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## Finding inverse functions (Algebra 2 level)

Current time:0:00Total duration:7:12

# Finding inverse functions: quadratic

CCSS Math: HSF.BF.B.4, HSF.BF.B.4a

## Video transcript

We've got the function f of x
is equal to x plus 2 squared plus 1, and we've constrained
our domain that x has to be greater than or equal
to negative 2. That's where we've
defined our function. And we want to
find its inverse. And I'll leave you to think
about why we had to constrain it to x being a greater than
or equal to negative 2. Wouldn't it have been possible
to find the inverse if we had just left it as the
full parabola? I'll leave you -- or maybe I'll
make a future video about that. But let's just figure
out the inverse here. So, like we've said in
the first video, in the introduction to inverses, we're
trying to find a mapping. Or, if we were to say that y --
if we were to say that y is equal to x plus 2
squared plus 1. This is the function you give
me an x and it maps to y. We want to go the other way. We want to take, I'll give you
a y and then map it to an x. So what we do is, we
essentially just solve for x in terms of y. So let's do that one
step at a time. So, the first thing to do, we
could subtract 1 from both sides of this equation. y minus 1 is equal to
x plus 2 squared. And now to solve here,
you might want to take the square root. And that actually will be
the correct thing to do. But it's very important to
think about whether you want to take the positive or
the negative square root at this step. So we've constrained our domain
to x is greater than or equal to negative 2. So this value right here, x
plus 2, if x is always greater than or equal to negative
2, x plus 2 will always be greater than or equal to 0. So this expression right here,
this right here is positive. This is positive. So we have a positive squared. So if we really want to get to
the x plus 2 in the appropriate domain, we want to take
the positive square root. And in the next video or the
video after that, we'll solve an example where you want to
take the negative square root. So we're going to take
theundefined positive square root, or just the principal
root, which is just the square root sign, of both sides. So you get the square root of y
minus 1 is equal to x plus 2. And one thing I should have
remembered to do is, from the beginning we had
a constraint on x. We had for x is greater than
or equal to negative 2. But what constraint
could we have on y? If you look at the graph right
here, x is greater than equal to negative 2. But what's why? What is the range of y-values
that we can get here? Well, if you just look at
the graph, y will always be greater than or equal to 1. And that just comes from the
fact that this term right here is always going to be
greater than or equal to 0. So the minimum value that the
function could take on is 1. So we could say for x is
greater than or equal to negative 2, and we could add
that y is always going to be greater than or equal to 1. y is always greater
than or equal to 1. The function is always
greater than or equal to that right there. To 1. And the reason why I want to
write it at the stage is because, you know, later
on, we're going to swap the the x's and y's. So let's just leave that there. So here we haven't explicitly
solved for x and y. But we can write for y is
greater than or equal to 1, this is going to be the domain
for our inverse, so to speak. And so here we can keep
it for y is greater than or equal to 1. This y constraint's
going to matter more. Because over here,
the domain is x. But for the inverse, the domain
is going to be the y-value. And then, let's see. We have the square root of y
minus 1 is equal to x plus 2. Now we can subtract
2 from both sides. We get the square root of y
minus 1 minus 2, is equal to x for y is greater
than or equal to 1. And so we've solved
for x in terms of y. Or, we could say, let me just
write it the other way. We could say, x is equal to,
I'm just swapping this. x is equal to the square root of y
minus one minus 2, for y is greater than or equal to one. So you see, now, the way
we've written it out. y is the input into the
function, which is going to be the inverse of that function. x the output. x is
now the range. So we could even rewrite
this as f inverse of y. That's what x is, is equal to
the square root of y minus 1 minus 2, for y is greater
than or equal to 1. And this is the
inverse function. We could say this
is our answer. But many times, people want
the answer in terms of x. And we know we could
put anything in here. If we put an a here, we
take f inverse of a. It'll become the square root
of a minus 1 minus 2, 4. Well, assuming a is greater
than or equal to 1. But we could put an x in here. So we can just rename
the the y for x. So we could just do
a renaming here. So we can just rename y for x. And then we would get -- let
me scroll down a little bit. We would f inverse of x. I'll highlight it here. Just to show you, we're
renaming y with x. You could rename it with
anything really, is equal to the square root of x minus 1. Of x minus 1. Minus 2 for, we have to
rename this to, for x being greater than or equal to 1. And so we now have our inverse
function as a function of x. And if we were to graph it,
let's try our best to graph it. Maybe the easiest thing to do
is to draw some points here. So the smallest value
x can take on is 1. If you put a 1 here,
you get a 0 here. So the point 1, negative 2,
is on our inverse graph. So 1, negative 2
is right there. And then if we go to 2,
let's see, 2 minus 1 is 1. The principle root
of that is 1. Minus 2. So it's negative 1, so
the point, 2, negative 1 is right there. And let's think about it. Let's see. If we did 5, I'm trying
take perfect squares. 5 minus 1 is 4, minus 2. So the point 5, 2 is,
let me make sure. 5 minus 1 is 4. Square root is 2. Minus 2 is 0. So the point 5, 0 is here. And so the inverse graph, it's
only defined for x greater than or equal to negative 1. So the inverse graph is going
to look something like this. It's going to look something
like, I started off well, and it got messy. So it's going to look
something like that. Just like that. And just like we saw, in the
first, the introduction to function inverses, these are
mirror images around the line y equals x. Let me graph y equals
x. y equals x. y equals x is that
line right there. Notice, they're mirror
images around that line. Over here, we map
the value 0 to 5. If x is 0, it gets mapped to 5. Here we go the other way. We're mapping 5 to -- we're
mapping 5 to the value 0. So that's why they're
mirror images. We've essentially
swapped the x and y.