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Current time:0:00Total duration:4:37

CCSS.Math: ,

- [Voiceover] So we're told
that h of x is equal to the negative cube root of
three x minus six plus 12. And what we wanna figure out
is, what is the inverse of h? So what is... What is h inverse of x
going to be equal to? And like always, pause the video and see if you could figure it out. Well in previous videos,
we've emphasized that what an inverse does is... A function will map from a
domain to a range and you can think of the inverse as
mapping back from that point in the range to where you started from. So one way to think about it
is, we want to come up with an expression that unwinds
whatever this does. So if we say that y, if we
say that y is equal to h of x, or we could say that y is
equal to the negative of the cube root of three
x minus six plus 12. This gives us our y. And you can think of y
as a member of the range. A member of the range in
terms of what our input is. In terms of a member of the domain. We wanna go the other way around, so what we could do is we
could try to solve for x. If we solve for x, we're
gonna have some expression that's a function of y. We're gonna have that being equal to x. And so that would be the inverse mapping. Another way you could do that, is you could just swap x
and y and then solve for y. But that's a little bit less intuitive that this is actually the inverse. So actually, let's just solve for x here. So the first thing we might want to do is, let's isolate this cube root on, let's say to the right hand side. So let's subtract 12 from both sides. And we would get y minus 12
is equal to the cube root of, it's actually the negative cube root. Don't wanna lose track of that. Negative cube root of three x minus six and then we subtracted 12
from both sides so that 12 is now, that 12 is now gone. And now what we would do, what
we could multiply both sides by negative one, that might
get rid of this negative here. So we multiply both sides by negative one. And then we multiply this
times a negative one. On the left hand side, we'll that's the same thing as 12 minus y. And on the right hand side,
we're gonna get the cube root of three x minus six. And now and this is gonna be a little bit algebraically hairy. We wanna cube both sides. So let's do that. So let's cube both sides. And actually it doesn't get
that algebraically hairy because I don't actually
have to figure what this, I don't have to expand it, I could just leave it as 12 minus y cubed. And so if we cube both
sides on the left hand side, we're just left with 12 minus y cubed. And on the right hand side,
well you take the cube of the cube root, you're
just gonna be left with what you originally had
under the cube root sign, I guess you could say. And now we wanna solve for x,
let's add six to both sides. So we're gonna get 12
minus y cubed plus six is equal to three x. Now we could divide both
sides by 3 and we're all done. Divide both sides by three and we get... We get x... Is equal to 12 minus y to the third power plus six over three. And so this, if you have a member of the, one way to think about it, if you have a member of the range y, this is going to map it back
to the x that would have gotten you to that member of the range. So this is the inverse
function so we could write, h inverse of y is equal to this business. 12 minus y cubed plus six over three. And like we said in previous videos, this choice of calling y the input, well it could be anything,
we could call that star. We could say h inverse of
star and we're just naming our input star is equal to 12 minus star cubed plus six over three. Or if we just want to call
the input x, we could just say h inverse of x and once again, this is just what we're calling the input, is equal to 12 minus y to the third plus six over three. Might be a little bit
confusing because now, in theory x could be considered
a member of the range and we're mapping back to
a member of the domain. But either way, we can
call the input function to a function partially anything. But there you have it, that
is our inverse function, that essentially unwinds what
our original function does.