Determining the Valid Domain for (x+2)/√(3x²-9)

Look at the following function:

x+23x29 \frac{x+2}{\sqrt{3x^2-9}}

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
00:10 Does the function have a domain? If it does, let's find out what it is!
00:15 Remember, a root must be for a positive number.
00:20 Let's set the equation equal to zero to find the solutions.
00:27 Alright, let's isolate the variable X.
00:39 When you take a root, there are always two solutions: one positive and one negative.
00:45 Let's draw a graph to help us find the domain.
00:51 Between these solutions, any value for X leads to a negative root.
00:58 And that's the solution to our question!

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

Look at the following function:

x+23x29 \frac{x+2}{\sqrt{3x^2-9}}

What is the domain of the function?

2

Step-by-step solution

To solve this problem, we must determine the domain of the function x+23x29 \frac{x+2}{\sqrt{3x^2-9}} .

To begin, the expression inside the square root, 3x29 3x^2 - 9 , must be greater than 0 for the square root to be real and the function to be defined. Thus, we set up the inequality:

3x29>0 3x^2 - 9 > 0

Next, solve this inequality for x x :

  • Start by factoring: 3x29=3(x23) 3x^2 - 9 = 3(x^2 - 3)
  • Set the factor equal to zero to find the critical points: x23=0 x^2 - 3 = 0
  • Solve for x x :
    • x2=3 x^2 = 3
    • x=±3 x = \pm \sqrt{3}

Now, determine the intervals where 3x29 3x^2 - 9 is positive. Consider the intervals defined by the critical points x=3 x = \sqrt{3} and x=3 x = -\sqrt{3} :

  • Interval 1: (,3)(-\infty, -\sqrt{3})
  • Interval 2: (3,3)(- \sqrt{3}, \sqrt{3})
  • Interval 3: (3,)(\sqrt{3}, \infty)

Test a value from each interval in the inequality 3x29>0 3x^2 - 9 > 0 :

  • For interval 1, test x=2 x = -2 :
    • 3(2)29=129=3>0 3(-2)^2 - 9 = 12 - 9 = 3 > 0 (True)
  • For interval 2, test x=0 x = 0 :
    • 3(0)29=9<0 3(0)^2 - 9 = -9 < 0 (False)
  • For interval 3, test x=2 x = 2 :
    • 3(2)29=129=3>0 3(2)^2 - 9 = 12 - 9 = 3 > 0 (True)

Thus, the function is defined for x x in the intervals (,3)(-\infty, -\sqrt{3}) and (3,)(\sqrt{3}, \infty).

This means that the domain of the function is:

x>3,x<3 x>\sqrt{3},x<-\sqrt{3}

Therefore, the solution to the problem is x>3,x<3 x>\sqrt{3},x<-\sqrt{3} .

3

Final Answer

x>3,x<3 x>\sqrt{3},x<-\sqrt{3}

Practice Quiz

Test your knowledge with interactive questions

Given the following function:

\( \frac{5-x}{2-x} \)

Does the function have a domain? If so, what is it?

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