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Q. 68

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*.

$f\left(x\right)={\left(1-x\right)}^{4}-2$

The sketch of the graph is

The sign chart is

The given function is $f\left(x\right)={(1-x)}^{4}-2.$

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

So,

$f\left(x\right)={\left(1-x\right)}^{4}-2\phantom{\rule{0ex}{0ex}}0={\left(1-x\right)}^{4}-2$

Let's examine the limits

$\underset{x\to \infty}{\mathrm{lim}}f\left(x\right)={(1-x)}^{4}-2\underset{x\to \infty}{\mathrm{lim}}f\left(x\right)=\infty \phantom{\rule{0ex}{0ex}}And\phantom{\rule{0ex}{0ex}}\underset{x\to -\infty}{\mathrm{lim}}f\left(x\right)={(1-x)}^{4}-2\phantom{\rule{0ex}{0ex}}\underset{x\to -\infty}{\mathrm{lim}}f\left(x\right)=\infty $

To sketch the sign chart, let's differentiate the equation to find $f\text{'}.$

So,

role="math" localid="1648472481521" $f\text{'}\left(x\right)=-4{\left(1-x\right)}^{3}\phantom{\rule{0ex}{0ex}}0=-4{\left(1-x\right)}^{3}\phantom{\rule{0ex}{0ex}}0\ne -4and{\left(1-x\right)}^{3}=0\phantom{\rule{0ex}{0ex}}x=1$

Testing the signs on both sides,

$f\text{'}\left(0\right)=-4{\left(1-0\right)}^{3}\phantom{\rule{0ex}{0ex}}f\text{'}\left(0\right)=-4\phantom{\rule{0ex}{0ex}}And\phantom{\rule{0ex}{0ex}}f\text{'}\left(2\right)=-4{\left(1-2\right)}^{3}\phantom{\rule{0ex}{0ex}}f\text{'}\left(2\right)=4$

Thus, $f\text{'}$ is positive on the interval $\left(1,\infty \right)$ and negative on the interval $\left(-\infty ,1\right).$ Hence the graph of *f *will be increasing on the positive intervals and decreasing on the negative intervals.

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

Let's differentiate again.

So,

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

Testing the signs on both sides,

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

Thus, $f\text{'}\text{'}$is positive. Hence *f *will be concave up everyplace.

The sign chart is

The graph of the function is

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