91Ó°ÊÓ

When dealing with fractions in mathematical functions, one crucial rule is that the denominator can never be zero. This is because dividing by zero is undefined, meaning it doesn't result in a finite or meaningful number.

For the function given, namely \[ f(s) = \frac{\sqrt{s-1}}{s-4} \]we look specifically at the denominator, which is \( s-4 \).

To avoid division by zero, set the denominator equal to zero and solve for the variable:

• Equation: \( s-4 = 0 \)
• Solution: \( s = 4 \)

Thus, the number 4 is a "no-go" in the context of this function because it would make the denominator zero, causing the function to be undefined at \( s = 4 \). This point is excluded from the domain of the function.

In mathematical functions involving a square root, you must ensure that the radicand—the expression under the square root—is not negative. Square roots with negative radicands are not real numbers and aren't defined in basic real-number arithmetic.

For the function \[ f(s) = \frac{\sqrt{s-1}}{s-4} \]the radicand is \( s-1 \).

To ensure it stays non-negative, set it up as an inequality:

• Equation: \( s-1 \geq 0 \)
• Solution: \( s \geq 1 \)

This tells us that for \( s = 1 \) and values greater than 1, the expression under the square root becomes zero or positive, keeping the function in real numbers without any mathematical hiccups.

To establish the domain of the function, you need to consider both conditions: avoiding a zero denominator and a non-negative radicand. Each condition places its own restriction on the variable \( s \).

From our previous analysis, we know that:

• The denominator restriction ensures \( s eq 4 \)
• The radicand condition ensures \( s \geq 1 \)

The domain of the function, therefore, merges these restrictions. It includes all numbers greater than or equal to 1, but excludes the number 4, as:

• Overall Domain: \( s \geq 1 \) and \( s eq 4 \)

Hence, writing this in interval notation, the domain is \[ [1, 4) \cup (4, \infty) \]This effectively covers values starting at 1 up to, but not including, 4 and then continues from just above 4 to infinity. This visualization ensures clarity, avoiding any misunderstandings about what values \( s \) can actually take.