Answer :
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]
