Determine the Domain of: √(3x²+7)/12

Q: Why doesn't the square root create any restrictions here?

The expression $ 3x^2 + 7 $ is always positive! Since $ x^2 ≥ 0 $ for any real number, we have $ 3x^2 ≥ 0 $, and adding 7 makes it at least 7.

Q: What's the smallest value under the square root?

The minimum occurs when $ x^2 = 0 $ (at x = 0), giving us $ 3(0) + 7 = 7 $. Since 7 > 0, we can take its square root!

Q: Does the denominator 12 affect the domain?

No! The denominator is a positive constant (12), so it never equals zero. Only the numerator with the square root determines the domain restrictions.

Q: How do I check if a quadratic expression is always positive?

For $ ax^2 + b $ where a > 0: if b ≥ 0, it's always positive. Here, a = 3 > 0 and b = 7 > 0, so $ 3x^2 + 7 > 0 $ for all x.

Q: What would make this function have a restricted domain?

If we had something like $ \sqrt{3x^2 - 7} $ instead, we'd need $ 3x^2 - 7 ≥ 0 $, which would restrict x to certain intervals.

Domain Determination with Square Root Functions

Look at the following function:

$\frac{\sqrt{3x^2+7}}{12}$

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?

00:04 The domain in the denominator is to avoid division by 0

00:07 Therefore, from the denominator's perspective there is no domain restriction, let's check the numerator

00:12 A root must be for a positive number

00:15 Let's isolate X

00:25 The square of any number is always positive

00:33 Therefore there is no solution that makes the root negative

00:39 Thus 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

Look at the following function:

$\frac{\sqrt{3x^2+7}}{12}$

What is the domain of the function?

Step-by-step solution

To find the domain of the function $\frac{\sqrt{3x^2+7}}{12}$ , we must ensure that the function is defined for all real numbers.

Step 1: Evaluate the expression under the square root, $3x^2 + 7$ , which must be non-negative. Since it's a quadratic expression in the form of $ax^2 + b$ , compute for any potential zero or negative range.

Step 2: Notice that $3x^2 + 7 \geq 0$ for all $x$ because $3x^2 \geq 0$ for any real number $x$ and adding 7 makes this entire expression always positive (i.e., tends upwards away from zero).

Step 3: As the denominator $12$ is a positive constant, it imposes no additional restrictions on the domain. Thus, the function is defined wherever the numerator is defined.

Conclusion: Since there's no $x$ that makes $3x^2 + 7 < 0$ , the function is defined for all real numbers.

This means the domain of the function is all real numbers, confirmed by choice number 1: $\text{All real numbers}$ .

Final Answer

All real numbers

Key Points to Remember

Essential concepts to master this topic

Domain Rule: Expression under square root must be non-negative
Analysis: For $3x^2 + 7$ , minimum value is 7 when x = 0
Verification: Test any x-value: $3(0)^2 + 7 = 7 > 0$ ✓

Common Mistakes

Avoid these frequent errors

Setting the expression under the square root equal to zero
Don't solve $3x^2 + 7 = 0$ and think it restricts the domain! This equation has no real solutions, which confirms all values work. Always remember that square roots need the expression to be ≥ 0, and $3x^2 + 7$ is always positive.

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 doesn't the square root create any restrictions here?

The expression $3x^2 + 7$ is always positive! Since $x^2 ≥ 0$ for any real number, we have $3x^2 ≥ 0$ , and adding 7 makes it at least 7.

What's the smallest value under the square root?

The minimum occurs when $x^2 = 0$ (at x = 0), giving us $3(0) + 7 = 7$ . Since 7 > 0, we can take its square root!

Does the denominator 12 affect the domain?

No! The denominator is a positive constant (12), so it never equals zero. Only the numerator with the square root determines the domain restrictions.

How do I check if a quadratic expression is always positive?

For $ax^2 + b$ where a > 0: if b ≥ 0, it's always positive. Here, a = 3 > 0 and b = 7 > 0, so $3x^2 + 7 > 0$ for all x.

What would make this function have a restricted domain?

If we had something like $\sqrt{3x^2 - 7}$ instead, we'd need $3x^2 - 7 ≥ 0$ , which would restrict x to certain intervals.

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