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 2.7, Problem 78E
To determine

To find: The domain in which the given function becomes one-to-one and find the inverse of the function with the restricted domain.

Expert Solution

Answer to Problem 78E

The domain in which the given function is one-to-one is [3,] and the inverse of the function is k1(x)=x+3 and domain is [ 0,] .

Explanation of Solution

Given information:

  k(x)=|x3|

  Precalculus: Mathematics for Calculus - 6th Edition, Chapter 2.7, Problem 78E

Concept used: A function with domain A is called a one-to-one function if no two elements of A have the same image, i.e.

  f(x1)f(x2)    Whenever x1x2

If f be a one-to-one function with domain A and range B. Then its inverse function f1 has domain B and range A and is defined by

  f1(y)=x      f(x)=y

Calculation:

From the given graph of the function, it is clear that the value of h(x) don’t repeat right hand side of x=3 . So, in the domain [3,] the given function is one-to-one.

  k(x)=|x3|k(x)=x3         3xy=x3              0yx=y+3k1(y)=y+3      0y

Therefore, the inverse of the given function with restricted domain can be written as

  k1(x)=x+3     [ 0,]

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