Finding the Domain of Radical Function √3x² + 3 / 9

Q: Why can't we have negative numbers under a square root?

In the real number system, square roots of negative numbers are undefined. That's why we need $ 3x^2 + 3 \geq 0 $ to ensure our function produces real outputs.

Q: How do I know if x² + 1 is always positive?

Since $ x^2 \geq 0 $ for any real number, adding 1 gives us $ x^2 + 1 \geq 1 $. The minimum value is 1 when x = 0, so it's never negative!

Q: What if the denominator was zero instead of 9?

Great question! If we had division by zero, we'd need to exclude those x-values from the domain. But since 9 ≠ 0, the denominator doesn't affect our domain here.

Q: Do all radical functions have restricted domains?

Not always! Some expressions under square roots (like $ x^2 + 1 $) are always positive, giving us all real numbers as the domain. Others do restrict the domain.

Domain of Radical Functions with Always-Positive Expressions

Given the following function:

$\frac{\sqrt{3x^2+3}}{9}$

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?

00:03 The domain in the denominator is to avoid division by 0

00:08 Therefore, from the denominator's perspective there is no domain restriction, let's check the numerator

00:11 A root must be for a positive number

00:16 Let's equate to 0 to find the solutions

00:24 Let's isolate X

00:32 Any number squared will always be positive, therefore there is no solution

00:41 The function exists for all X, 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

Given the following function:

$\frac{\sqrt{3x^2+3}}{9}$

What is the domain of the function?

Step-by-step solution

To determine the domain of the function $\frac{\sqrt{3x^2+3}}{9}$ , we need to ensure that the expression under the square root is non-negative:

$3x^2 + 3 \geq 0$

Simplifying this inequality, we can factor it:

Factor out the common term: $3(x^2 + 1) \geq 0$ .
Since $3$ is a positive constant, we focus on $x^2 + 1 \geq 0$ .
The term $x^2$ is always non-negative, hence $x^2 + 1$ is always positive for any real number $x$ , as the smallest value it can take, when $x = 0$ , is 1.

Thus, $x^2 + 1$ is never negative, making the expression under the square root always non-negative.

Therefore, the domain of the function is all real numbers.

Emphasizing the conclusion: The entire domain of this function is all real numbers.

Final Answer

The entire domain

Key Points to Remember

Essential concepts to master this topic

Domain Rule: Expression under square root must be non-negative
Analysis Technique: Factor $3x^2 + 3 = 3(x^2 + 1)$ to simplify
Verification: Check that $x^2 + 1 \geq 1$ for all real numbers ✓

Common Mistakes

Avoid these frequent errors

Setting the radicand equal to zero instead of greater than or equal to zero
Don't solve 3x² + 3 = 0 for exact values = missing the entire domain! This only finds where the expression equals zero, not where it's non-negative. Always use the inequality 3x² + 3 ≥ 0 to find all valid x-values.

Practice Quiz

Test your knowledge with interactive questions

$\frac{6}{x+5}=1$

What is the field of application of the equation?

FAQ

Everything you need to know about this question

Why can't we have negative numbers under a square root?

In the real number system, square roots of negative numbers are undefined. That's why we need $3x^2 + 3 \geq 0$ to ensure our function produces real outputs.

How do I know if x² + 1 is always positive?

Since $x^2 \geq 0$ for any real number, adding 1 gives us $x^2 + 1 \geq 1$ . The minimum value is 1 when x = 0, so it's never negative!

What if the denominator was zero instead of 9?

Great question! If we had division by zero, we'd need to exclude those x-values from the domain. But since 9 ≠ 0, the denominator doesn't affect our domain here.

Do all radical functions have restricted domains?

Not always! Some expressions under square roots (like $x^2 + 1$ ) are always positive, giving us all real numbers as the domain. Others do restrict the domain.

How can I check my domain answer is correct?

Pick a few test values from your domain and substitute them into the original function. If you get real number outputs, you're on the right track! Also verify any boundary points work.

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