Determine the Domain of the Function: Analyzing (2x+2)/√(x-16)

Q: Why can't x equal 16 if the square root of 0 is defined?

While $ \sqrt{0} = 0 $ is mathematically valid, having zero in the denominator makes the entire fraction undefined. Division by zero is never allowed in mathematics!

Q: What's the difference between x > 16 and x ≥ 16?

x > 16 means x cannot equal 16, while x ≥ 16 means x can equal 16. Since our denominator becomes zero when x = 16, we must use the strict inequality x > 16.

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

The domain $ x > 16 $ in interval notation is (16, ∞). Use a parenthesis at 16 because 16 is not included in the domain.

Q: What if the numerator also had restrictions?

Always find all restrictions separately, then take their intersection. For $ \frac{2x+2}{\sqrt{x-16}} $, only the denominator creates restrictions since 2x + 2 is defined for all real numbers.

Domain Restrictions with Square Root Denominators

Look at the following function:

$\frac{2x+2}{\sqrt{x-16}}$

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 A root must be for a positive number greater than 0

00:09 Let's isolate X

00:15 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:

$\frac{2x+2}{\sqrt{x-16}}$

What is the domain of the function?

Step-by-step solution

To determine the domain of the function $\frac{2x+2}{\sqrt{x-16}}$ , follow these steps:

Step 1: Identify the constraint imposed by the square root in the denominator.
Step 2: Solve the inequality $x-16 > 0$ .
Step 3: Interpret the solution in terms of the domain.

Let's proceed:

Step 1: The function $\frac{2x+2}{\sqrt{x-16}}$ has a square root in the denominator. For the square root to be defined in the real number system and prevent division by zero, the expression under the square root, $x-16$ , must be greater than zero.

Step 2: Solve the inequality:

$x - 16 > 0$

Add 16 to both sides:

$x > 16$

Step 3: The solution $x > 16$ means that the domain of the function is all real numbers greater than 16.

Therefore, the domain of the function is $x > 16$ , which corresponds to choice 2.

Final Answer

$x > 16$

Key Points to Remember

Essential concepts to master this topic

Square Root Rule: Expression under square root must be positive
Technique: Solve x - 16 > 0 to get x > 16
Check: Test x = 17: √(17-16) = √1 = 1 ✓

Common Mistakes

Avoid these frequent errors

Using ≥ instead of > for square root denominators
Don't write x ≥ 16 when there's a square root in the denominator = division by zero at x = 16! When x = 16, we get √(16-16) = √0 = 0 in the denominator, making the function undefined. Always use x > 16 to exclude the zero point.

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 16 if the square root of 0 is defined?

While $\sqrt{0} = 0$ is mathematically valid, having zero in the denominator makes the entire fraction undefined. Division by zero is never allowed in mathematics!

What's the difference between x > 16 and x ≥ 16?

x > 16 means x cannot equal 16, while x ≥ 16 means x can equal 16. Since our denominator becomes zero when x = 16, we must use the strict inequality x > 16.

How do I write the domain in interval notation?

The domain $x > 16$ in interval notation is (16, ∞). Use a parenthesis at 16 because 16 is not included in the domain.

What if the numerator also had restrictions?

Always find all restrictions separately, then take their intersection. For $\frac{2x+2}{\sqrt{x-16}}$ , only the denominator creates restrictions since 2x + 2 is defined for all real numbers.

Can I simplify the numerator first?

You can factor 2x + 2 = 2(x + 1), but this doesn't change the domain restrictions. The square root in the denominator is still the only constraint that matters.

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