• :00Days
  • :00Hours
  • :00Mins
  • 00Seconds
A new era for learning is coming soonSign up for free
Log In Start studying!

Select your language

Suggested languages for you:
Answers without the blur. Sign up and see all textbooks for free! Illustration

Q. 83

Expert-verified
Calculus
Found in: Page 276
Calculus

Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

Answers without the blur.

Just sign up for free and you're in.

Illustration

Short Answer

In Exercises 83–86, 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

The possible graph of f is

See the step by step solution

Step by Step Solution

Step 1. Given Information.

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

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, 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''(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.

Step 4. Sketch the graph of function f.   

The graph of the function is

Most popular questions for Math Textbooks

Icon

Want to see more solutions like these?

Sign up for free to discover our expert answers
Get Started - It’s free

Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.

Sign up for free
94% of StudySmarter users get better grades.