To find if the inverse of the function exists: For the interval ( ∞ , 0 ) , f ( x ) = x 2 Let us assume that if f ( − a ) = f ( − b ) Then, ( − a ) 2 = ( − b ) 2 − a = ± b Since, the interval is ( ∞ , 0 ) , hence − a = - b Thus, the inverse of the function f ( x ) = x 2 exists. For the interval [ 0 , ∞ ) , f ( x ) = x Let us assume that if f ( a ) = f ( b ) Then, a = b Thus, the inverse of the function f ( x ) = x exists. Hence, we get that the inverse of the function f ( x ) = { − x 2 w h e n x < 0 a n d x w h e n x ≥ 0 } exists. Inverse of the function: For the interval ( ∞ , 0 ) , f ( x ) = x 2 Replacing f ( x ) with y , we get y = x 2 On switching y and x , we get x = y 2 Solving for y , we get y 2 = x y = x Replacing y with f − 1 ( x ) , we get f − 1 ( x ) = x For the interval [ 0 , ∞ ) , f ( x ) = x Replacing f ( x ) with y , we get y = x On switching y and x , we get x = y Solving for y , we get y = x Replacing y with f − 1 ( x ) , we get f − 1 ( x ) = x Thus, Inverse of function is f − 1 ( x ) = { x w h e n x < 0 a n d x w h e n x ≥ 0 }

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Answer 1

Question

Chapter 1, Problem 57CRE

b.

To determine

To find if the inverse of the function exists:

For the interval $(\infty, 0)$ , $f (x) = x^{2}$

Let us assume that if $f (- a) = f (- b)$

Then, ${(- a)}^{2} = {(- b)}^{2}$

$- a = \pm b$

Since, the interval is $(\infty, 0)$ , hence $- a = - b$

Thus, the inverse of the function $f (x) = x^{2}$ exists.

For the interval $[0, \infty)$ , $f (x) = x$

Let us assume that if $f (a) = f (b)$

Then, $a = b$

Thus, the inverse of the function $f (x) = x$ exists.

Hence, we get that the inverse of the function $f (x) = {- x^{2} w h e n x < 0 a n d x w h e n x \geq 0}$ exists.

Inverse of the function:

For the interval $(\infty, 0)$ , $f (x) = x^{2}$

Replacing $f (x)$ with $y$ , we get

$y = x^{2}$

On switching $y$ and $x$ , we get

$x = y^{2}$

Solving for $y$ , we get

$y^{2} = x$

$y = \sqrt{x}$

Replacing $y$ with $f^{- 1} (x)$ , we get

$f^{- 1} (x) = \sqrt{x}$

For the interval $[0, \infty)$ , $f (x) = x$

Replacing $f (x)$ with $y$ , we get

$y = x$

On switching $y$ and $x$ , we get

$x = y$

Solving for $y$ , we get

$y = x$

Replacing $y$ with $f^{- 1} (x)$ , we get

$f^{- 1} (x) = x$

Thus, Inverse of function is $f^{- 1} (x) = {\sqrt{x} w h e n x < 0 a n d x w h e n x \geq 0}$

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Answer 2

Question

Chapter 1, Problem 57CRE

b.

To determine

To find if the inverse of the function exists:

For the interval $(\infty, 0)$ , $f (x) = x^{2}$

Let us assume that if $f (- a) = f (- b)$

Then, ${(- a)}^{2} = {(- b)}^{2}$

$- a = \pm b$

Since, the interval is $(\infty, 0)$ , hence $- a = - b$

Thus, the inverse of the function $f (x) = x^{2}$ exists.

For the interval $[0, \infty)$ , $f (x) = x$

Let us assume that if $f (a) = f (b)$

Then, $a = b$

Thus, the inverse of the function $f (x) = x$ exists.

Hence, we get that the inverse of the function $f (x) = {- x^{2} w h e n x < 0 a n d x w h e n x \geq 0}$ exists.

Inverse of the function:

For the interval $(\infty, 0)$ , $f (x) = x^{2}$

Replacing $f (x)$ with $y$ , we get

$y = x^{2}$

On switching $y$ and $x$ , we get

$x = y^{2}$

Solving for $y$ , we get

$y^{2} = x$

$y = \sqrt{x}$

Replacing $y$ with $f^{- 1} (x)$ , we get

$f^{- 1} (x) = \sqrt{x}$

For the interval $[0, \infty)$ , $f (x) = x$

Replacing $f (x)$ with $y$ , we get

$y = x$

On switching $y$ and $x$ , we get

$x = y$

Solving for $y$ , we get

$y = x$

Replacing $y$ with $f^{- 1} (x)$ , we get

$f^{- 1} (x) = x$

Thus, Inverse of function is $f^{- 1} (x) = {\sqrt{x} w h e n x < 0 a n d x w h e n x \geq 0}$

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Answer 3

Question

Chapter 1, Problem 57CRE

b.

To determine

To find if the inverse of the function exists:

For the interval $(\infty, 0)$ , $f (x) = x^{2}$

Let us assume that if $f (- a) = f (- b)$

Then, ${(- a)}^{2} = {(- b)}^{2}$

$- a = \pm b$

Since, the interval is $(\infty, 0)$ , hence $- a = - b$

Thus, the inverse of the function $f (x) = x^{2}$ exists.

For the interval $[0, \infty)$ , $f (x) = x$

Let us assume that if $f (a) = f (b)$

Then, $a = b$

Thus, the inverse of the function $f (x) = x$ exists.

Hence, we get that the inverse of the function $f (x) = {- x^{2} w h e n x < 0 a n d x w h e n x \geq 0}$ exists.

Inverse of the function:

For the interval $(\infty, 0)$ , $f (x) = x^{2}$

Replacing $f (x)$ with $y$ , we get

$y = x^{2}$

On switching $y$ and $x$ , we get

$x = y^{2}$

Solving for $y$ , we get

$y^{2} = x$

$y = \sqrt{x}$

Replacing $y$ with $f^{- 1} (x)$ , we get

$f^{- 1} (x) = \sqrt{x}$

For the interval $[0, \infty)$ , $f (x) = x$

Replacing $f (x)$ with $y$ , we get

$y = x$

On switching $y$ and $x$ , we get

$x = y$

Solving for $y$ , we get

$y = x$

Replacing $y$ with $f^{- 1} (x)$ , we get

$f^{- 1} (x) = x$

Thus, Inverse of function is $f^{- 1} (x) = {\sqrt{x} w h e n x < 0 a n d x w h e n x \geq 0}$

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Answer 4

Question

Chapter 1, Problem 57CRE

b.

To determine

To find if the inverse of the function exists:

For the interval $(\infty, 0)$ , $f (x) = x^{2}$

Let us assume that if $f (- a) = f (- b)$

Then, ${(- a)}^{2} = {(- b)}^{2}$

$- a = \pm b$

Since, the interval is $(\infty, 0)$ , hence $- a = - b$

Thus, the inverse of the function $f (x) = x^{2}$ exists.

For the interval $[0, \infty)$ , $f (x) = x$

Let us assume that if $f (a) = f (b)$

Then, $a = b$

Thus, the inverse of the function $f (x) = x$ exists.

Hence, we get that the inverse of the function $f (x) = {- x^{2} w h e n x < 0 a n d x w h e n x \geq 0}$ exists.

Inverse of the function:

For the interval $(\infty, 0)$ , $f (x) = x^{2}$

Replacing $f (x)$ with $y$ , we get

$y = x^{2}$

On switching $y$ and $x$ , we get

$x = y^{2}$

Solving for $y$ , we get

$y^{2} = x$

$y = \sqrt{x}$

Replacing $y$ with $f^{- 1} (x)$ , we get

$f^{- 1} (x) = \sqrt{x}$

For the interval $[0, \infty)$ , $f (x) = x$

Replacing $f (x)$ with $y$ , we get

$y = x$

On switching $y$ and $x$ , we get

$x = y$

Solving for $y$ , we get

$y = x$

Replacing $y$ with $f^{- 1} (x)$ , we get

$f^{- 1} (x) = x$

Thus, Inverse of function is $f^{- 1} (x) = {\sqrt{x} w h e n x < 0 a n d x w h e n x \geq 0}$

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Answer 5

Chapter 1, Problem 57CRE

b.

To determine

To find if the inverse of the function exists:

For the interval $(\infty, 0)$ , $f (x) = x^{2}$

Let us assume that if $f (- a) = f (- b)$

Then, ${(- a)}^{2} = {(- b)}^{2}$

$- a = \pm b$

Since, the interval is $(\infty, 0)$ , hence $- a = - b$

Thus, the inverse of the function $f (x) = x^{2}$ exists.

For the interval $[0, \infty)$ , $f (x) = x$

Let us assume that if $f (a) = f (b)$

Then, $a = b$

Thus, the inverse of the function $f (x) = x$ exists.

Hence, we get that the inverse of the function $f (x) = {- x^{2} w h e n x < 0 a n d x w h e n x \geq 0}$ exists.

Inverse of the function:

For the interval $(\infty, 0)$ , $f (x) = x^{2}$

Replacing $f (x)$ with $y$ , we get

$y = x^{2}$

On switching $y$ and $x$ , we get

$x = y^{2}$

Solving for $y$ , we get

$y^{2} = x$

$y = \sqrt{x}$

Replacing $y$ with $f^{- 1} (x)$ , we get

$f^{- 1} (x) = \sqrt{x}$

For the interval $[0, \infty)$ , $f (x) = x$

Replacing $f (x)$ with $y$ , we get

$y = x$

On switching $y$ and $x$ , we get

$x = y$

Solving for $y$ , we get

$y = x$

Replacing $y$ with $f^{- 1} (x)$ , we get

$f^{- 1} (x) = x$

Thus, Inverse of function is $f^{- 1} (x) = {\sqrt{x} w h e n x < 0 a n d x w h e n x \geq 0}$

Answer 6

Chapter 1, Problem 57CRE

b.

To determine

To find if the inverse of the function exists:

For the interval $(\infty, 0)$ , $f (x) = x^{2}$

Let us assume that if $f (- a) = f (- b)$

Then, ${(- a)}^{2} = {(- b)}^{2}$

$- a = \pm b$

Since, the interval is $(\infty, 0)$ , hence $- a = - b$

Thus, the inverse of the function $f (x) = x^{2}$ exists.

For the interval $[0, \infty)$ , $f (x) = x$

Let us assume that if $f (a) = f (b)$

Then, $a = b$

Thus, the inverse of the function $f (x) = x$ exists.

Hence, we get that the inverse of the function $f (x) = {- x^{2} w h e n x < 0 a n d x w h e n x \geq 0}$ exists.

Inverse of the function:

For the interval $(\infty, 0)$ , $f (x) = x^{2}$

Replacing $f (x)$ with $y$ , we get

$y = x^{2}$

On switching $y$ and $x$ , we get

$x = y^{2}$

Solving for $y$ , we get

$y^{2} = x$

$y = \sqrt{x}$

Replacing $y$ with $f^{- 1} (x)$ , we get

$f^{- 1} (x) = \sqrt{x}$

For the interval $[0, \infty)$ , $f (x) = x$

Replacing $f (x)$ with $y$ , we get

$y = x$

On switching $y$ and $x$ , we get

$x = y$

Solving for $y$ , we get

$y = x$

Replacing $y$ with $f^{- 1} (x)$ , we get

$f^{- 1} (x) = x$

Thus, Inverse of function is $f^{- 1} (x) = {\sqrt{x} w h e n x < 0 a n d x w h e n x \geq 0}$

Answer 7

Chapter 1, Problem 57CRE

b.

To determine

To find if the inverse of the function exists:

For the interval $(\infty, 0)$ , $f (x) = x^{2}$

Let us assume that if $f (- a) = f (- b)$

Then, ${(- a)}^{2} = {(- b)}^{2}$

$- a = \pm b$

Since, the interval is $(\infty, 0)$ , hence $- a = - b$

Thus, the inverse of the function $f (x) = x^{2}$ exists.

For the interval $[0, \infty)$ , $f (x) = x$

Let us assume that if $f (a) = f (b)$

Then, $a = b$

Thus, the inverse of the function $f (x) = x$ exists.

Hence, we get that the inverse of the function $f (x) = {- x^{2} w h e n x < 0 a n d x w h e n x \geq 0}$ exists.

Inverse of the function:

For the interval $(\infty, 0)$ , $f (x) = x^{2}$

Replacing $f (x)$ with $y$ , we get

$y = x^{2}$

On switching $y$ and $x$ , we get

$x = y^{2}$

Solving for $y$ , we get

$y^{2} = x$

$y = \sqrt{x}$

Replacing $y$ with $f^{- 1} (x)$ , we get

$f^{- 1} (x) = \sqrt{x}$

For the interval $[0, \infty)$ , $f (x) = x$

Replacing $f (x)$ with $y$ , we get

$y = x$

On switching $y$ and $x$ , we get

$x = y$

Solving for $y$ , we get

$y = x$

Replacing $y$ with $f^{- 1} (x)$ , we get

$f^{- 1} (x) = x$

Thus, Inverse of function is $f^{- 1} (x) = {\sqrt{x} w h e n x < 0 a n d x w h e n x \geq 0}$

Answer 8

Expert Solution & Answer

Answer 9

Expert Solution & Answer

Answer 10

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Answer 11

Ch. 1.1 - Prob. 1PQ Ch. 1.1 - Prob. 2PQ Ch. 1.1 - Prob. 3PQ Ch. 1.1 - Prob. 4PQ Ch. 1.1 - Prob. 5PQ Ch. 1.1 - Prob. 6PQ Ch. 1.1 - Prob. 7PQ Ch. 1.1 - Prob. 8PQ Ch. 1.1 - Prob. 9PQ Ch. 1.1 - Prob. 10PQ

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