Consider the following nonlinear system. Work Exercises 75 –80 in...

Question

Consider the following nonlinear system. Work Exercises 75 –80 in order.
y = | x - 1 |
y = x^2 - 4

Use the definition of absolute value to write y = | x - 1 | as a piecewise-defined function.

Accepted Answer

Hello everybody. I hope you're doing right today. Today we're gonna be looking at this question that states right? Y equals the absolute value of X minus three as a piecewise defined function using the definition of the absolute function. Now, the answer traces given are a, the absolute value of X minus three equals a big squiggly bracket where X, where we have a positive X if X is squared and, or equals to zero and a negative X, if X is less than zero B, the same as letter A. But with the terms being if X is less than zero on top and if X is greater than, or equal to zero on the bottom C, the absolute value of X minus three equal to a big square degree bracket where we have X minus three if X is greater than, or equal to three on top and on the bottom, negative X plus three, if X is less than three and D the same as C but switching the conditions. So we would have if X is less than three on top and if X is greater than, or equal to three on the bottom. Now when I see this question, I see that they're emphasizing the absolute function. So what is we have Y equals the absolute value of X minus three. So now we have to think what's the absolute function of this? Well, the absolute function here would be the absolute value of X, right? So that's the absolute function without the minus three. So the absolute value of X, how do we write that as a piecewise defying function? Well, we would write the absolute value of X is equal to and then an open big squiggy bracket. And on top, we would have a positive X. So here we have to think about what numbers would we, if we plug into the X and the absolute value, what would happen? So we would get a positive X if our X is greater than or equal to zero. So if we put any number, that's any positive number inside absolute brackets, then a positive number would come out, that number would stay the same, therefore, that X would stay the same. So if we plug in 123 and so on, the answer would be 123 and so on the same number. However, now let's think of a different condition. Let's say if X is less than zero. So if we plug in negative one, negative two, negative three in inside an absolute bracket, we know absolute brackets will give us a positive answer. Therefore, if we plug in negative one into an absolute bracket. The answer will be positive one. If we plug in negative two, the answer will be positive too. So what, what are we doing? Well, if you think about it, you're multiplying each X, each negative X by negative one so that you can get a positive answer. So we would have negative X if X is less than zero and just imagine you're basically multiplying by a negative. So that answer would be, would turn positive once you take it out of the absolute value bars. So this is how we would write the absolute function as a piecewise defined function. So now we have to apply that same logic to the function that we had, which was Y equals absolute value of X minus three. So we had the absolute value of X minus three equals and once again, we'll have an open squiggly bracket. And now let's just apply the same idea. So if, if X is greater than or equal to three, right? So this is what would happen if, if we, we chose three here, because if we plug in three to our, to our equation X minus three, that is where we reach zero and any value above three will be above zero, which is our first condition. In the absolute function, we have to be greater than or equal to zero. That's why in this option where we have X minus three and instead the, the number we pick is greater than equal to three. Now, we know when that happens, we, we just have the same, so the, or the same style function, so we would stay the same, we would just have X minus three again. Now let's look at our other function. So if X is less than, or E or, or is less than three, so not equal to just if X is less than three, this is the second option that we saw with our absolute function. So if X is less than three here, we would be at less than zero. Therefore, we would have to multiply it by a negative one so that we can get a positive answer at the end. And if we multiply X minus three times negative one or, or multiply it by a negative one, we would have negative X plus three. So basically, we're just flipping the signs of our function here. And that is how we would write the X, the absolute value of X minus three as a piecewise defined function. We would have a big squiggly bracket opening on top, we would have X minus three if X is greater than, or equal to three. And at the bottom, we would have negative X plus three if axis is less than three. So now we just look at our answer choices and we see that answer choice C matches with our answer. So that would be the answer to this question. So, I hope that was helpful and until next time.

Consider the following nonlinear system. Work Exercises 75 –80 in order. y = | x - 1 | y = x^2 - 4 Use the definition of absolute value to write y = | x - 1 | as a piecewise-defined function.

Key Concepts

Absolute Value Function

Piecewise Functions

Nonlinear Systems

Watch next