Questions: Find the domain of the function. f(x) = sqrt(x-3) / (x+10)(x-11) [3, infinity) (-10,11) union (11, infinity) (11, infinity) (-10,3] union [3, infinity) [3,11) union (11, infinity)

Solution Steps

To find the domain of the function \( f(x) = \frac{\sqrt{x-3}}{(x+10)(x-11)} \), we need to consider the constraints imposed by the square root and the denominator. The expression under the square root, \( x-3 \), must be non-negative, so \( x \geq 3 \). Additionally, the denominator cannot be zero, so \( x \neq -10 \) and \( x \neq 11 \). Combining these conditions, the domain is \( [3, 11) \cup (11, \infty) \).

The numerator of the function is \(\sqrt{x-3}\). The expression under the square root, \(x-3\), must be non-negative for the square root to be defined. Therefore, we have:

\[ x - 3 \geq 0 \]

Solving this inequality gives:

\[ x \geq 3 \]

Thus, the domain of the numerator is \([3, \infty)\).

The denominator of the function is \((x+10)(x-11)\). This expression is undefined when either factor is zero. Therefore, we need to find the values of \(x\) that make the denominator zero:

1. \(x + 10 = 0 \Rightarrow x = -10\)
2. \(x - 11 = 0 \Rightarrow x = 11\)

Thus, the denominator is undefined at \(x = -10\) and \(x = 11\).

The function \(f(x)\) is defined where both the numerator is defined and the denominator is non-zero. Therefore, we combine the conditions:

• \(x \geq 3\) (from the numerator)
• \(x \neq -10\) and \(x \neq 11\) (from the denominator)

The domain of the function is the intersection of these conditions:

• From \(x \geq 3\), we have \([3, \infty)\).
• Excluding \(x = 11\) from \([3, \infty)\), we get \([3, 11) \cup (11, \infty)\).

Final Answer