Derivatives and graphical behavior: In the next chapter we will see that we can get a lot of information about the graph of a function \(f\) by looking at the signs of \(f(x)\) and its first and second derivatives. Let's do this for the function \(f(x)=x^{3}-3x^{2}-9x+27\)

a. Find the roots of \(f\), and then determine the intervals on which \(f\), and then determine the intervals on which \(f\) is positive or negative.

b. Find the roots of \(f'\), and then determine the intervals on which \(f'\) is positive or negative.

c. Find the roots of \(f''\), and then determine the intervals on which \(f"\) is positive or negative.

d. The graph of \(f\) will be above the \(x-\)axis when \(f(x)\) is positive and below the \(x-\)axis when \(f(x)\) is negative. Moreover, the graph of \(f\) will be increasing when \(f'\) is positive and decreasing when \(f'\) is negative. Finally the graph of \(f\) will be concave up when \(f"\) is positive and concave down when \(f"\) is negative. Given this information and your answers from part (a)-(c), sketch a careful labeled graph of \(f\).

Step by step solution

01

Part a. Step 1. Explanation

\(f(x)=x^{3}-3x^{2}-9x+27\)

\(f(x)=(x-3)^{2}(x+3)\)

let \(f(x)=0\)

\((x-3)^{2}(x+3)=0\)

Critical points are \(3,-3\)

Now,

Interval	Sign of \('f'\)
\((-\infty,-3)\)	negative
\((-3,3)\)	positive
\((3,\infty )\)	positive

02

Part b. Step 1. Calculation

\(f(x)=x^{3}-3x^{2}-9x+27\)

\(f'(x)=3x^{2}-6x-9\)

\(f'(x)=3x^{2}-9x+3x-9\)

\(f'(x)=3(x-3)(x+1)\)

let \(f'(x)=0\)

critical points are \(3,-1\)

Now,

Interval	Sign of \('f'\)
\((-\infty,-1)\)	negative
\((-1,3)\)	negative
\((3,\infty )\)	positive

03

Part c. Step 1. Calculation

\(f(x)=x^{3}-3x^{2}-9x+27\)

\(f'(x)=3x^{2}-6x-9\)

\(f'(x)=6x-6\)

\(f'(x)=6(x-1\)

let \(f''(x)=0\)

critical point is \(1\)

Now,

Interval	Sign of \('f''(x)'\)
\((-\infty,1)\)	negative
\((1,\infty)\)	positive

04

Part d. Step 1. Calculation

\(f(x)=x^{3}-3x^{2}-9x+27\)

\(f'(x)=3x^{2}-6x-9\)

\(f''(x)=6x-6\)

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!