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Answers without the blur. Sign up and see all textbooks for free! Q. 69

Expert-verified Found in: Page 276 ### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861 # 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. $f\left(x\right)={x}^{3}\left(x+2\right)$

The sign chart is The sketch of the graph is See the step by step solution

## Step 1. Given Information.

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

## Step 2. Finding the roots.

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

So,

$f\left(x\right)={x}^{3}\left(x+2\right)\phantom{\rule{0ex}{0ex}}0={x}^{3}\left(x+2\right)\phantom{\rule{0ex}{0ex}}x=0andx+2=0\phantom{\rule{0ex}{0ex}}x=-2$

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

$f\left(-3\right)={\left(-3\right)}^{3}\left(-3+2\right)\phantom{\rule{0ex}{0ex}}f\left(-3\right)=27\phantom{\rule{0ex}{0ex}}\mathrm{Now},f\left(-1\right)={\left(-1\right)}^{3}\left(-1+2\right)\phantom{\rule{0ex}{0ex}}f\left(-1\right)=-1\phantom{\rule{0ex}{0ex}}\mathrm{Now},f\left(1\right)={\left(1\right)}^{3}\left(1+2\right)\phantom{\rule{0ex}{0ex}}f\left(1\right)=3$

## 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,

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

Testing the signs on both sides,

$f\text{'}\left(-2\right)=4{\left(-2\right)}^{3}+6{\left(-2\right)}^{2}\phantom{\rule{0ex}{0ex}}f\text{'}\left(-2\right)=-8\phantom{\rule{0ex}{0ex}}Now,f\text{'}\left(-1\right)=4{\left(-1\right)}^{3}+6{\left(-1\right)}^{2}\phantom{\rule{0ex}{0ex}}f\text{'}\left(-1\right)=2\phantom{\rule{0ex}{0ex}}Now,f\text{'}\left(1\right)=4{\left(1\right)}^{3}+6{\left(1\right)}^{2}\phantom{\rule{0ex}{0ex}}f\text{'}\left(1\right)=10$

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 f will be increasing on the positive intervals and decreasing on the negative intervals.

Let's differentiate again.

So,

$f\text{'}\text{'}\left(x\right)=12{x}^{2}+12x\phantom{\rule{0ex}{0ex}}0=12x\left(x+1\right)\phantom{\rule{0ex}{0ex}}0=x\mathrm{and}x+1=0\phantom{\rule{0ex}{0ex}}x=-1$

Testing the sign on both sides,

role="math" localid="1648474789089" $f\text{'}\text{'}\left(-2\right)=12{\left(-2\right)}^{2}+12\left(-2\right)\phantom{\rule{0ex}{0ex}}f\text{'}\text{'}\left(-2\right)=24\phantom{\rule{0ex}{0ex}}Andf\text{'}\text{'}\left(-0.5\right)=12{\left(-0.5\right)}^{2}+12\left(-0.5\right)\phantom{\rule{0ex}{0ex}}f\text{'}\text{'}\left(-0.5\right)=-3\phantom{\rule{0ex}{0ex}}Andf\text{'}\text{'}\left(1\right)=12{\left(1\right)}^{2}+12\left(1\right)\phantom{\rule{0ex}{0ex}}f\text{'}\text{'}\left(1\right)=24$

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 f will 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 $f\left(x\right)={x}^{3}\left(x+2\right)\mathrm{as}x\to ±\infty .$

$\underset{x\to \infty }{\mathrm{lim}}f\left(x\right)=\infty \phantom{\rule{0ex}{0ex}}\underset{x\to -\infty }{\mathrm{lim}}f\left(x\right)=\infty$

## Step 7. Sketch the graph of function f.  ### Want to see more solutions like these? 