To find the domain of the function $f(x) = \frac{1}{\sqrt{6x - 6}}$, we need to consider the restrictions that the square root and the denominator impose on the function.

Step 1: Non-negativity of the Square Root Argument

The expression inside the square root, $6x - 6$, must be non-negative because the square root of a negative number is not a real number. Therefore, we set up the inequality: $6x - 6 \geq 0$

Step 2: Solve the Inequality

Solve the inequality for $x$: $6x \geq 6$ $x \geq 1$

Step 3: Denominator Cannot Be Zero

The denominator $\sqrt{6x - 6}$ must not be zero, because division by zero is undefined. Therefore, we set up the condition: $\sqrt{6x - 6} \neq 0$ This implies: $6x - 6 \neq 0$ $6x \neq 6$ $x \neq 1$

Step 4: Combine the Conditions

From Step 2, we found $x \geq 1$. However, from Step 3, we found that $x$ cannot be equal to 1. Thus, the domain is all values of $x$ greater than 1.

Final Answer

The domain of $f(x) = \frac{1}{\sqrt{6x - 6}}$ is: $\boxed{(1, \infty)}$

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Related Questions:

1. What is the domain of $f(x) = \frac{1}{\sqrt{4x + 2}}$?
2. How do you find the domain of a function with both a square root and a logarithm?
3. What is the range of $f(x) = \frac{1}{\sqrt{6x - 6}}$?
4. How do you determine the domain of a function involving a cube root?
5. Can you find the domain of $g(x) = \sqrt{x^2 - 4x + 3}$?

Tip: Always start by checking if the function involves square roots or denominators when determining its domain.