Domain Of Functions: Radical Expressions And Fractions

Hey guys! Today, we're diving into the exciting world of finding the domain of some pretty interesting functions. Specifically, we're going to tackle functions that involve radical expressions (those square roots and such) and fractions. These types of functions have certain restrictions on their domain, and figuring out what those are is like solving a fun puzzle! So, grab your thinking caps, and let's get started!

Understanding the Domain

Before we jump into the nitty-gritty of the problems, let's make sure we're all on the same page about what the domain actually is. Simply put, the domain of a function is the set of all possible input values (usually x-values) that will produce a valid output (a real y-value). Think of it like this: the domain is the list of ingredients you can use in your mathematical recipe without causing the recipe to explode (or, in mathematical terms, become undefined).

For the types of functions we're looking at today, there are two main things we need to watch out for:

1. Square roots: You can't take the square root of a negative number (in the real number system, anyway). So, anything under a square root must be greater than or equal to zero.
2. Fractions: You can't divide by zero. So, the denominator of a fraction can never be zero.

Keeping these two rules in mind is crucial for finding the domain of our functions. We're essentially looking for the x-values that satisfy these conditions.

Part a: ${ y = \sqrt{\frac{17 - 15x - 2x^2}{x + 3}} }$

Okay, let's dive into our first function: ${ y = \sqrt{\frac{17 - 15x - 2x^2}{x + 3}} }$. This looks a little intimidating, but don't worry, we'll break it down step by step. Remember, our goal is to find all the x-values that make this function happy (i.e., produce a real output).

Step 1: Address the Square Root

First things first, we have a square root. That means the expression inside the square root must be greater than or equal to zero. So, we have the following inequality:

${\frac{17 - 15x - 2x^2}{x + 3} \ge 0}$

This inequality is the heart of the problem. We need to figure out when this fraction is non-negative.

Step 2: Factor the Quadratic

To make things easier, let's factor the quadratic expression in the numerator: 17 - 15x - 2x². Factoring this quadratic gives us -(2x + 17)(x - 1). So, our inequality now looks like this:

${\frac{-(2x + 17)(x - 1)}{x + 3} \ge 0}$

Step 3: Find the Critical Points

The critical points are the values of x that make either the numerator or the denominator equal to zero. These are the points where the expression can change sign. From our factored inequality, the critical points are:

• 2x + 17 = 0 => x = -17/2 = -8.5
• x - 1 = 0 => x = 1
• x + 3 = 0 => x = -3

These three critical points divide the number line into four intervals: (-∞, -8.5], [-8.5, -3), (-3, 1], and [1, ∞). Notice the parentheses and brackets. We use brackets for the values where the expression can be equal to zero (numerator) and parentheses for the values where the expression is undefined (denominator).

Step 4: Test Intervals

Now, we need to test a value from each interval in our inequality to see if the expression is positive or negative. This will tell us which intervals satisfy our inequality.

• Interval (-∞, -8.5): Let's try x = -9. ${\frac{-(2(-9) + 17)(-9 - 1)}{-9 + 3} = \frac{-(-1)(-10)}{-6} = \frac{-10}{-6} > 0}$ So, this interval satisfies the inequality.
• Interval (-8.5, -3): Let's try x = -4. ${\frac{-(2(-4) + 17)(-4 - 1)}{-4 + 3} = \frac{-(9)(-5)}{-1} = -45 < 0}$ This interval does not satisfy the inequality.
• Interval (-3, 1): Let's try x = 0. ${\frac{-(2(0) + 17)(0 - 1)}{0 + 3} = \frac{-(17)(-1)}{3} = \frac{17}{3} > 0}$ This interval satisfies the inequality.
• Interval (1, ∞): Let's try x = 2. ${\frac{-(2(2) + 17)(2 - 1)}{2 + 3} = \frac{-(21)(1)}{5} = -\frac{21}{5} < 0}$ This interval does not satisfy the inequality.

Step 5: State the Domain

Based on our interval testing, the solution to the inequality is (-∞, -8.5] ∪ (-3, 1]. This means the domain of the function ${ y = \sqrt{\frac{17 - 15x - 2x^2}{x + 3}} }$ is:

Domain: x ∈ (-∞, -8.5] ∪ (-3, 1]

We've successfully navigated the first function! See? It's like a puzzle, and we just found all the pieces that fit.

Part b: ${ y = \sqrt{\frac{7 - x}{\sqrt{4x^2 - 19x + 12}}} }$

Now, let's move on to the second function: ${ y = \sqrt{\frac{7 - x}{\sqrt{4x^2 - 19x + 12}}} }$. This one looks even more interesting because we have a square root inside a square root! But the same principles apply – we just need to be extra careful with our conditions.

Step 1: Address the Outer Square Root

As before, the expression inside the outer square root must be greater than or equal to zero. This gives us the inequality:

${\frac{7 - x}{\sqrt{4x^2 - 19x + 12}} \ge 0}$

Step 2: Address the Inner Square Root

Now, let's think about the inner square root, ${ \sqrt{4x^2 - 19x + 12} }$. Not only does the expression inside this square root have to be non-negative, but since it's in the denominator, it also can't be zero. So, we have two conditions:

1. 4x² - 19x + 12 > 0 (Strictly greater than zero because it's in the denominator)
2. The entire fraction ${ \frac{7 - x}{\sqrt{4x^2 - 19x + 12}} }$ must be greater than or equal to zero.

Step 3: Solve the Quadratic Inequality

Let's tackle the quadratic inequality first: 4x² - 19x + 12 > 0. We need to factor the quadratic. Factoring gives us (4x - 3)(x - 4) > 0. Now, we find the critical points:

• 4x - 3 = 0 => x = 3/4 = 0.75
• x - 4 = 0 => x = 4

These critical points divide the number line into three intervals: (-∞, 0.75), (0.75, 4), and (4, ∞). We test a value from each interval:

• Interval (-∞, 0.75): Let's try x = 0. (4(0) - 3)(0 - 4) = (-3)(-4) = 12 > 0. This interval satisfies the inequality.
• Interval (0.75, 4): Let's try x = 1. (4(1) - 3)(1 - 4) = (1)(-3) = -3 < 0. This interval does not satisfy the inequality.
• Interval (4, ∞): Let's try x = 5. (4(5) - 3)(5 - 4) = (17)(1) = 17 > 0. This interval satisfies the inequality.

So, the solution to the quadratic inequality 4x² - 19x + 12 > 0 is x ∈ (-∞, 0.75) ∪ (4, ∞).

Step 4: Consider the Numerator

Now, let's go back to the original inequality: ${ \frac{7 - x}{\sqrt{4x^2 - 19x + 12}} \ge 0 }$. We've already figured out when the denominator is positive. Now we need to consider the numerator, 7 - x. For the entire fraction to be non-negative, we need:

7 - x ≥ 0 => x ≤ 7

Step 5: Combine the Conditions

We have three conditions to satisfy:

1. x ∈ (-∞, 0.75) ∪ (4, ∞) (from the inner square root and denominator)
2. x ≤ 7 (from the numerator)

We need to find the intersection of these intervals. Graphing these intervals on a number line can be very helpful. We're looking for the regions where both conditions are true.

Step 6: State the Domain

Combining these conditions, we find that the domain of the function ${ y = \sqrt{\frac{7 - x}{\sqrt{4x^2 - 19x + 12}}} }$ is:

Domain: x ∈ (-∞, 0.75) ∪ (4, 7]

And there you have it! We've conquered another domain problem, even with that tricky square root inside a square root. You guys are amazing!

Key Takeaways

Let's recap the key ideas we've learned today:

• Domain Restrictions: Remember the two main restrictions: square roots (must be non-negative) and fractions (denominator can't be zero).
• Critical Points: Find the values that make the numerator or denominator zero. These are your critical points.
• Interval Testing: Test values in each interval created by the critical points to determine the sign of the expression.
• Combining Conditions: When dealing with multiple restrictions (like our second example), make sure you find the intersection of all the solution sets.

Finding the domain of functions can seem challenging at first, but with practice and a systematic approach, you'll become domain-finding pros! Keep practicing, and don't be afraid to ask questions. You've got this!