Single Variable Calculus, Chapter 4, 4.5, Section 4.5, Problem 6

Use the guidelines of curve sketching to sketch the curve. $y = x(x+2)^3$

The guidelines of Curve Sketching
A. Domain.
We know that $f(x)$ is a polynomial having a domain of $(-\infty, \infty)$

B. Intercepts.
Solving for $y$-intercept, when $x=0$.
$ y = (0)(0+2)^3 = 0$
Solving for $x$-intercept, when $y = 0$.
$0 = x(x+2)^3$
We have, $x = 0 $ and $x +2 = 0$
The $x$-intercepts are, $x = 0 $ and $x = -2$


C. Symmetry.
The function is not symmetric to $y$-axis and origin by using symmetry test.


D. Asymptotes.
None, the function has no denominator.


E. Intervals of Increase or Decrease.
If we take the derivative of $f(x)$, by using Product and Chain Rule.

$
\begin{equation}
\begin{aligned}
f'(x) &= x \cdot 3 (x+2)^2 + (x+2)^3 \cdot (1)\\
\\
f'(x) &= (x+2)^2 (3x + (x+2))\\
\\
f'(x) &= (x+2)^2(4x+2)
\end{aligned}
\end{equation}
$

When $f'(x) = 0$, $0 = (x+2)^2(4x+2)$
We have, $0 = (x+2)^2$ and $4x + 2 = 0$
The critical numbers are, $x = -2$ and $\displaystyle x = \frac{-1}{2}$

So, the intervals of increase or decrease are...

$
\begin{array}{|c|c|c|}
\hline\\
\text{Interval} & f'(x) & f\\
\hline\\
x < -2 & + & \text{increasing on } (-\infty, -2)\\
\hline\\
-2 < x < \frac{-1}{2} & - & \text{decreasing on } (-2, \frac{-1}{2})\\
\hline\\
x > \frac{-1}{2} & + & \text{increasing on } (\frac{-1}{2}, \infty)\\
\hline
\end{array}
$

F. Local Maximum and Minimum Values.
Since $f'(x)$ changes from positive to negative at $x = -2$, $f(-2) = 0$ is a local maximum. On the other hand, since $f'(x)$ changes from negative to positive at $\displaystyle x = \frac{-1}{2}, f \left( \frac{-1}{2} \right) = -1.6875$ is a local minimum.

G. Concavity and Points of Inflection.

$
\begin{equation}
\begin{aligned}
\text{if } f'(x) &= (x+2)^2 (4x+2), \text{ then }\\
\\
f''(x) &= 2(x+2)(4x+2) + (x+2)^2 (4)\\
\\
\\
\\
\text{which can be simplified as, }\\
\\
f''(x) &= 12(x^2 + 3x+2)\\
\\
\text{when } f''(x) &= 0, \\
\\
0 &= 12(x^2 + 3x +2)
\end{aligned}
\end{equation}
$

The inflection point is at $x = -1$ and $x = -2$
Thus, the concavity can be determined by dividing the interval to...

$
\begin{array}{|c|c|c|}
\hline\\
\text{Interval} & f''(x) & \text{Concavity}\\
\hline\\
x < - 2 & + & \text{Upward}\\
\hline\\
-2 < x < -1 & - & \text{Downward}\\
\hline\\
x > -1 & + & \text{Upward}\\
\hline
\end{array}
$



H. Sketch the Graph.