Answer :
Certainly! Let's analyze the function [tex]\( f(x) = \sqrt{49 - x^2} + \sqrt{x^2 - 1} \)[/tex].
To understand this function better, we need to consider the domains of the two square root functions separately, since the square roots impose constraints on the values that [tex]\( x \)[/tex] can take.
### Step 1: Determine the Domain
1. First Square Root Term: [tex]\(\sqrt{49 - x^2}\)[/tex]
- For this term to be real, the radicand (the expression inside the square root) must be non-negative.
- That is, [tex]\( 49 - x^2 \geq 0 \)[/tex].
- Solving for [tex]\( x \)[/tex]:
[tex]\[ 49 \geq x^2 \][/tex]
[tex]\[ x^2 \leq 49 \][/tex]
[tex]\[ -7 \leq x \leq 7 \][/tex]
2. Second Square Root Term: [tex]\(\sqrt{x^2 - 1}\)[/tex]
- For this term to be real, the radicand must also be non-negative.
- That is, [tex]\( x^2 - 1 \geq 0 \)[/tex].
- Solving for [tex]\( x \)[/tex]:
[tex]\[ x^2 \geq 1 \][/tex]
[tex]\[ x \leq -1 \quad \text{or} \quad x \geq 1 \][/tex]
### Step 2: Combine the Domain Restrictions
We need [tex]\( x \)[/tex] to satisfy both conditions simultaneously:
- From [tex]\(\sqrt{49 - x^2}\)[/tex]: [tex]\(-7 \leq x \leq 7\)[/tex]
- From [tex]\(\sqrt{x^2 - 1}\)[/tex]: [tex]\( x \leq -1 \quad \text{or} \quad x \geq 1 \)[/tex]
Combining these, the valid domain for both functions is:
[tex]\[ x \in [-7, -1] \cup [1, 7] \][/tex]
### Step 3: Evaluate [tex]\( f(x) \)[/tex]
Given that [tex]\( f(x) = \sqrt{49 - x^2} + \sqrt{x^2 - 1} \)[/tex], we can substitute specific values of [tex]\( x \)[/tex] within the valid domain to observe how the function behaves. Here, evaluating at some selected points gives us insight:
1. At [tex]\( x = -7 \)[/tex]:
[tex]\[ f(-7) = \sqrt{49 - (-7)^2} + \sqrt{(-7)^2 - 1} \][/tex]
[tex]\[ f(-7) = \sqrt{49 - 49} + \sqrt{49 - 1} \][/tex]
[tex]\[ f(-7) = \sqrt{0} + \sqrt{48} \][/tex]
[tex]\[ f(-7) = 0 + 4\sqrt{3} \][/tex]
[tex]\[ f(-7) = 4\sqrt{3} \][/tex]
2. At [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = \sqrt{49 - 1^2} + \sqrt{1^2 - 1} \][/tex]
[tex]\[ f(1) = \sqrt{49 - 1} + \sqrt{1 - 1} \][/tex]
[tex]\[ f(1) = \sqrt{48} + \sqrt{0} \][/tex]
[tex]\[ f(1) = 4\sqrt{3} + 0 \][/tex]
[tex]\[ f(1) = 4\sqrt{3} \][/tex]
3. At [tex]\( x = 7 \)[/tex]:
[tex]\[ f(7) = \sqrt{49 - 7^2} + \sqrt{7^2 - 1} \][/tex]
[tex]\[ f(7) = \sqrt{49 - 49} + \sqrt{49 - 1} \][/tex]
[tex]\[ f(7) = \sqrt{0} + \sqrt{48} \][/tex]
[tex]\[ f(7) = 0 + 4\sqrt{3} \][/tex]
[tex]\[ f(7) = 4\sqrt{3} \][/tex]
### Summary
The function [tex]\( f(x) = \sqrt{49 - x^2} + \sqrt{x^2 - 1} \)[/tex] is defined for [tex]\( x \in [-7, -1] \cup [1, 7] \)[/tex], and its specific values depend on [tex]\( x \)[/tex] within the domain. For instance, at [tex]\( x = -7 \)[/tex], [tex]\( x = 1 \)[/tex], and [tex]\( x = 7 \)[/tex], we find that [tex]\( f(x) \)[/tex] evaluates to [tex]\( 4 \sqrt{3} \)[/tex]. Each computation respects the constraint imposed by the square root conditions, ensuring that the function remains real and well-defined.
To understand this function better, we need to consider the domains of the two square root functions separately, since the square roots impose constraints on the values that [tex]\( x \)[/tex] can take.
### Step 1: Determine the Domain
1. First Square Root Term: [tex]\(\sqrt{49 - x^2}\)[/tex]
- For this term to be real, the radicand (the expression inside the square root) must be non-negative.
- That is, [tex]\( 49 - x^2 \geq 0 \)[/tex].
- Solving for [tex]\( x \)[/tex]:
[tex]\[ 49 \geq x^2 \][/tex]
[tex]\[ x^2 \leq 49 \][/tex]
[tex]\[ -7 \leq x \leq 7 \][/tex]
2. Second Square Root Term: [tex]\(\sqrt{x^2 - 1}\)[/tex]
- For this term to be real, the radicand must also be non-negative.
- That is, [tex]\( x^2 - 1 \geq 0 \)[/tex].
- Solving for [tex]\( x \)[/tex]:
[tex]\[ x^2 \geq 1 \][/tex]
[tex]\[ x \leq -1 \quad \text{or} \quad x \geq 1 \][/tex]
### Step 2: Combine the Domain Restrictions
We need [tex]\( x \)[/tex] to satisfy both conditions simultaneously:
- From [tex]\(\sqrt{49 - x^2}\)[/tex]: [tex]\(-7 \leq x \leq 7\)[/tex]
- From [tex]\(\sqrt{x^2 - 1}\)[/tex]: [tex]\( x \leq -1 \quad \text{or} \quad x \geq 1 \)[/tex]
Combining these, the valid domain for both functions is:
[tex]\[ x \in [-7, -1] \cup [1, 7] \][/tex]
### Step 3: Evaluate [tex]\( f(x) \)[/tex]
Given that [tex]\( f(x) = \sqrt{49 - x^2} + \sqrt{x^2 - 1} \)[/tex], we can substitute specific values of [tex]\( x \)[/tex] within the valid domain to observe how the function behaves. Here, evaluating at some selected points gives us insight:
1. At [tex]\( x = -7 \)[/tex]:
[tex]\[ f(-7) = \sqrt{49 - (-7)^2} + \sqrt{(-7)^2 - 1} \][/tex]
[tex]\[ f(-7) = \sqrt{49 - 49} + \sqrt{49 - 1} \][/tex]
[tex]\[ f(-7) = \sqrt{0} + \sqrt{48} \][/tex]
[tex]\[ f(-7) = 0 + 4\sqrt{3} \][/tex]
[tex]\[ f(-7) = 4\sqrt{3} \][/tex]
2. At [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = \sqrt{49 - 1^2} + \sqrt{1^2 - 1} \][/tex]
[tex]\[ f(1) = \sqrt{49 - 1} + \sqrt{1 - 1} \][/tex]
[tex]\[ f(1) = \sqrt{48} + \sqrt{0} \][/tex]
[tex]\[ f(1) = 4\sqrt{3} + 0 \][/tex]
[tex]\[ f(1) = 4\sqrt{3} \][/tex]
3. At [tex]\( x = 7 \)[/tex]:
[tex]\[ f(7) = \sqrt{49 - 7^2} + \sqrt{7^2 - 1} \][/tex]
[tex]\[ f(7) = \sqrt{49 - 49} + \sqrt{49 - 1} \][/tex]
[tex]\[ f(7) = \sqrt{0} + \sqrt{48} \][/tex]
[tex]\[ f(7) = 0 + 4\sqrt{3} \][/tex]
[tex]\[ f(7) = 4\sqrt{3} \][/tex]
### Summary
The function [tex]\( f(x) = \sqrt{49 - x^2} + \sqrt{x^2 - 1} \)[/tex] is defined for [tex]\( x \in [-7, -1] \cup [1, 7] \)[/tex], and its specific values depend on [tex]\( x \)[/tex] within the domain. For instance, at [tex]\( x = -7 \)[/tex], [tex]\( x = 1 \)[/tex], and [tex]\( x = 7 \)[/tex], we find that [tex]\( f(x) \)[/tex] evaluates to [tex]\( 4 \sqrt{3} \)[/tex]. Each computation respects the constraint imposed by the square root conditions, ensuring that the function remains real and well-defined.
