Single Variable Calculus, Chapter 4, 4.3, Section 4.3, Problem 42

Sketch the curve of $y = x ^3 - 3a^2x+2a^3$; where $a$ is a positive constant.
If $y = x^3 - 3a^2 x + 2a^3$, then

$
\begin{equation}
\begin{aligned}
y' &= 3x^2 - 3a^2\\
y'' &= 6x
\end{aligned}
\end{equation}
$


Solving for critical numbers, when $y'=0$

$
\begin{equation}
\begin{aligned}
0 & = 3x^2 - 3a^2\\
3x^2 &= 3a^2\\
x &= \sqrt{a^2}\\
x &= \pm a
\end{aligned}
\end{equation}
$

Hence, we can divide the interval by;

$
\begin{array}{|c|c|c|}
\hline\\
\text{Interval} & f'(x) & f\\
\hline\\
x < -a & + & \text{increasing on } (-\infty,-a)\\
\hline\\
-a < x < a & - & \text{decreasing on } (-a,a)\\
\hline\\
x > a & + & \text{increasing on } (a,\infty)\\
\hline
\end{array}
$


Solving for the inflection point, when $y'' = 0$

$
\begin{equation}
\begin{aligned}
y'' = 0 &= 6x\\
x &= 0&
\end{aligned}
\end{equation}
$


$
\begin{array}{|c|c|c|}
\hline\\
\text{Interval} & f'(x) & f\\
\hline\\
x < 0 & + & \text{Upward } \\
\hline\\
x > 0 & - & \text{Downward }\\
\hline
\end{array}
$


For the coordinates of the graph,

$
\begin{equation}
\begin{aligned}
\text{when } x &= a, &&& \text{when } x &= -a,\\
\\
f(a) &= a^3 - 3a^2(a) + 2a^3 &&& f(-a) &= (-a)^3 - 3a^2 (-a) + 2a^3\\
\\
f(a) &= 0 \text{ (Local Minimum)} &&& f(-a) &= 4a^3 \text{ (Local Maximum)}\\
\\
\text{when } x &= 0,\\
\\
f(0) &= (0)^3 - 3a^2 (0) + 2a^3\\
\\
f(0) &= 2a^3
\end{aligned}
\end{equation}
$

Therefore, we can illustrate the function as