Determine the Domain of the Rational Function 12/(8x-4)

Q: Why can't the denominator be zero?

Division by zero is undefined in mathematics! When the denominator equals zero, the function has no value at that point, creating a vertical asymptote on the graph.

Q: What does 'x ≠ 1/2' actually mean?

This means x can be any real number except 1/2. You can use 0.49, 0.51, -100, or 1000 - just not exactly $ \frac{1}{2} $!

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

The domain is $ (-\infty, \frac{1}{2}) \cup (\frac{1}{2}, \infty) $. The union symbol ∪ combines the two intervals, and parentheses show that 1/2 is excluded.

Q: What if the denominator has multiple terms like this one?

Treat it like any equation! Set 8x - 4 = 0 and solve step by step. Add 4 to both sides, then divide by 8 to get your restricted value.

Q: Can a rational function have more than one restricted value?

Absolutely! If the denominator factors into $ (x-2)(x+3) $, then both x = 2 and x = -3 would be excluded from the domain.

Rational Function Domain with Zero Denominators

Given the following function:

$\frac{12}{8x-4}$

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

00:07 So let's find the solution that makes the denominator zero

00:10 Let's isolate X

00:26 Let's factor 8 into 4 and 2

00:29 Let's reduce what we can

00:32 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:

$\frac{12}{8x-4}$

What is the domain of the function?

Step-by-step solution

To find the domain of the function $\frac{12}{8x-4}$ , we must determine when the denominator equals zero and exclude these values.

Step 1: Set the denominator equal to zero and solve for $x$ :

$8x - 4 = 0$

Step 2: Solve the equation $8x - 4 = 0$ for $x$ :

Add 4 to both sides: $8x = 4$

Divide both sides by 8: $x = \frac{4}{8} = \frac{1}{2}$

Step 3: The value $x = \frac{1}{2}$ is where the denominator becomes zero, so this value is excluded from the domain.

Therefore, the domain of the function is all real numbers except $x = \frac{1}{2}$ .

The domain of the function is $\boxed{x \ne \frac{1}{2}}$ .

Final Answer

$x\ne\frac{1}{2}$

Key Points to Remember

Essential concepts to master this topic

Domain Rule: Exclude values that make the denominator equal zero
Technique: Set 8x - 4 = 0, then solve: x = 1/2
Check: Substitute x = 1/2: 8(1/2) - 4 = 0 ✓

Common Mistakes

Avoid these frequent errors

Setting the entire fraction equal to zero instead of just the denominator
Don't set 12/(8x-4) = 0 to find domain restrictions = this finds zeros, not restrictions! This gives you when the function equals zero, not when it's undefined. Always set only the denominator equal to zero to find domain restrictions.

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 the denominator be zero?

Division by zero is undefined in mathematics! When the denominator equals zero, the function has no value at that point, creating a vertical asymptote on the graph.

What does 'x ≠ 1/2' actually mean?

This means x can be any real number except 1/2. You can use 0.49, 0.51, -100, or 1000 - just not exactly $\frac{1}{2}$ !

How do I write the domain in interval notation?

The domain is $(-\infty, \frac{1}{2}) \cup (\frac{1}{2}, \infty)$ . The union symbol ∪ combines the two intervals, and parentheses show that 1/2 is excluded.

What if the denominator has multiple terms like this one?

Treat it like any equation! Set 8x - 4 = 0 and solve step by step. Add 4 to both sides, then divide by 8 to get your restricted value.

Can a rational function have more than one restricted value?

Absolutely! If the denominator factors into $(x-2)(x+3)$ , then both x = 2 and x = -3 would be excluded from the domain.

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