Problem 9 - College Algebra Essentials

Question

Find the domain of each function. $$f(x)=\frac{1}{x+7}+\frac{3}{x-9}$$

Step-by-Step Solution

Answer

The domain of the function \(f(x)=\frac{1}{x+7}+\frac{3}{x-9}\) is \((-\infty, -7) \cup (-7, 9) \cup (9, \infty)\).

Step 1: Identify the Zeroes of the Denominator

In the given function \(f(x)=\frac{1}{x+7}+\frac{3}{x-9}\), the denominators are \(x+7\) and \(x-9\). For the domain, x cannot be a value that makes the denominator equal to zero. So, set \(x+7=0\) and \(x-9=0\) and solve for x.

Step 2: Solve the Equation x+7=0

Subtract 7 from both sides of the equation \(x+7=0\). This gives \(x=-7\). That means the function is not defined when x equals -7.

Step 3: Solve the Equation x-9=0

Add 9 to both sides of the equation \(x-9=0\). This gives \(x=9\). So the function is also not defined when x equals 9.

Step 4: Write Down the Domain of the Function

The domain of \(f(x)=\frac{1}{x+7}+\frac{3}{x-9}\) is all real numbers except x=-7 and x=9. In interval notation, this would be \((-\infty, -7) \cup (-7, 9) \cup (9, \infty)\).