Finding the Domain of the Function: Simplifying 9/(3x+1/2)

Q: Why can't the denominator be zero?

Division by zero is undefined in mathematics! When the denominator equals zero, the function has no meaningful value, so we must exclude those x-values from the domain.

Q: How do I solve 3x + 1/2 = 0?

First, subtract $ \frac{1}{2} $ from both sides: $ 3x = -\frac{1}{2} $. Then divide by 3: $ x = -\frac{1}{6} $. This is the value to exclude from the domain!

Q: What does the ≠ symbol mean?

The symbol ≠ means "not equal to." When we write $ x \ne -\frac{1}{6} $, we're saying x can be any real number except $ -\frac{1}{6} $.

Q: How do I write the complete domain?

You can write it as: "All real numbers except $ x = -\frac{1}{6} $" or in interval notation: $ (-\infty, -\frac{1}{6}) \cup (-\frac{1}{6}, \infty) $.

Q: What if the denominator has multiple terms?

The same rule applies! Set the entire denominator equal to zero and solve. Even with expressions like $ x^2 - 4 $, you'd solve $ x^2 - 4 = 0 $ to find restrictions.

Domain of Rational Functions with Fractions

Look at the following function:

$\frac{9}{3x+\frac{1}{2}}$

What is its domain?

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:06 Does the function have a domain? Let's find out!

00:10 Remember, dividing by zero is not allowed in math.

00:14 So, let's discover when the denominator becomes zero.

00:18 To do that, we'll isolate X. Ready? Here we go!

00:36 Multiply by the reciprocal. Remember, it's numerator with numerator and denominator with denominator.

00:45 Great job! And that's how we find the solution.

Step-by-step written solution

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Understand the problem

Look at the following function:

$\frac{9}{3x+\frac{1}{2}}$

What is its domain?

Step-by-step solution

To find the domain of the function $\frac{9}{3x+\frac{1}{2}}$ , we need to determine for which values of $x$ the expression is defined.

1. The function is of the form $\frac{9}{3x+\frac{1}{2}}$ . A fraction is undefined if its denominator is zero.

2. Set the denominator of the function equal to zero: $3x + \frac{1}{2} = 0$ .

3. Solve the equation for $x$ :

First, subtract $\frac{1}{2}$ from both sides: $3x = -\frac{1}{2}$ .
Next, divide both sides by 3 to solve for $x$ : $x = -\frac{1}{2} \times \frac{1}{3} = -\frac{1}{6}$ .

4. The solution $x = -\frac{1}{6}$ indicates that at this value, the denominator becomes zero, making the function undefined. Therefore, $x \ne -\frac{1}{6}$ .

Thus, the domain of the function is all real numbers except $x = -\frac{1}{6}$ .

The correct answer is $x \ne -\frac{1}{6}$ , which matches choice 2: $x\ne-\frac{1}{6}$ .

Final Answer

$x\ne-\frac{1}{6}$

Key Points to Remember

Essential concepts to master this topic

Rule: Domain excludes values that make the denominator equal zero
Technique: Set $3x + \frac{1}{2} = 0$ and solve to get $x = -\frac{1}{6}$
Check: Substitute $x = -\frac{1}{6}$ : $3(-\frac{1}{6}) + \frac{1}{2} = 0$ ✓

Common Mistakes

Avoid these frequent errors

Setting the numerator equal to zero
Don't set the numerator 9 equal to zero to find domain restrictions = this finds where the function equals zero, not where it's undefined! The numerator being zero doesn't affect the domain. Always set only the denominator equal to zero to find domain restrictions.

Practice Quiz

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

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 meaningful value, so we must exclude those x-values from the domain.

How do I solve 3x + 1/2 = 0?

First, subtract $\frac{1}{2}$ from both sides: $3x = -\frac{1}{2}$ . Then divide by 3: $x = -\frac{1}{6}$ . This is the value to exclude from the domain!

What does the ≠ symbol mean?

The symbol ≠ means "not equal to." When we write $x \ne -\frac{1}{6}$ , we're saying x can be any real number except $-\frac{1}{6}$ .

How do I write the complete domain?

You can write it as: "All real numbers except $x = -\frac{1}{6}$ " or in interval notation: $(-\infty, -\frac{1}{6}) \cup (-\frac{1}{6}, \infty)$ .

What if the denominator has multiple terms?

The same rule applies! Set the entire denominator equal to zero and solve. Even with expressions like $x^2 - 4$ , you'd solve $x^2 - 4 = 0$ to find restrictions.

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