Answer :
To determine the domain of the function [tex]\( f(x) = \sqrt{(x + 4)(6 - x)} + \frac{2}{\sqrt{|x| - 3}} \)[/tex], we need to find all [tex]\( x \)[/tex]-values for which the expression under the square root and the denominator in the fraction are valid and defined. Here’s the step-by-step process:
### 1. Condition from the First Square Root:
The first part of the function is [tex]\( \sqrt{(x + 4)(6 - x)} \)[/tex]. For the square root to be defined, the expression inside it must be non-negative:
[tex]\[ (x + 4)(6 - x) \geq 0 \][/tex]
To solve [tex]\( (x + 4)(6 - x) \geq 0 \)[/tex], we find the roots of the quadratic expression.
The roots are:
[tex]\[ x + 4 = 0 \implies x = -4 \][/tex]
[tex]\[ 6 - x = 0 \implies x = 6 \][/tex]
Now, we consider the intervals determined by these roots: [tex]\( (-\infty, -4) \)[/tex], [tex]\( (-4, 6) \)[/tex], and [tex]\( (6, \infty) \)[/tex].
We check the sign of [tex]\( (x + 4)(6 - x) \)[/tex] in each of these intervals:
- For [tex]\( x \)[/tex] in [tex]\( (-\infty, -4) \)[/tex]: Both [tex]\( x + 4 \)[/tex] and [tex]\( 6 - x \)[/tex] are negative, so the product is positive.
- For [tex]\( x \)[/tex] in [tex]\( (-4, 6) \)[/tex]: The sign of [tex]\( (x + 4) \)[/tex] and [tex]\( (6 - x) \)[/tex] will both make one positive and one negative, so the product is negative.
- For [tex]\( x \)[/tex] in [tex]\( (6, \infty) \)[/tex]: Both [tex]\( x + 4 \)[/tex] and [tex]\( 6 - x \)[/tex] are positive, so the product is positive.
Therefore, [tex]\( (x + 4)(6 - x) \geq 0 \)[/tex] has solutions [tex]\( x \in [-4, 6] \)[/tex].
### 2. Condition from the Second Square Root and Fraction:
The second part of the function is [tex]\( \frac{2}{\sqrt{|x| - 3}} \)[/tex]. For this expression to be defined, the denominator must not be zero and the radicand (the term inside the square root) must be positive:
[tex]\[ |x| - 3 > 0 \][/tex]
This simplifies to:
[tex]\[ |x| > 3 \][/tex]
Which implies:
[tex]\[ x > 3 \quad \text{or} \quad x < -3 \][/tex]
### 3. Combined Conditions:
We must find the intersection of the intervals from both conditions:
1. [tex]\( x \in [-4, 6] \)[/tex]
2. [tex]\( x > 3 \)[/tex] or [tex]\( x < -3 \)[/tex]
Combining these:
- From [tex]\( x \in [-4, 6] \)[/tex] and [tex]\( x > 3 \)[/tex], we get [tex]\( 3 < x \leq 6 \)[/tex].
- From [tex]\( x \in [-4, 6] \)[/tex] and [tex]\( x < -3 \)[/tex], we get [tex]\( -4 \leq x < -3 \)[/tex].
Thus, the domain is:
[tex]\[ [-4, -3) \cup (3, 6] \][/tex]
### Conclusion:
The domain of the function is [tex]\( [-4, -3) \cup (3, 6] \)[/tex]. This corresponds to option (3).
[tex]\[ \boxed{[-4, -3) \cup (3, 6]} \][/tex]
### 1. Condition from the First Square Root:
The first part of the function is [tex]\( \sqrt{(x + 4)(6 - x)} \)[/tex]. For the square root to be defined, the expression inside it must be non-negative:
[tex]\[ (x + 4)(6 - x) \geq 0 \][/tex]
To solve [tex]\( (x + 4)(6 - x) \geq 0 \)[/tex], we find the roots of the quadratic expression.
The roots are:
[tex]\[ x + 4 = 0 \implies x = -4 \][/tex]
[tex]\[ 6 - x = 0 \implies x = 6 \][/tex]
Now, we consider the intervals determined by these roots: [tex]\( (-\infty, -4) \)[/tex], [tex]\( (-4, 6) \)[/tex], and [tex]\( (6, \infty) \)[/tex].
We check the sign of [tex]\( (x + 4)(6 - x) \)[/tex] in each of these intervals:
- For [tex]\( x \)[/tex] in [tex]\( (-\infty, -4) \)[/tex]: Both [tex]\( x + 4 \)[/tex] and [tex]\( 6 - x \)[/tex] are negative, so the product is positive.
- For [tex]\( x \)[/tex] in [tex]\( (-4, 6) \)[/tex]: The sign of [tex]\( (x + 4) \)[/tex] and [tex]\( (6 - x) \)[/tex] will both make one positive and one negative, so the product is negative.
- For [tex]\( x \)[/tex] in [tex]\( (6, \infty) \)[/tex]: Both [tex]\( x + 4 \)[/tex] and [tex]\( 6 - x \)[/tex] are positive, so the product is positive.
Therefore, [tex]\( (x + 4)(6 - x) \geq 0 \)[/tex] has solutions [tex]\( x \in [-4, 6] \)[/tex].
### 2. Condition from the Second Square Root and Fraction:
The second part of the function is [tex]\( \frac{2}{\sqrt{|x| - 3}} \)[/tex]. For this expression to be defined, the denominator must not be zero and the radicand (the term inside the square root) must be positive:
[tex]\[ |x| - 3 > 0 \][/tex]
This simplifies to:
[tex]\[ |x| > 3 \][/tex]
Which implies:
[tex]\[ x > 3 \quad \text{or} \quad x < -3 \][/tex]
### 3. Combined Conditions:
We must find the intersection of the intervals from both conditions:
1. [tex]\( x \in [-4, 6] \)[/tex]
2. [tex]\( x > 3 \)[/tex] or [tex]\( x < -3 \)[/tex]
Combining these:
- From [tex]\( x \in [-4, 6] \)[/tex] and [tex]\( x > 3 \)[/tex], we get [tex]\( 3 < x \leq 6 \)[/tex].
- From [tex]\( x \in [-4, 6] \)[/tex] and [tex]\( x < -3 \)[/tex], we get [tex]\( -4 \leq x < -3 \)[/tex].
Thus, the domain is:
[tex]\[ [-4, -3) \cup (3, 6] \][/tex]
### Conclusion:
The domain of the function is [tex]\( [-4, -3) \cup (3, 6] \)[/tex]. This corresponds to option (3).
[tex]\[ \boxed{[-4, -3) \cup (3, 6]} \][/tex]
