Single Variable Calculus, Chapter 4, 4.5, Section 4.5, Problem 2

Use the guidelines of curve sketching to sketch the curve. $y = x^3 + 6x^2 + 9x$

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


B. Intercepts.
Solving for $y$-intercept, when $x=0$.
$y = 0^3 + 6(0)^2 + 9(0) = 0$
Solving for $x$-intercept, when $y = 0$.

$\begin{equation} \begin{aligned} 0 &= x^3 + 6x^2 + 9x\\ \\ 0 &= x(x^2 + 6x + 9)\\ \\ x &= 0 \text{ and } x^2 + 6x + 9 = 0 \qquad \Longleftarrow \text{(By using Quadratic Formula)} \end{aligned} \end{equation}$

The $x$-intercept are, $x = 0$ and $x = -3$


C. Symmetry.
The function is not symmetric to both $y$-axis and origin.


D. Asymptotes.
None.


E. Intervals of Increase or Decrease.
If we take the derivative of $f(x)$, we have $y' = 3x^2 + 12x + 9$
When $y' =0$, $\quad 0 = 3x^2 + 12x + 9$
The critical numbers are, $x = -1$ and $x = -3$
So, the intervals of increase or decrease are.

$\begin{array}{|c|c|c|} \hline\\ \text{Interval} & f'(x) & f\\ \hline\\ x < - 3 & + & \text{increasing on } (-\infty, -3)\\ \hline\\ -3 < x < -1 & - & \text{decreasing on } (-3,-1)\\ \hline\\ x > -1 & + & \text{increasing on } (-1, \infty)\\ \hline \end{array}$


F. Local Maximum and Minimum Values.
since $f'(x)$ changes from positive to negative at $x = 3$, then $f(-3) = 0$ is a local maximum. On the other hand, since $f'(x)$ changes from negative to positive of $x = -1$, then $f(-1) = -4$ is a local minimum.

G. Concavity and Points of Inflection.

$\begin{equation} \begin{aligned} \text{if } f'(x) &= 3x^2 + 12x + 9, \text{ then}\\ \\ f''(x) &= 6x + 12 \end{aligned} \end{equation}$

when $f''(x) = 0$, the inflection points is at $x = -2$
Thus, the concavity can be determined by divding the inteval to...

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


H. Sketch the Curve