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Q. 83

Expert-verified
Found in: Page 276

### Calculus

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

See the step by step solution

## Step 1. Given Information.

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

## Step 2. Apply the first derivative test.

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.

## Step 3. Finding inflection points.

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.

## Step 4. Sketch the graph of function f.

The graph of the function is