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Q. 83

Expert-verifiedFound in: Page 276

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

In Exercises *83–86*, use the given derivative $f\text{'}$ to find any local extrema and inflection points of *f* and sketch a possible graph without first finding a formula for *f*.

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

The possible graph of *f* is

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

To find the local extrema the first derivative of the function must be zero.

So,

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

Thus, by the first derivative test, *f *has a local minimum at $x=0.$The function has no local maxima.

An inflection point occurs when $f\text{'}\text{'}\left(x\right)=0.$

To find the inflection points, use the second derivative test.

So,

$f\text{'}\text{'}\left(x\right)=3{x}^{2}-6x+3\phantom{\rule{0ex}{0ex}}0=3{x}^{2}-6x+3\phantom{\rule{0ex}{0ex}}0=3\left({x}^{2}-2x+1\right)\phantom{\rule{0ex}{0ex}}0=\left(x-1\right)\left(3x-3\right)\phantom{\rule{0ex}{0ex}}x=1$

Thus, the function has an inflection point at $x=1.$ It is positive everywhere. Hence the graph of *f * will be concave up everywhere.

The graph of the function is

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