Domain Of Functions: Solving Square Root Equations

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Hey guys! Let's dive into finding the domain of functions, especially those involving square roots. It might sound intimidating, but trust me, it's totally manageable once you get the hang of it. We're going to break down two examples step-by-step, so you'll be a pro in no time! So, let's find the domain of the function: a) y=12x−3x2y=\sqrt{12x-3x^2}; b) y=12x2−12x+18y=\frac{1}{\sqrt{2x^2-12x+18}}.

Understanding Domain

Before we jump into the problems, let's quickly recap what the domain of a function actually means. Simply put, the domain is the set of all possible input values (usually x) that will produce a valid output (y). In other words, it's all the x values that you can plug into the function without causing any mathematical catastrophes.

Why is this important? Well, some functions have restrictions on their inputs. For example, you can't take the square root of a negative number (at least not in the realm of real numbers), and you can't divide by zero. These restrictions limit the domain of the function. Identifying these limitations is crucial for understanding the function's behavior and ensuring accurate calculations. When dealing with square roots, the expression inside the square root must be greater than or equal to zero. For fractions, the denominator cannot be zero. These are the key considerations when determining the domain.

Part a: Finding the Domain of y=12x−3x2y=\sqrt{12x-3x^2}

Okay, let's tackle the first function: y=12x−3x2y=\sqrt{12x-3x^2}. The key here is to remember our golden rule for square roots: the expression inside the square root must be greater than or equal to zero. Why? Because the square root of a negative number isn't a real number. So, we need to make sure 12x−3x212x - 3x^2 is non-negative. Think of this as setting up a protective barrier around our function, making sure it stays in the realm of real numbers. To do this, we'll set up an inequality:

12x−3x2≥012x - 3x^2 \geq 0

Now, let's solve this inequality. First, we can factor out a 3x3x from the left side:

3x(4−x)≥03x(4 - x) \geq 0

Great! Now we have a product of two factors, 3x3x and (4−x)(4 - x), that must be greater than or equal to zero. To figure out when this happens, we'll use a sign chart. A sign chart is a visual tool that helps us determine the sign of an expression over different intervals. It's like a map that guides us through the solution.

Creating the Sign Chart

  1. Find the zeros: The zeros are the values of x that make each factor equal to zero. So, we set each factor equal to zero and solve:

    • 3x=0⇒x=03x = 0 \Rightarrow x = 0
    • 4−x=0⇒x=44 - x = 0 \Rightarrow x = 4

    These zeros, 0 and 4, are our critical points. They divide the number line into three intervals: (−∞,0)(-\infty, 0), (0,4)(0, 4), and (4,∞)(4, \infty).

  2. Create the chart: Draw a number line and mark the zeros (0 and 4) on it. Then, create a table with columns for each factor (3x3x and 4−x4 - x) and for the product 3x(4−x)3x(4 - x).

    Interval 3x3x 4−x4 - x 3x(4−x)3x(4 - x)
    (−∞,0)(-\infty, 0)
    (0,4)(0, 4)
    (4,∞)(4, \infty)
  3. Determine the signs: Pick a test value within each interval and plug it into each factor to determine its sign. Then, multiply the signs of the factors to get the sign of the product.

    • For (−∞,0)(-\infty, 0), let's pick x=−1x = -1:

      • 3x=3(−1)=−33x = 3(-1) = -3 (negative)
      • 4−x=4−(−1)=54 - x = 4 - (-1) = 5 (positive)
      • 3x(4−x)=(−)×(+)=−3x(4 - x) = (-) \times (+) = - (negative)
    • For (0,4)(0, 4), let's pick x=2x = 2:

      • 3x=3(2)=63x = 3(2) = 6 (positive)
      • 4−x=4−2=24 - x = 4 - 2 = 2 (positive)
      • 3x(4−x)=(+)×(+)=+3x(4 - x) = (+) \times (+) = + (positive)
    • For (4,∞)(4, \infty), let's pick x=5x = 5:

      • 3x=3(5)=153x = 3(5) = 15 (positive)
      • 4−x=4−5=−14 - x = 4 - 5 = -1 (negative)
      • 3x(4−x)=(+)×(−)=−3x(4 - x) = (+) \times (-) = - (negative)
  4. Fill in the chart: Now we can complete the sign chart:

    Interval 3x3x 4−x4 - x 3x(4−x)3x(4 - x)
    (−∞,0)(-\infty, 0) - + -
    (0,4)(0, 4) + + +
    (4,∞)(4, \infty) + - -

Interpreting the Sign Chart

We're looking for the intervals where 3x(4−x)≥03x(4 - x) \geq 0. This means we want the intervals where the product is positive or zero. From the sign chart, we see that 3x(4−x)3x(4 - x) is positive in the interval (0,4)(0, 4). It's also equal to zero at x=0x = 0 and x=4x = 4. Therefore, the domain of the function y=12x−3x2y = \sqrt{12x - 3x^2} is the closed interval [0,4][0, 4].

In simpler terms: The function is defined (i.e., gives a real number output) only when x is between 0 and 4, inclusive. Any x value outside this range would result in taking the square root of a negative number, which is a no-go in the real number system.

Part b: Finding the Domain of y=12x2−12x+18y=\frac{1}{\sqrt{2x^2-12x+18}}

Alright, let's move on to the second function: y=12x2−12x+18y=\frac{1}{\sqrt{2x^2-12x+18}}. This one's a bit trickier because we have a square root in the denominator. Remember our rules? We can't have a negative number inside a square root, and we definitely can't divide by zero. So, we have two conditions to satisfy:

  1. 2x2−12x+182x^2 - 12x + 18 must be greater than zero (to avoid the square root of a negative number).
  2. 2x2−12x+18\sqrt{2x^2 - 12x + 18} must not be equal to zero (to avoid division by zero).

Combining these two conditions, we can say that 2x2−12x+182x^2 - 12x + 18 must be strictly greater than zero. No equal to, just greater than. This ensures we're dealing with a positive number inside the square root and a non-zero denominator. So, let's set up the inequality:

2x2−12x+18>02x^2 - 12x + 18 > 0

Solving the Inequality

First, we can simplify the inequality by dividing both sides by 2:

x2−6x+9>0x^2 - 6x + 9 > 0

Now, let's factor the quadratic expression on the left side. You might recognize this as a perfect square trinomial:

(x−3)2>0(x - 3)^2 > 0

This is fantastic! We have a squared term, (x−3)2(x - 3)^2. Think about it: any real number squared is always non-negative (either positive or zero). So, (x−3)2(x - 3)^2 will always be greater than or equal to zero. But remember, we need it to be strictly greater than zero. When is (x−3)2(x - 3)^2 equal to zero? Only when x=3x = 3.

Determining the Domain

Therefore, (x−3)2>0(x - 3)^2 > 0 for all values of x except x=3x = 3. This means our domain includes all real numbers except 3. We can express this in interval notation as:

(−∞,3)∪(3,∞)(-\infty, 3) \cup (3, \infty)

In plain English: The function is defined for any x value you can think of, as long as it's not 3. If you plug in 3, you'll end up with a zero in the denominator, which is a big no-no.

Wrapping Up

So, there you have it! We've successfully found the domains of two functions involving square roots. The key takeaway here is to identify the restrictions imposed by the square roots and denominators and then set up inequalities to find the valid input values. Remember to consider both the non-negativity of the expression inside the square root and the non-zero nature of the denominator.

Finding the domain of a function is a fundamental concept in mathematics, and mastering it will help you understand the behavior and limitations of various functions. Keep practicing, and you'll become a domain-finding wizard in no time!