Question

Sketch careful, labeled graphs of each function f in Exercises 63–82 by hand, without consulting a calculator or graphing utility. As part of your work, make sign charts for the signs, roots, and undefined points of $f,f\text{'},andf\text{'}\text{'},$ and examine any relevant limits so that you can describe all key points and behaviors of f.

$f\left(x\right)=x^3(x+2)$

Short answer

The sign chart is

The sketch of the graph is

Step-by-step solution

Step 1. Given Information. 

The given function is$f\left(x\right)=x^3(x+2).$

Step 2. Finding the roots.  

To find the roots we will put the given function equal to zero.

So,

$$\begin{gathered} f\left(x\right)=x^3(x+2) \\ 0=x^3(x+2) \\ x=0andx+2=0 \\ x=-2 \end{gathered}$$

Therefore, the given function have roots at$x=0,-2.$

Step 3. Testing the signs of f.  

To sketch the sign chart, let's test the signs on both sides.

For f

$$\begin{gathered} f(-3)={\left(-3\right)}^3\left(-3+2\right) \\ f(-3)=27 \\ \mathrm{Now},f(-1)={\left(-1\right)}^3\left(-1+2\right) \\ f(-1)=-1 \\ \mathrm{Now},f\left(1\right)={\left(1\right)}^3\left(1+2\right) \\ f\left(1\right)=3 \end{gathered}$$

Step 4. Testing the signs.  

Now, let's test the sign for $f\text{'}andf\text{'}\text{'}.$

Let's differentiate the equation to find $f\text{'}.$

So,

$$\begin{gathered} f\text{'}\left(x\right)=4x^3+6x^2 \\ 0=4x^3+6x^2 \\ 0=2x^2\left(2x+3\right) \\ x=0\mathrm{and}0=2x+3 \\ x=-\frac{3}{2} \end{gathered}$$

Testing the signs on both sides,

$$\begin{gathered} f\text{'}(-2)=4{\left(-2\right)}^3+6{\left(-2\right)}^2 \\ f\text{'}(-2)=-8 \\ Now,f\text{'}(-1)=4{\left(-1\right)}^3+6{\left(-1\right)}^2 \\ f\text{'}(-1)=2 \\ Now,f\text{'}\left(1\right)=4{\left(1\right)}^3+6{\left(1\right)}^2 \\ f\text{'}\left(1\right)=10 \end{gathered}$$

Thus, $f\text{'}$is negative on the interval $\left(-\infty ,-\frac{3}{2}\right)$and positive on the interval $\left(-\frac{3}{2},\infty \right).$ Hence the graph of fwill be increasing on the positive intervals and decreasing on the negative intervals.

Let's differentiate again.

So,

$$\begin{gathered} f\text{'}\text{'}\left(x\right)=12x^2+12x \\ 0=12x\left(x+1\right) \\ 0=x\mathrm{and}x+1=0 \\ x=-1 \end{gathered}$$

Testing the sign on both sides,

role="math" localid="1648474789089" $$\begin{gathered} f\text{'}\text{'}(-2)=12{\left(-2\right)}^2+12\left(-2\right) \\ f\text{'}\text{'}(-2)=24 \\ Andf\text{'}\text{'}(-0.5)=12{\left(-0.5\right)}^2+12\left(-0.5\right) \\ f\text{'}\text{'}(-0.5)=-3 \\ Andf\text{'}\text{'}\left(1\right)=12{\left(1\right)}^2+12\left(1\right) \\ f\text{'}\text{'}\left(1\right)=24 \end{gathered}$$

Thus, $f\text{'}\text{'}$is positive on the interval role="math" localid="1648474872810" $\left(-\infty ,-1\right)and\left(0,\infty \right)$and negative on the interval $\left(-1,0\right).$ Hence, the graph of fwill be concave up on the positive interval and concave down on the negative interval. Inflection point at$x=-1,0.$

Step 5. Sketch the sign chart. 

The sign chart is

Step 6. Examine the relevant limit.

Let's examine the limits of $f\left(x\right)=x^3\left(x+2\right)\mathrm{as}x\to \pm \infty .$

$$\begin{gathered} \underset{x\to \infty }{\lim }f\left(x\right)=\infty \\ \underset{x\to -\infty }{\lim }f\left(x\right)=\infty \end{gathered}$$

Step 7. Sketch the graph of function f.