{1}{3}\\
\\
\frac{3x^2 }{3} + \frac{6x}{3} + \frac{1}{3} &= \frac{0}{3}
&& \text{Simplify the equation}\\
\\
x^2 + 2x + \frac{1}{3} &= 0
&& \text{Add } \frac{-1}{3} \text{ to each sides}\\
\\
x^2 + 2x &= -\frac{1}{3}
&& \text{To make the left side a perfect square, we use completing the square}\\
\\
x^2 + 2x + 1 &= \frac{-1}{3} + 1
&& \text{Simplify the equation}\\
\\
(x+1)^2 &= \frac{2}{3}
&& \text{Take the square root of both sides}\\
\\
\sqrt{(x+1)^2} &= \pm \sqrt{\frac{2}{3}}
&& \text{Add -1 to each sides}\\
\\
x &= \pm \sqrt{\frac{2}{3}} - 1
&& \text{Get the LCD}\\
\\
x &= \frac{\pm \sqrt{2} - \sqrt{3}}{\sqrt{3}}
&& \text{Multiply both numerator and denominator by } \sqrt{3}\\
\\
x &= \frac{\pm \sqrt{2} - \sqrt{3}}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}}
&& \text{Simplify the equation}\\
\\
x &= \frac{-3 \pm \sqrt{6}}{3}
\end{aligned}
\end{equation}
$


The values of $x$ are $\displaystyle x = \frac{-3 +\sqrt{6}}{3}$ and $\displaystyle x = \frac{-3-\sqrt{6}}{3}$