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In Exercises $29-38,$ for the function determined by the given equation (a) determine its domain, (b) sketch its graph, (c) determine its range, (d) show the function is one-to-one, (e) sketch the graph of its inverse function, (f) determine the domain of the inverse function, (g) determine the range of the inverse function, (h) find the equation of the inverse function and (i) verify that $f^{-1}(f(x))=x$ and $f\left(f^{-1}(x)\right)=x$.$$f(x)=\sqrt{6-2 x}$$

(a) $x \leq 3$(c) $y \geq 0$(f) $x \geq 0$(g) $y \leq 3$(h) $f^{-1}(x)=\left(6-x^{2}\right) / 2$

Algebra

Chapter 4

Exponential and Logarithmic Functions

Section 1

Inverse Functions

Harvey Mudd College

Idaho State University

Lectures

06:35

In Exercises $29-38,$ for …

11:51

07:08

06:32

08:30

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06:28

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01:16

In Problems $29-38,$ the f…

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In Exercises $31-36,$ (a) …

02:29

04:08

03:58

02:47

In Exercises $23-30,$ (a) …

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So first, if we want to find the domain of this function here, Uh, since we're taking the square root of something, we know we cannot plug negative numbers into that. So we want the inside of this to be greater than or equal to zero so we could add two X divide by two. And that gives us three less than or equal to X. Uh, you can leave your domain like that or you can write in the interval notation, which will be from negative infinity to three. So I just like to write it in this notation here, so I'll leave it like that now for the graph. If we're going to do this by hand, I would think it would be a good idea to go ahead and first plot with the square root of X looks like. And then we can go from there. So, yeah, effects square root of X because that's the parent function of it. We have 01 and four and these are like the nice values released the first three nice values. So that would be zero. This would be one, and that would be too. And so now we need to figure out Well, what happens when we make this transformation. So since everything is on the inside of this, then everything is going to be a change in the X value. So, uh, first we're going to shift six and left. So this is all changes to the X value. So we're really just going to subtract six. And actually, we should just see, subtract, subtract six. And then, since we're multiplying by negative two, we're going to divide by negative two. So the Y values are the outputs stay the same. So it's to be 01 and two and so let's see. Zero gets taken to. So subtract six negative six. Divided by two. That would be three. One would be negative. Five divided by two. That would be 2.5 and then four minus six is negative. Two divided by negative two is one. Um, and I'm just gonna plug zero in also, um, since we can get our y intercept And that would be Route six. Uh, so let's go ahead and graph this now. So Route six is less than three, so I'll just need to go up to three. At least let me make that a little bit more to scale over here. Okay, so let's go in and plot these. Uh, so we have 302.51 12 and then zero Route six, which would be probably around, like here or something. And then we can just go ahead and connect these like that. So this is going to be why is he go to f of X now? We can go ahead and get all of our points for our inverse function by just switching the X and Y values. So remember, if this is X y, And over here, this is supposed to be y X. So if we have X f inverse of X now, we just switch these. So this is going to be 03 Let me actually skip this down a little bit, So 03 12.5, uh, to one. And then route 60 So we could just take these and plug those into 03 is going to be here. 12.5, uh, would be around like, here. 21 is there. And then Route six is kind of like this. And so this is going to be Why is he go to F in verse of X? Okay, Uh, now, to show this is 1 to 1. Oh, actually, I forgot to do the range. So the range if we come over here and remember, we're looking at f of X still. So the green line, Well, that's just going to be zero going upwards. So zero to infinity. Now, to show this is 1 to 1. Uh, we could go ahead and just look at the green line and say, Well, that does pass the horizontal line test. So that is 1 to 1. Or we can look at the derivative of this as well. And that's what I'm going to do. Um, just to kind of be a little bit more rigorous about. I guess so. If we look at the derivative, so remember, this is actually to a one half power, so we use power rules. Would be one half six minus two x rays, the negative one half, and then we take the derivative on the inside, which is going to be negative, too. So these two's cancel out with each other, and then we're going to be left with negative one over the square root of six minus two X. Now, the square root only outputs positive numbers. If we divide that into something negative, that means this will always be less than zero, Which implies this is always decreasing, which is one of the ways we can say a function is 1 to 1. Okay, uh, now, to get the range in the domain of the inverse, we don't even need to look at the graph over here. All we need to do is come over here and switch our range in our domain or the original function. So the domain of F inverse is supposed to be the range of our original. So this is going to be zero, uh, two infinity. Um, I have zero to infinity, and then the range of FN verse is supposed to be the domain of our original function, which we have up here being negative. Infinity to three. So, yeah, And if you look at the graph over there, you can see how that matches up as well. Um, but again, all we need to do is come over here and look at this Now. The next thing we need to do is to find what is our inverse. So remember, the first thing we're gonna do is replace this with why, uh and then interchange the X and y so this would be X is equal to the square root of six minus two y. And then we want to solve for y so we would square each side So the X squared is equal to six minus two. Why, we would add to why over and then subtract X squared. So that would give two. Why is even to six minus x word and the dividers side by two that gives y is equal to six minus x squared over two. Um, let me just make sure they didn't do anything weird. Yeah, that looks good. So this is f inverse of X. But there's one thing that we need to make note of notice how this here is defined for all real values of X. But our domain over here says X has to be greater than or equal to zero. So we can just go ahead and slap that on here as well. So x greater than or equal to zero. Because otherwise, if we just say the inverse is, uh, six minus X squared over two. Then when we come up here and look at this graph of the red, it would actually look like this here. So we have that blue portion as well. But we don't have that in this case. So yeah, just throwing that on there is kind of good, um to do. Now we want to actually show that these are in versus by the compositional rules. So I'll go ahead and first you f inverse of act the facts. So we're going to take a FedEx and plugging into FM verse, so this would be six minus f of X squared all over to. So now we can go ahead and plug in ffx, so be six minus the square root of six minus two x squared all over two. Now, these square roots cancel out, and then the sixes will also cancel. And so, in the numerator that will leave us with negative negative two x over two. So notice these negatives, cancel the two's cancel, and we're just going to be left with X, which is what we were looking for. Uh, now or the other way around. So for f of f inverse of X. So we're gonna take f inverse and plug it into F. Let me see what f was again. So it'll be the square root of six minus two f inverse effects. And doing it this way will need to be a little bit careful when we do it, and I'll kind of explain why when we get to that point. So let's go out and plug that in first. So six minus two, Uh, and then it is time six minus X squared all over to so first notice, these two's cancel and the sixes cancel. And then we're going to have the square root of negative negative X squared. So the negatives cancel, and then we have the square root of X squared. And remember, when we do the square root of X squared, we don't just get X, we get the absolute value of X. So now the reason why we can just go ahead and say, Well, this is equal to X is because if we look here, we're starting on the inside of this. So the domain of this composition is actually the domain of a inverse and the domain of F inverse is 02 infinity. So we would say, since the domain of F inverse is zero to infinity and the absolute value of X if we were to kind of break the South is X if X is greater than or equal to zero and negative X if X is strictly less than zero. So that's why we can just go ahead and drop this. So this is something important to kind of make a note of. And this is why the, um domain is also important because otherwise this absolute value of X gives us something that is not by objective around. It's not a 1 to 1 function. Yeah, so then this one also checks out as well.

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