Questions: Section 4.5 - Rational Functions and Their Graphs Question 8 Find the domain of the given function. Enter the solution using interval notation. p(x)=-8/(1+x^2) Domain: Find the domain of the given function. Enter the solution using interval notation. g(x)=-1/(64-x^2) Domain:

Solution Steps

To find the domain of a rational function, we need to determine the values of \( x \) for which the denominator is not equal to zero. For the given functions, we will set the denominators equal to zero and solve for \( x \). The domain will be all real numbers except those that make the denominator zero.

1. For \( p(x) = -\frac{8}{1 + x^2} \):

   • The denominator is \( 1 + x^2 \).
   • Set \( 1 + x^2 \neq 0 \).
   • Solve for \( x \).

2. For \( g(x) = -\frac{1}{64 - x^2} \):

   • The denominator is \( 64 - x^2 \).
   • Set \( 64 - x^2 \neq 0 \).
   • Solve for \( x \).

For the function \( p(x) = -\frac{8}{1 + x^2} \), the denominator is \( 1 + x^2 \). This expression is never equal to zero for any real number \( x \) since \( x^2 \geq 0 \) for all \( x \). Therefore, the domain of \( p(x) \) is all real numbers.

Thus, we can express the domain in interval notation as: \[ \text{Domain of } p(x): \quad (-\infty, \infty) \]

For the function \( g(x) = -\frac{1}{64 - x^2} \), we need to find when the denominator \( 64 - x^2 \) is not equal to zero. Setting the denominator to zero gives: \[ 64 - x^2 = 0 \implies x^2 = 64 \implies x = \pm 8 \] Thus, the values \( x = -8 \) and \( x = 8 \) are excluded from the domain. Therefore, the domain of \( g(x) \) can be expressed in interval notation as: \[ \text{Domain of } g(x): \quad (-\infty, -8) \cup (-8, 8) \cup (8, \infty) \]

Final Answer