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Q. 69

Expert-verifiedFound in: Page 276

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*.

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

The sign chart is

The sketch of the graph is

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

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

So,

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

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

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

For *f*

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

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,

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

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

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 $x=-1,0.$

The sign chart is

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

$\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 $

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