Questions: Find the domain of the following rational function. R(x)=8(x^2-3x-70)/(9(x^2-100)) The domain is (Type your answer in interval notation. Use integers or fractions for any numbers in the expression.)

Find the domain of the following rational function.
R(x)=8(x^2-3x-70)/(9(x^2-100))

The domain is 
(Type your answer in interval notation. Use integers or fractions for any numbers in the expression.)
Transcript text: Find the domain of the following rational function. \[ R(x)=\frac{8\left(x^{2}-3 x-70\right)}{9\left(x^{2}-100\right)} \] The domain is $\square$ (Type your answer in interval notation. Use integers or fractions for any numbers in the expression.)
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Solution

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Solution Steps

To find the domain of the rational function \( R(x) = \frac{8(x^2 - 3x - 70)}{9(x^2 - 100)} \), we need to determine the values of \( x \) for which the denominator is not zero. The domain will be all real numbers except those that make the denominator zero.

  1. Identify the denominator of the function: \( 9(x^2 - 100) \).
  2. Set the denominator equal to zero and solve for \( x \): \( 9(x^2 - 100) = 0 \).
  3. Solve the equation \( x^2 - 100 = 0 \) to find the values of \( x \) that are not in the domain.
  4. Express the domain in interval notation, excluding the values found in step 3.
Step 1: Identify the Denominator

The given rational function is: \[ R(x) = \frac{8(x^2 - 3x - 70)}{9(x^2 - 100)} \] The denominator of the function is: \[ 9(x^2 - 100) \]

Step 2: Set the Denominator Equal to Zero

To find the values of \( x \) that are not in the domain, we set the denominator equal to zero: \[ 9(x^2 - 100) = 0 \]

Step 3: Solve for \( x \)

Solving the equation \( x^2 - 100 = 0 \): \[ x^2 - 100 = 0 \] \[ x^2 = 100 \] \[ x = \pm 10 \] The values \( x = -10 \) and \( x = 10 \) make the denominator zero.

Step 4: Express the Domain in Interval Notation

The domain of the function is all real numbers except \( x = -10 \) and \( x = 10 \). In interval notation, this is: \[ (-\infty, -10) \cup (-10, 10) \cup (10, \infty) \]

Final Answer

\[ \boxed{(-\infty, -10) \cup (-10, 10) \cup (10, \infty)} \]

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