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Chapter 3: Q. 83 (page 276)

In Exercises 83鈥�86, use the given derivative f' to find any local extrema and inflection points of f and sketch a possible graph without first finding a formula for f.

f'x=x3-3x2+3x

Short Answer

Expert verified

The possible graph of f is

Step by step solution

01

Step 1. Given Information.

The given derivative function isf'(x)=x3-3x2+3x.

02

Step 2. Apply the first derivative test.

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

So,

f'(x)=x3-3x2+3x0=x3-3x2+3x0=xx2-3x+3x=0

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

03

Step 3. Finding inflection points.

An inflection point occurs when f''(x)=0.

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

So,

f''(x)=3x2-6x+30=3x2-6x+30=3x2-2x+10=x-13x-3x=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.

04

Step 4. Sketch the graph of function f.   

The graph of the function is

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Most popular questions from this chapter

Find the critical points of each function f .Then use a graphing utility to determine whether f has a local minimum, a local maximum, or neither at each of these critical points.

fx=sexx

Find the possibility graph of its derivative f'.

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