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Determine whether or not the function determined by the given equation is one-to-one.$$s(x)=-2 x^{2}+6, x \leq 0$$

Yes

Algebra

Chapter 4

Exponential and Logarithmic Functions

Section 1

Inverse Functions

Missouri State University

Harvey Mudd College

University of Michigan - Ann Arbor

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So if we want to determine if this is going to be 1 to 1 function, what we want to do is plug into arbitrary values into it, set them equal, and then see if we can show that a is equal to be No. If we were to just look at a quadratic normally essentially especially a quadratic where it has no middle term is going to look something along the lines of, like this here actually kind of sketch for this one actually Looks like look, maybe something along these lines here. So normally this would not be 1 to 1. But since X is less than or equal to zero, then that means over here this would only be the left side of it. So in this case, it would be 1 to 1. Well, let's actually be a little bit more rigorous when we're showing this. But we do need to keep this in mind that all these input values we have our going to the zero or negative. Otherwise, we're going to end up with, um, some issues for the numbers. So let's go ahead and start plugging things in, so we want to first plug in, um, A and B. So B s of A is equal to s a B, so that would give us negative to a squared. Plus six is equal to negative two B squared, plus six. So if we were to add, subtract and move everything around, we're going to end up with a squared. Is it to be squared? So now, um, we need to be a little bit rigorous about how we do this. So I'm going to first take the square root on each side, and then I'm going to write this in the following way where it's the absolute value of a is equal to the absolute value of Be like that. So let's write out what these peace or what these are as piecewise functions. So on the left side, Well, this is going to be, um a if a is greater than or equal to zero, Or actually I'll just say, uh, strictly greater than zero. And it's going to be negative. A A if a is less than or equal to zero. And then on the right side here, this is going to be B. If B is strictly larger than zero and then negative B. If B is less than or equal to zero now, this is when this X, less than or equal to zero comes into play. So this is now going to be equal to negative A on the left and negatively on the right. And this is since X is less than or equal to zero. And now we can just divide each side by negative A or multiply by negative eight by just negative one. And then that's going to give us a Is it good to be which that implies that what was the bearer when we started with s so implies S of X is 1 to 1, and then you can put your little foot box and smiley face because you're glad you're able to prove it. But so again, the thing that is important and you should kind of make note of this is this axe is less than zero. Because if not, then we end up with, like a is equal to plus or minus B and then if it's plus or minus B, then we have multiple different outputs. So, yeah, you should, like, make a note of this when you're kind of proving it. Otherwise, you don't necessarily get A is always equal to be

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