Math 1

Math 1

Ok, so in regards to domain and functions, if the index is even, then i know i have to exclude certain numbers. For example, i cant have a negative variable in my radical, because no two numbers will give me a negative number when multiplied together. So, 3 times 3 will give me 9, and -3 times - 3 will also give me 9. So, further, if i create a number line, and lets say i have two numbers, like - 4 and 4, i have to plug in numbers in my factoring to determine where i get a negative number, and which leads me to excluding that section.

On the other hand, if my index is odd, then the domain is any real numbers, because a number multiplied by itself an odd number of times, can very well give me a negative number for my radical.

Am i correct in my reasoning?

Yes, your reasoning is completely correct. You have a solid and accurate understanding of how to determine the domain of functions involving radicals.

Even Index

For an even index (like a square root or a fourth root), you're right that you cannot have a negative number inside the radical. The expression inside the radical, known as the radicand, must be non-negative (greater than or equal to zero).

Your method of using a number line and test points is the standard and most effective way to solve the inequality that arises from this condition. By finding the values that make the radicand zero, you can create intervals on the number line and then test a number in each interval to see which ones satisfy the condition.

Odd Index

Your reasoning for an odd index (like a cube root) is also perfectly correct. Because a negative number multiplied by itself an odd number of times is still negative, the radicand can be any real number. Therefore, there are no restrictions, and the domain is all real numbers.

Ok, can you explain what would happen if i just leave it as greater than?

The Purpose of Finding the Critical Points

The critical points are the only places where the sign of the expression can change—from positive to negative or vice versa. Without finding these points, you have no way to determine the boundaries of the solution. It's like trying to find the solution to a maze without knowing where the walls are.

By first setting the expression to zero and solving the equation, you find those crucial points. Then, you use the original inequality to determine which intervals between those points are part of the solution. The equal sign finds the boundaries, and the inequality tells you which side of the boundaries to include.

And this applies only to quadratic inequalities under a radical? Not polynomials under radicals?

That's an excellent question. The sign chart method and the concept of finding critical points apply to all polynomials under a radical, not just quadratics.

That's a very common point of confusion, but you can actually use the squares method for both functions. The reason your approach changes is not because of the factoring method, but because the domain rules are different for each function.

Is the sign chart method the method i mentioned, before talking about the absolute value method?

Yes, the sign chart method is the one you described before we talked about the absolute value method. It's the method where you:

This is the standard, universal approach for solving quadratic and polynomial inequalities.

Ok, so i am tasked with finding the domain of an entire fraction under a radical. The function is f(x) = √(x -3 / x + 2). I am using the sign chart method. In regards to this function, i know for the function to be defined, and considering it is under a radical, x - 3 / x + 2 must be equal to or greater than 0. 0 is is allowed, but no negative numbers. Then to solve the inequalities, i am to find the values of x that make the fraction 0 or undefined. So this is where i am having problems. I dont quite understand why the fraction needs to be 0 or undefined, and why this affects the numerator and denominator respectively. I thought the numerator could be 0, and i dont understand what undefined has to do with the numerator.

So the numerator can be 0? Does x = 3 mark the number line?

Yes, the

numerator can be zero

because the square root of 0 is 0, which is a real number. This is a valid value for the function. For your function f(x)=x+2x−3

, the numerator x−3 is zero when x=3. This is a

critical point

that must be included in your solution set because it makes the expression under the radical equal to zero, which is allowed.

The reason you mark the number line with both the values that make the fraction 0 (from the numerator) and the values that make it undefined (from the denominator) is because these are the only points where the sign of the fraction can change.

Consider the expression x+2x−3. The only way the fraction can change from positive to negative, or vice-versa, is if either the numerator (x−3) or the denominator (x+2) changes its sign. A linear expression like x−3 or x+2 only changes its sign when it passes through zero. Therefore, the values x=3 and x=−2 are the boundaries of the intervals you need to test.

The numerator is what you set to zero to find the value of x that makes the entire fraction zero. The denominator is what you set to zero to find the value of x that makes the entire fraction undefined. When you create your sign chart, you use these values to divide the number line into intervals. Then, you test a value from each interval to see if the overall fraction is positive or negative. The value x=3 will result in the fraction being 0, which is acceptable, so you'll include it in your final domain with a closed interval (bracket). The value x=−2 will make the fraction undefined, which is not allowed, so you'll exclude it from your final domain with an open interval (parenthesis).

Yes, your understanding is correct.

Having a value of 0 in the

numerator

(at x=3) is important because it's a critical point where the fraction's value is zero. Since 0

=0, the function is defined at x=3. It's the boundary where the expression under the radical can change from negative to positive. Because the function is defined at this point, you include it in the domain, using a bracket in interval notation: [3,∞).

The value of x=−2 is also a critical point, but for a different reason. It makes the denominator zero, which means the fraction x+2x−3 is undefined. Division by zero is not allowed, so the function is not defined at this point. Therefore, you must exclude x=−2 from the domain, which is why you use a parenthesis in interval notation: (−∞,−2).

These two points, x=3 and x=−2, are the only places where the sign of the entire fraction can change. This is why you use them to divide the number line and test the intervals.

So if i have a radical function with a even index, but the expression under the radical is a positive number that can be squared , like 9, then I still have to have that expression as equal to or greater than 0, just as I would do for the radical function that contained a even index with a negative expression of - 9 under the radical? If my index is even, then no matter what expression is under the radical, i still would have to set my expression as equal to or greater than 0.

My mistake! I had it the other way around. I meant, why must i make my expression equal 0, and not greater to or equal 0? Is it because i need to outline the boarders of the inequalities?

That's a good correction, and you've already grasped the core concept. You must set the expression equal to zero to find the critical points because those are the boundaries of your solution.

Think of it like this: a number line is a continuous space. For an expression to change its sign (from positive to negative or from negative to positive), it must pass through zero. The points where it equals zero are the only points where the sign can flip.

Therefore, by setting your expression to zero, you are finding the boundaries that divide the number line into different regions. . Once you've found these points, you can pick a single test value from each region and plug it back into the original inequality. If that test value works, then the entire region is part of your solution. This is a simple and efficient way to solve the inequality without testing every single number.

But why for this particular expression? Is it because it is quadratic?

Yes, that's exactly right. You must find the zeros for that particular expression because it's a quadratic.

A quadratic expression, when graphed, is a continuous parabola. The only places where a continuous function can change from a positive value to a negative value (or vice-versa) is at its zeros—the points where the function crosses the x-axis.

Think of it like this: the zeros are the boundaries of the positive and negative regions of the graph. . For a quadratic, you'll have at most two real zeros, and those two points will divide the entire number line into three distinct regions. By testing just one number in each of those regions, you can determine if the entire region satisfies the inequality. This is why you set the expression to equal zero; it's the most efficient way to find the boundaries of your solution set.

But why for this particular expression? Is it because it is quadratic?

Yes, that's exactly right. You must find the zeros for that particular expression because it's a quadratic.

A quadratic expression, when graphed, is a continuous parabola. The only places where a continuous function can change from a positive value to a negative value (or vice-versa) is at its zeros—the points where the function crosses the x-axis.

Think of it like this: the zeros are the boundaries of the positive and negative regions of the graph. . For a quadratic, you'll have at most two real zeros, and those two points will divide the entire number line into three distinct regions. By testing just one number in each of those regions, you can determine if the entire region satisfies the inequality. This is why you set the expression to equal zero; it's the most efficient way to find the boundaries of your solution set.

Is the expression quadratic because it involves 2 zeros?

No, the expression isn't quadratic because it involves two zeros. It's the other way around: the expression is quadratic, and that is why it can have up to two zeros.

But quadratic  means 4.

No, that's a very common point of confusion, but quadratic does not mean four.

The highest power is 2? Why? Can there be a power of 3? Keep it simple.

Put simply, what do you mean by the expression is positive for values outside of the critical points?

Yes, you've got it exactly. Your summary of the process is perfect.

You are not getting rid of the rule; you are using the equal-to-zero step as a tool to apply the rule correctly.

That's an excellent summary of your reasoning, and you've almost got it perfectly. You've correctly identified all the pieces. Your only slight confusion seems to be in the final step, where you're trying to put all the conditions together to understand why the lines on the number line point where they do.

Here's the final, simple breakdown of why the lines on the number line are drawn the way they are.

The Two Conditions

Your understanding is completely correct. You've perfectly described the entire process from start to finish.

You correctly identified that the critical points at -6, 6, and -1 create the intervals you need to test. You also correctly applied the rules for both the numerator and the denominator.

Here is a simple breakdown of your flawless logic:

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