At Westonci.ca, we connect you with experts who provide detailed answers to your most pressing questions. Start exploring now! Connect with a community of experts ready to provide precise solutions to your questions on our user-friendly Q&A platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
Sagot :
To solve this problem, we need to make the function [tex]\( f(x)=x^2+3 \)[/tex] one-to-one by restricting its domain and then find its inverse.
First, let's understand why the function [tex]\( f(x)=x^2+3 \)[/tex] is not one-to-one on its natural domain. By definition, a function is one-to-one if every value in its domain maps to a unique value in its range.
Consider [tex]\( f(x) = x^2 + 3 \)[/tex]:
- For [tex]\( x = -2 \)[/tex], [tex]\( f(-2) = 4 + 3 = 7 \)[/tex]
- For [tex]\( x = 2 \)[/tex], [tex]\( f(2) = 4 + 3 = 7 \)[/tex]
Since different values of [tex]\( x \)[/tex] (specifically, [tex]\( -2 \)[/tex] and [tex]\( 2 \)[/tex]) produce the same value of [tex]\( f(x) \)[/tex] (which is [tex]\( 7 \)[/tex]), the function is not one-to-one over the entire real number line.
To make [tex]\( f(x) \)[/tex] one-to-one, we need to restrict its domain.
### Exploring Possible Restrictions
Let's evaluate the given options one by one:
#### 1. Domain: [tex]\( x \leq 0 \)[/tex]; Inverse: [tex]\( f^{-1}(x)=\sqrt{x-3} \)[/tex]
- When [tex]\( x \leq 0 \)[/tex], [tex]\( f(x) \leq 3 \)[/tex].
- However, [tex]\( \sqrt{x-3} \)[/tex] is only defined for [tex]\( x \geq 3 \)[/tex], leading to a contradiction since [tex]\( x-3 \)[/tex] is non-negative only if [tex]\( x \geq 3 \)[/tex].
#### 2. Domain: [tex]\( x \geq 0 \)[/tex]; Inverse: [tex]\( f^{-1}(x)=\sqrt{x+3} \)[/tex]
- With [tex]\( x \geq 0 \)[/tex], [tex]\( f(x) \geq 3 \)[/tex].
- However, the inverse should undo the function [tex]\( f(x) = x^2 + 3 \)[/tex]. Here, [tex]\( \sqrt{x+3} \)[/tex] does not represent this inversion correctly, as [tex]\( x + 3 \)[/tex] should not be the term under the square root.
#### 3. Domain: [tex]\( x \geq 0 \)[/tex]; Inverse: [tex]\( f^{-1}(x)=\sqrt{x-3} \)[/tex]
- When [tex]\( x \geq 0 \)[/tex], [tex]\( f(x) \geq 3 \)[/tex].
- For the inverse, we start with [tex]\( y = x^2 + 3 \)[/tex], and solving for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex] gives [tex]\( x = \sqrt{y-3} \)[/tex]. Therefore, this works.
#### 4. Domain: [tex]\( x \geq 3 \)[/tex]; Inverse: [tex]\( f^{-1}(x)=\sqrt{x-3} \)[/tex]
- If the domain is [tex]\( x \geq 3 \)[/tex], we're not starting from the logical definition of the quadratic expression's domain restriction from [tex]\( x\geq0 \)[/tex].
### Conclusion
The correct way to restrict the domain so that the function is one-to-one and find its inverse is:
- Restricted domain: [tex]\( x \geq 0 \)[/tex]
- Inverse function: [tex]\( f^{-1}(x)=\sqrt{x-3} \)[/tex]
Thus, the valid solution is:
[tex]\[ \text{Restricted domain: } x \geq 0, \quad f^{-1}(x) = \sqrt{x-3} \][/tex]
First, let's understand why the function [tex]\( f(x)=x^2+3 \)[/tex] is not one-to-one on its natural domain. By definition, a function is one-to-one if every value in its domain maps to a unique value in its range.
Consider [tex]\( f(x) = x^2 + 3 \)[/tex]:
- For [tex]\( x = -2 \)[/tex], [tex]\( f(-2) = 4 + 3 = 7 \)[/tex]
- For [tex]\( x = 2 \)[/tex], [tex]\( f(2) = 4 + 3 = 7 \)[/tex]
Since different values of [tex]\( x \)[/tex] (specifically, [tex]\( -2 \)[/tex] and [tex]\( 2 \)[/tex]) produce the same value of [tex]\( f(x) \)[/tex] (which is [tex]\( 7 \)[/tex]), the function is not one-to-one over the entire real number line.
To make [tex]\( f(x) \)[/tex] one-to-one, we need to restrict its domain.
### Exploring Possible Restrictions
Let's evaluate the given options one by one:
#### 1. Domain: [tex]\( x \leq 0 \)[/tex]; Inverse: [tex]\( f^{-1}(x)=\sqrt{x-3} \)[/tex]
- When [tex]\( x \leq 0 \)[/tex], [tex]\( f(x) \leq 3 \)[/tex].
- However, [tex]\( \sqrt{x-3} \)[/tex] is only defined for [tex]\( x \geq 3 \)[/tex], leading to a contradiction since [tex]\( x-3 \)[/tex] is non-negative only if [tex]\( x \geq 3 \)[/tex].
#### 2. Domain: [tex]\( x \geq 0 \)[/tex]; Inverse: [tex]\( f^{-1}(x)=\sqrt{x+3} \)[/tex]
- With [tex]\( x \geq 0 \)[/tex], [tex]\( f(x) \geq 3 \)[/tex].
- However, the inverse should undo the function [tex]\( f(x) = x^2 + 3 \)[/tex]. Here, [tex]\( \sqrt{x+3} \)[/tex] does not represent this inversion correctly, as [tex]\( x + 3 \)[/tex] should not be the term under the square root.
#### 3. Domain: [tex]\( x \geq 0 \)[/tex]; Inverse: [tex]\( f^{-1}(x)=\sqrt{x-3} \)[/tex]
- When [tex]\( x \geq 0 \)[/tex], [tex]\( f(x) \geq 3 \)[/tex].
- For the inverse, we start with [tex]\( y = x^2 + 3 \)[/tex], and solving for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex] gives [tex]\( x = \sqrt{y-3} \)[/tex]. Therefore, this works.
#### 4. Domain: [tex]\( x \geq 3 \)[/tex]; Inverse: [tex]\( f^{-1}(x)=\sqrt{x-3} \)[/tex]
- If the domain is [tex]\( x \geq 3 \)[/tex], we're not starting from the logical definition of the quadratic expression's domain restriction from [tex]\( x\geq0 \)[/tex].
### Conclusion
The correct way to restrict the domain so that the function is one-to-one and find its inverse is:
- Restricted domain: [tex]\( x \geq 0 \)[/tex]
- Inverse function: [tex]\( f^{-1}(x)=\sqrt{x-3} \)[/tex]
Thus, the valid solution is:
[tex]\[ \text{Restricted domain: } x \geq 0, \quad f^{-1}(x) = \sqrt{x-3} \][/tex]
Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.
