To find the domain of the function [tex]\( f(x) = \sqrt{\frac{77}{x^2 - x - 56}} \)[/tex], we need to ensure that the expression inside the square root is defined and non-negative. This means two things:
1. The denominator [tex]\( x^2 - x - 56 \)[/tex] must be non-zero because division by zero is undefined.
2. The expression inside the square root must be positive for the square root to be a real number (in other words, it must be greater than zero because the square root of zero is zero and hence does not affect the function being non-negative).
Let's proceed step by step:
### Step 1: Setting the Denominator to Non-zero
The denominator [tex]\( x^2 - x - 56 \)[/tex] should not be equal to zero:
[tex]\[
x^2 - x - 56 \neq 0
\][/tex]
We solve the equation [tex]\( x^2 - x - 56 = 0 \)[/tex] for [tex]\( x \)[/tex]:
[tex]\[
x^2 - x - 56 = 0
\][/tex]
Factoring the quadratic equation:
[tex]\[
(x - 8)(x + 7) = 0
\][/tex]
So, the solutions are:
[tex]\[
x = 8 \quad \text{or} \quad x = -7
\][/tex]
This means that the values [tex]\( x = 8 \)[/tex] and [tex]\( x = -7 \)[/tex] make the denominator zero, thus these values should be excluded from the domain.
### Step 2: Ensuring the Expression Inside the Square Root is Positive
Next, we need the expression inside the square root, [tex]\( \frac{77}{x^2 - x - 56} \)[/tex], to be positive:
[tex]\[
\frac{77}{x^2 - x - 56} > 0
\][/tex]
Since [tex]\( 77 \)[/tex] is a positive constant, the inequality reduces to:
[tex]\[
x^2 - x - 56 > 0
\][/tex]
We analyze the quadratic inequality by considering the roots at [tex]\( x = 8 \)[/tex] and [tex]\( x = -7 \)[/tex]. We need to determine in which intervals the quadratic expression is positive. We test the intervals divided by the critical points [tex]\( x = -7 \)[/tex] and [tex]\( x = 8 \)[/tex]:
1. For [tex]\( x < -7 \)[/tex]:
Choose a test point, [tex]\( x = -8 \)[/tex]:
[tex]\[
(-8)^2 - (-8) - 56 = 64 + 8 - 56 = 16 > 0
\][/tex]
Thus, the expression is positive in the interval [tex]\(( -\infty, -7 )\)[/tex].
2. For [tex]\( -7 < x < 8 \)[/tex]:
Choose a test point, [tex]\( x = 0 \)[/tex]:
[tex]\[
0^2 - 0 - 56 = -56 < 0
\][/tex]
Thus, the expression is negative in the interval [tex]\(( -7, 8 )\)[/tex].
3. For [tex]\( x > 8 \)[/tex]:
Choose a test point, [tex]\( x = 9 \)[/tex]:
[tex]\[
9^2 - 9 - 56 = 81 - 9 - 56 = 16 > 0
\][/tex]
Thus, the expression is positive in the interval [tex]\(( 8, \infty )\)[/tex].
### Conclusion
Combining the intervals where the expression is positive and excluding the points where the denominator is zero, the domain of [tex]\( f(x) \)[/tex] is:
[tex]\[
(-\infty, -7) \cup (8, \infty)
\][/tex]
Hence, the domain of [tex]\( f(x) \)[/tex] in interval notation is:
[tex]\[
(-\infty, -7) \cup (8, \infty)
\][/tex]
