Consider the function.
[tex]f(x)=x^2+3[/tex]
Which answer pairs a possible domain restriction that if placed upon [tex]f(x)[/tex] would impact [tex]f^{-1}(x)[/tex] as shown?
A. [tex]f(x)[/tex] domain: [tex]x \geq 3[/tex]
[tex]f^{-1}(x)[/tex] domain: [tex]x \geq 3[/tex]
B. [tex]f(x)[/tex] domain: [tex]x \geq 3[/tex]
[tex]f^{-1}(x)[/tex] range: [tex]y \geq 3[/tex]
C. [tex]f(x)[/tex] domain: [tex]x \geq 0[/tex]
[tex]f^{-1}(x)[/tex] domain: [tex]x \geq 0[/tex]
D. [tex]f(x)[/tex] domain: [tex]x \geq 0[/tex]
[tex]f^{-1}(x)[/tex] range: [tex]y \geq 0[/tex]
Answer (1)
Find the inverse function f − 1 ( x ) = ± x − 3 .
Analyze each option by considering the domain restriction on f ( x ) and its impact on the range of f ( x ) , which becomes the domain of f − 1 ( x ) .
Option 1 is incorrect because if x ≥ 3 , then the domain of f − 1 ( x ) is x ≥ 12 , not x ≥ 3 .
Option 4 is correct because if x ≥ 0 , then the range of f ( x ) is f ( x ) ≥ 3 , so the domain of f − 1 ( x ) is x ≥ 3 , and the range of f − 1 ( x ) is y ≥ 0 .
The correct answer is f ( x ) domain: x ≥ 0 , f − 1 ( x ) range: y ≥ 0 .
Explanation
Understanding the Problem We are given the function f ( x ) = x 2 + 3 and asked to determine which domain restriction on f ( x ) would impact the domain or range of its inverse function f − 1 ( x ) . We will analyze each option to see if the given restriction on f ( x ) leads to the stated impact on f − 1 ( x ) .
Finding the Inverse Function First, let's find the inverse function f − 1 ( x ) . To do this, we replace f ( x ) with y , so y = x 2 + 3 . Then we swap x and y to get x = y 2 + 3 . Solving for y , we have y 2 = x − 3 , so y = ± x − 3 . Therefore, f − 1 ( x ) = ± x − 3 .
Analyzing Each Option Now, let's analyze each option:
Option 1: f ( x ) domain: x ≥ 3 , f − 1 ( x ) domain: x ≥ 3 If f ( x ) has the domain x ≥ 3 , then the range of f ( x ) is f ( x ) ≥ ( 3 ) 2 + 3 = 12 . This means the domain of f − 1 ( x ) would be x ≥ 12 . Since the given domain of f − 1 ( x ) is x ≥ 3 , this option is incorrect.
Option 2: f ( x ) domain: x ≥ 3 , f − 1 ( x ) range: y ≥ 3 If f ( x ) has the domain x ≥ 3 , then the range of f ( x ) is f ( x ) ≥ ( 3 ) 2 + 3 = 12 . This means the domain of f − 1 ( x ) would be x ≥ 12 . Also, since x ≥ 3 , f − 1 ( x ) = x − 3 , so the range of f − 1 ( x ) is y ≥ 0 . Since the given range of f − 1 ( x ) is y ≥ 3 , this option is incorrect.
Option 3: f ( x ) domain: x ≥ ?, f − 1 ( x ) domain: x ≥ 0 The domain of f − 1 ( x ) is x ≥ 3 because the expression inside the square root must be non-negative, i.e., x − 3 ≥ 0 . So, x ≥ 3 . Therefore, it is not possible for the domain of f − 1 ( x ) to be x ≥ 0 . This option is incorrect.
Option 4: f ( x ) domain: x ≥ 0 , f − 1 ( x ) range: y ≥ 0 If f ( x ) has the domain x ≥ 0 , then the range of f ( x ) is f ( x ) ≥ ( 0 ) 2 + 3 = 3 . This means the domain of f − 1 ( x ) would be x ≥ 3 . Since f ( x ) has the domain x ≥ 0 , we take the positive square root for the inverse function, so f − 1 ( x ) = x − 3 . The range of f − 1 ( x ) is y ≥ 0 . This option is correct.
Final Answer Therefore, the correct answer is: f ( x ) domain: x ≥ 0 f − 1 ( x ) range: y ≥ 0
Examples
Understanding the domain and range of functions and their inverses is crucial in many real-world applications. For example, in cryptography, the domain and range of encryption and decryption functions must be carefully chosen to ensure that the original message can be recovered without ambiguity. Similarly, in engineering, when designing a system, it's important to understand the limitations of the input (domain) and the possible outputs (range) to ensure the system operates correctly and safely. By understanding these concepts, you can better analyze and design systems that are reliable and efficient.
