Determine the Domain: Analyzing (5x+15)/(10x+1/2) for Validity

Q: Why do I only worry about the denominator being zero?

Division by zero is undefined in mathematics! When the denominator equals zero, the function has no value at that point. The numerator can be zero (that just makes the whole function equal zero), but a zero denominator breaks the function completely.

Q: How do I solve equations with fractions like this one?

When you have $ 10x + \frac{1}{2} = 0 $, multiply everything by 2 to clear the fraction: $ 20x + 1 = 0 $. Then solve normally!

Q: What does the domain notation mean exactly?

The domain is all real numbers except the values that make the denominator zero. We write $ x \ne -\frac{1}{20} $ to show that $ x $ can be any real number except $ -\frac{1}{20} $.

Rational Function Domains with Fractional Terms

Given the following function:

$\frac{5x+15}{10x+\frac{1}{2}}$

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? And if so, what is it?

00:04 To find the domain, remember that division by 0 is not allowed

00:09 Therefore let's see what solution makes the denominator zero

00:14 Let's isolate X

00:45 Let's multiply by the reciprocal

00:53 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:

$\frac{5x+15}{10x+\frac{1}{2}}$

What is the domain of the function?

Step-by-step solution

To find the domain of the function $\frac{5x+15}{10x+\frac{1}{2}}$ , we must ensure the denominator is not zero.

The critical expression to consider is the denominator:

$10x + \frac{1}{2}$

Let's solve the equation:

Set the denominator equal to zero: $10x + \frac{1}{2} = 0$ .
To clear the fraction, multiply everything by 2: $2(10x) + 2\left(\frac{1}{2}\right) = 0$ .
This simplifies to: $20x + 1 = 0$ .
Subtract 1 from both sides: $20x = -1$ .
Divide by 20: $x = -\frac{1}{20}$ .

Thus, the function is undefined when $x = -\frac{1}{20}$ . Consequently, the domain of the function is all real numbers except $x = -\frac{1}{20}$ .

Therefore, the solution to the problem is $x \ne -\frac{1}{20}$ .

Final Answer

$x\ne-\frac{1}{20}$

Key Points to Remember

Essential concepts to master this topic

Domain Rule: Denominator cannot equal zero in any rational function
Technique: Set $10x + \frac{1}{2} = 0$ and solve for restricted value
Check: Substitute $x = -\frac{1}{20}$ : denominator becomes 0 ✓

Common Mistakes

Avoid these frequent errors

Setting the numerator equal to zero instead of denominator
Don't solve $5x + 15 = 0$ to find domain restrictions = gives wrong excluded value! The numerator being zero just means the function equals zero, not that it's undefined. Always set the denominator equal to zero to find domain restrictions.

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 do I only worry about the denominator being zero?

Division by zero is undefined in mathematics! When the denominator equals zero, the function has no value at that point. The numerator can be zero (that just makes the whole function equal zero), but a zero denominator breaks the function completely.

How do I solve equations with fractions like this one?

When you have $10x + \frac{1}{2} = 0$ , multiply everything by 2 to clear the fraction: $20x + 1 = 0$ . Then solve normally!

What does the domain notation mean exactly?

The domain is all real numbers except the values that make the denominator zero. We write $x \ne -\frac{1}{20}$ to show that $x$ can be any real number except $-\frac{1}{20}$ .

Can a rational function have more than one restricted value?

Yes! If the denominator factors into multiple terms, each factor that equals zero gives a different restriction. For example, $\frac{1}{(x-2)(x+3)}$ excludes both $x = 2$ and $x = -3$ .

How can I check if my domain answer is correct?

Substitute your excluded value back into the original denominator. If it makes the denominator equal exactly zero, then you found the right restriction! For $x = -\frac{1}{20}$ : $10(-\frac{1}{20}) + \frac{1}{2} = -\frac{1}{2} + \frac{1}{2} = 0$ ✓

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