BuyFind

Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071
BuyFind

Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

Solutions

Chapter 1, Problem 6RCC

(A)

To determine

To find:

The graph of the equation of the form an=b

Expert Solution

Explanation of Solution

Given:

  b=an,ontheintervalaxb

Concept used:

Let n be a positive integer greater than 1 and let a be a real number

  1. If a=0,thenan=0
  2. If a>0, then an is the positive real number b such that bn=a

Calculation:

The equation of the form an=b

Where p and q are positive integers and x0

  an=b

  y=(x)1/3

Comparing the equation b=ap/q

Where p=1,q=3 ( p and q are positive integers)

  y=x

  y=(x)1/2

Comparing the equation y=xp/q

Where p=1,q=2 ( p and q are positive integers)

  y=x

  y=(x)1/1

Comparing the equation y=xp/q

Where p=1,q=1 ( p and q are positive integers)

  b=a2

  b=(a)2/1

(B)

To determine

To find:

The graph of the equation of the form b=ap/q

Expert Solution

Explanation of Solution

Given:

  y=x3,y=x,y=x,y=x2,y=x3ontheinterval0x2

Concept used:

Let n be a positive integer greater than 1 and let a be a real number

  1. If a=0,thenan=0
  2. If a>0, then an is the positive real number b such that bn=a

Calculation:

The equation of the form y=xp/q

Where p and q are positive integers and x0

  y=x3

  y=(x)1/3

Comparing the equation y=xp/q

Where p=1,q=3 ( p and q are positive integers)

  y=x

  y=(x)1/2

Comparing the equation y=xp/q

Where p=1,q=2 ( p and q are positive integers)

  y=x

  y=(x)1/1

Comparing the equation y=xp/q

Where p=1,q=1 ( p and q are positive integers)

  y=x2

  y=(x)2/1

Comparing the equation y=xp/q

Where p=2,q=1 ( p and q are positive integers)

  y=x3

  y=(x)3/1

Comparing the equation y=xp/q

Where p=3,q=1 ( p and q are positive integers)

(C)

To determine

To find:

The graph of the equation of the form y=xp/q

Expert Solution

Explanation of Solution

Given:

  b=an,y=x3ontheintervalaxb

Concept used:

Let n be a positive integer greater than 1 and let a be a real number

  1. If a=0,thenan=0
  2. If a>0, then an is the positive real number b such that bn=a

Calculation:

The equation of the form b=ap/q

Where p and q are positive integers and x0

  b=(a)1/3

Comparing the equation b=ap/q

Where p=1,q=3 ( p and q are positive integers)

  b=a

  b=(a)1/2

  b=a3

  b=(a)3/1

Comparing the equation y=xp/q

Where p=3,q=1 ( p and q are positive integers)

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