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We've got the function f of x is equal to x plus 2 squared

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plus 1, and we've constrained our domain that x has to be

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greater than or equal to negative 2.

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That's where we've defined our function.

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And we want to find its inverse.

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And I'll leave you to think about why we had to constrain

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it to x being a greater than or equal to negative 2.

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Wouldn't it have been possible to find the inverse if we had

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just left it as the full parabola?

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I'll leave you -- or maybe I'll make a future video about that.

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But let's just figure out the inverse here.

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So, like we've said in the first video, in the

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introduction to inverses, we're trying to find a mapping.

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Or, if we were to say that y -- if we were to say that y is

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equal to x plus 2 squared plus 1.

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This is the function you give me an x and it maps to y.

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We want to go the other way.

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We want to take, I'll give you a y and then map it to an x.

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So what we do is, we essentially just solve

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for x in terms of y.

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So let's do that one step at a time.

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So, the first thing to do, we could subtract 1 from both

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sides of this equation.

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y minus 1 is equal to x plus 2 squared.

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And now to solve here, you might want to

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take the square root.

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And that actually will be the correct thing to do.

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But it's very important to think about whether you want

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to take the positive or the negative square

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root at this step.

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So we've constrained our domain to x is greater than or

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equal to negative 2.

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So this value right here, x plus 2, if x is always greater

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than or equal to negative 2, x plus 2 will always be

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greater than or equal to 0.

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So this expression right here, this right here is positive.

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This is positive.

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So we have a positive squared.

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So if we really want to get to the x plus 2 in the appropriate

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domain, we want to take the positive square root.

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And in the next video or the video after that, we'll solve

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an example where you want to take the negative square root.

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So we're going to take theundefined positive square

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root, or just the principal root, which is just the square

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root sign, of both sides.

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So you get the square root of y minus 1 is equal to x plus 2.

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And one thing I should have remembered to do is, from

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the beginning we had a constraint on x.

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We had for x is greater than or equal to negative 2.

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But what constraint could we have on y?

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If you look at the graph right here, x is greater than

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equal to negative 2.

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But what's why?

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What is the range of y-values that we can get here?

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Well, if you just look at the graph, y will always be

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greater than or equal to 1.

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And that just comes from the fact that this term right

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here is always going to be greater than or equal to 0.

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So the minimum value that the function could take on is 1.

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So we could say for x is greater than or equal to

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negative 2, and we could add that y is always going to be

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greater than or equal to 1.

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y is always greater than or equal to 1.

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The function is always greater than or equal

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to that right there.

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To 1.

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And the reason why I want to write it at the stage is

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because, you know, later on, we're going to swap

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the the x's and y's.

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So let's just leave that there.

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So here we haven't explicitly solved for x and y.

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But we can write for y is greater than or equal to 1,

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this is going to be the domain for our inverse, so to speak.

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And so here we can keep it for y is greater

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than or equal to 1.

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This y constraint's going to matter more.

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Because over here, the domain is x.

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But for the inverse, the domain is going to be the y-value.

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And then, let's see.

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We have the square root of y minus 1 is equal to x plus 2.

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Now we can subtract 2 from both sides.

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We get the square root of y minus 1 minus 2, is equal

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to x for y is greater than or equal to 1.

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And so we've solved for x in terms of y.

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Or, we could say, let me just write it the other way.

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We could say, x is equal to, I'm just swapping this. x is

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equal to the square root of y minus one minus 2, for y is

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greater than or equal to one.

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So you see, now, the way we've written it out.

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y is the input into the function, which is going to be

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the inverse of that function.

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x the output. x is now the range.

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So we could even rewrite this as f inverse of y.

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That's what x is, is equal to the square root of y minus 1

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minus 2, for y is greater than or equal to 1.

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And this is the inverse function.

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We could say this is our answer.

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But many times, people want the answer in terms of x.

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And we know we could put anything in here.

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If we put an a here, we take f inverse of a.

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It'll become the square root of a minus 1 minus 2, 4.

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Well, assuming a is greater than or equal to 1.

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But we could put an x in here.

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So we can just rename the the y for x.

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So we could just do a renaming here.

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So we can just rename y for x.

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And then we would get -- let me scroll down a little bit.

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We would f inverse of x.

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I'll highlight it here.

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Just to show you, we're renaming y with x.

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You could rename it with anything really, is equal to

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the square root of x minus 1.

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Of x minus 1.

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Minus 2 for, we have to rename this to, for x being

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greater than or equal to 1.

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And so we now have our inverse function as a function of x.

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And if we were to graph it, let's try our best to graph it.

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Maybe the easiest thing to do is to draw some points here.

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So the smallest value x can take on is 1.

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If you put a 1 here, you get a 0 here.

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So the point 1, negative 2, is on our inverse graph.

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So 1, negative 2 is right there.

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And then if we go to 2, let's see, 2 minus 1 is 1.

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The principle root of that is 1.

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Minus 2.

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So it's negative 1, so the point, 2, negative

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1 is right there.

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And let's think about it.

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Let's see.

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If we did 5, I'm trying take perfect squares.

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5 minus 1 is 4, minus 2.

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So the point 5, 2 is, let me make sure.

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5 minus 1 is 4.

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Square root is 2.

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Minus 2 is 0.

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So the point 5, 0 is here.

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And so the inverse graph, it's only defined for x greater

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than or equal to negative 1.

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So the inverse graph is going to look something like this.

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It's going to look something like, I started off

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well, and it got messy.

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So it's going to look something like that.

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Just like that.

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And just like we saw, in the first, the introduction to

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function inverses, these are mirror images around

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the line y equals x.

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Let me graph y equals x. y equals x.

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y equals x is that line right there.

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Notice, they're mirror images around that line.

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Over here, we map the value 0 to 5.

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If x is 0, it gets mapped to 5.

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Here we go the other way.

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We're mapping 5 to -- we're mapping 5 to the value 0.

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So that's why they're mirror images.

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We've essentially swapped the x and y.

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