Find the Domain of 3/√(x-2): Square Root Function Analysis

Q: Why can't x equal 2 if the square root of 0 exists?

Great question! While $ \sqrt{0} = 0 $ is perfectly valid, having zero in the denominator makes the fraction undefined. Think of it like asking "what's 3 ÷ 0?" - it has no answer!

Q: What's the difference between x ≥ 2 and x > 2?

The symbol ≥ means "greater than or equal to" while > means "strictly greater than." Since x = 2 makes our denominator zero, we must exclude x = 2, so we use the strict inequality x > 2.

Q: How do I find what makes the square root undefined?

Set the expression under the square root less than zero. For $ \sqrt{x-2} $, solve x - 2

Q: Do I need to check both the square root AND the fraction?

Yes! First check that x - 2 ≥ 0 for the square root to exist. Then check that the denominator ≠ 0. Combine both conditions to get your final domain.

Q: Can the domain ever be empty for this type of function?

It's possible! If the conditions conflict (like needing x > 5 AND x

Domain Analysis with Square Root Restrictions

Look at the following function:

$\frac{3}{\sqrt{x-2}}$

What is the domain of the function?

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Does the function have a domain? If so, what is it?

00:04 A root must be for a positive number greater than 0

00:10 Let's isolate X

00:16 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Look at the following function:

$\frac{3}{\sqrt{x-2}}$

What is the domain of the function?

Step-by-step solution

To determine the domain of the function $\frac{3}{\sqrt{x-2}}$ , we need to consider the constraints imposed by the square root and the fraction.

First, the expression inside the square root must be non-negative: $x - 2 \geq 0$ . This simplifies to:

$x \geq 2$

However, because the expression $\sqrt{x-2}$ is in the denominator of the fraction, we must also ensure that it is not equal to zero, as division by zero is undefined. Therefore, we have:

$x - 2 \neq 0$

Solving this inequality gives:

$x \neq 2$

Together, the solution to both conditions is that $x$ must be greater than 2 to ensure the function is defined. Thus, the domain of the function is:

$x > 2$

Therefore, the correct choice for the domain is $x > 2$ .

Final Answer

$x > 2$

Key Points to Remember

Essential concepts to master this topic

Square Root Rule: Expression under radical must be non-negative
Denominator Restriction: $\sqrt{x-2} \neq 0$ means x-2 > 0
Check Domain: Test x=3: $\frac{3}{\sqrt{3-2}} = \frac{3}{1} = 3$ ✓

Common Mistakes

Avoid these frequent errors

Using x ≥ 2 instead of x > 2
Don't forget the denominator cannot equal zero! Using x ≥ 2 allows x = 2, which makes $\sqrt{x-2} = 0$ and creates division by zero. Always exclude values that make denominators zero.

Practice Quiz

Test your knowledge with interactive questions

$22(\frac{2}{x}-1)=30$

What is the domain of the equation above?

FAQ

Everything you need to know about this question

Why can't x equal 2 if the square root of 0 exists?

Great question! While $\sqrt{0} = 0$ is perfectly valid, having zero in the denominator makes the fraction undefined. Think of it like asking "what's 3 ÷ 0?" - it has no answer!

What's the difference between x ≥ 2 and x > 2?

The symbol ≥ means "greater than or equal to" while > means "strictly greater than." Since x = 2 makes our denominator zero, we must exclude x = 2, so we use the strict inequality x > 2.

How do I find what makes the square root undefined?

Set the expression under the square root less than zero. For $\sqrt{x-2}$ , solve x - 2 < 0 to get x < 2. These are the values that make the square root undefined.

Do I need to check both the square root AND the fraction?

Yes! First check that x - 2 ≥ 0 for the square root to exist. Then check that the denominator ≠ 0. Combine both conditions to get your final domain.

Can the domain ever be empty for this type of function?

It's possible! If the conditions conflict (like needing x > 5 AND x < 3 at the same time), then no values of x work and the domain would be empty. But that's not the case here.

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