Find the Domain of 12/√(4x-4): Square Root Function Analysis

Q: Why can't x equal 1 if the square root of 0 is 0?

While $ \sqrt{0} = 0 $, we have division by zero which is undefined! The function becomes $ \frac{12}{0} $, not just $ \sqrt{0} $.

Q: What's the difference between domain restrictions for √x and 1/√x?

For $ \sqrt{x} $, you need x ≥ 0. But for $ \frac{1}{\sqrt{x}} $, you need x > 0 because the denominator cannot be zero!

Q: How do I write the domain using interval notation?

The domain $ x > 1 $ in interval notation is (1, ∞). Use a parenthesis ( because x = 1 is not included in the domain.

Q: Can I simplify 4x - 4 before solving?

Yes! Factor out 4: $ 4x - 4 = 4(x - 1) $. Then $ 4(x - 1) > 0 $ gives $ x - 1 > 0 $, so x > 1.

Domain Analysis with Radical Denominators

Look at the following function:

$\frac{12}{\sqrt{4x-4}}$

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? And if so, what is it?

00:03 Root must be for a positive number greater than 0

00:10 Let's isolate X

00:22 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:

$\frac{12}{\sqrt{4x-4}}$

What is the domain of the function?

Step-by-step solution

To find the domain of the function $\frac{12}{\sqrt{4x-4}}$ , let's analyze the conditions necessary for the function to be defined.

The expression under the square root, $4x - 4$ , must be positive, as the square root of a negative number is not defined in the real numbers, and division by zero is undefined. Therefore, we need:

$4x - 4 > 0$

Solve this inequality step by step:

Add 4 to both sides: $4x - 4 + 4 > 0 + 4$ , which simplifies to $4x > 4$ .
Divide both sides by 4: $\frac{4x}{4} > \frac{4}{4}$ , which simplifies to $x > 1$ .

The inequality $x > 1$ describes the domain of the function.

Therefore, the domain of the function $\frac{12}{\sqrt{4x-4}}$ is $x > 1$ .

Final Answer

$x > 1$

Key Points to Remember

Essential concepts to master this topic

Rule: Expression under square root must be strictly positive
Technique: Set 4x - 4 > 0, solve to get x > 1
Check: Test x = 2: √(4(2)-4) = √4 = 2, function equals 6 ✓

Common Mistakes

Avoid these frequent errors

Using ≥ instead of > for the inequality
Don't write 4x - 4 ≥ 0 giving x ≥ 1 = includes x = 1 which makes denominator zero! When x = 1, we get 12/√0 which is undefined. Always use strict inequality > to exclude values that make the denominator 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 1 if the square root of 0 is 0?

While $\sqrt{0} = 0$ , we have division by zero which is undefined! The function becomes $\frac{12}{0}$ , not just $\sqrt{0}$ .

What's the difference between domain restrictions for √x and 1/√x?

For $\sqrt{x}$ , you need x ≥ 0. But for $\frac{1}{\sqrt{x}}$ , you need x > 0 because the denominator cannot be zero!

How do I write the domain using interval notation?

The domain $x > 1$ in interval notation is (1, ∞). Use a parenthesis ( because x = 1 is not included in the domain.

Can I simplify 4x - 4 before solving?

Yes! Factor out 4: $4x - 4 = 4(x - 1)$ . Then $4(x - 1) > 0$ gives $x - 1 > 0$ , so x > 1.

What if the problem asked for the range instead?

For the range, consider that as x approaches 1 from the right, the function approaches +∞, and as x approaches +∞, the function approaches 0. So the range is $(0, +∞)$ .

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