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### Course: Algebra (all content)>Unit 7

Lesson 20: Finding inverse functions (Algebra 2 level)

Sal finds the inverse of h(x)=-∛(3x-6)+12.

## Want to join the conversation?

• Can the answer not be put in standard form? I simplified (12-y)^3 to get a solution of

h^-1(x) = -1/3x^3 +12x^2 - 144x + 578

• Your answer is equivalent to Sal's. You did the hard work... he took the easy route.
• my question is, why do we call y x if we dont have to do so?
• Why didn't Sal simplify the equation to (12-x)^3+2?
• You can't apply the denominator to just the 6. The denominator applies to the entire expression.
You either keep it as: [ (12-x)^3+6] / 3
Or, you can change it to (12-x)^3 / 3 + 2
• Can u tell how will we know whether to take positive or negative root?
• Theres something called a principal root where its only positive. But usually in equations, it is +_ a square root.
(1 vote)
• I took the step 3x-6=(-y+12)^3 and applied the cube to the terms, which is not what Sal did, to get -y^3 +1728=3x-6. I solved it from there, then I compared my equation to Sal's equation on my graphing calculator, but the results were very different. Did I possibly type it wrong, or is my end result of x= -y^3 /3 +578 simply wrong?
• Your error is in how you cubed ( -y + 12 ) (which I'll write as 12 - y). You missed out lots of terms!
( 12 - y )^3 = ( 12 - y )( 12 - y ) ( 12 - y )
= ( 144 - 24y + y^2) ( 12 - y )
and then you also need to distribute each term in the first parentheses to each term in the last parentheses.
• Could you theoretically turn Sal's answer ( ) into (4-y/3)^3+2, by dividing everything by 3, or would that be wrong? Thanks in advance!
(1 vote)
• Remember that (12 − 𝑦)³ is a product, (12 −𝑦)(12 − 𝑦)(12 − 𝑦),
and when dividing a product by 3 we only divide one of its factors by 3, not all of its factors.

So, at best we can write
((12 − 𝑦)³ + 6)∕3 = (4 − 𝑦∕3)(12 − 𝑦)² + 2
• @ when Sal gives the final answer as ((12-y)^3 +6)/3 can't the 6 and the 3 cancel out and wouldn't the final answer be (12-y)^3 +2? Thanks in advance!
(1 vote)
• No, the denominator belongs to the entire fraction. You are applying the denominator to just the 6.
The only way to reduce would be to split the answer into 2 fractions: `(12-y)^3/3 + 6/3`
Then, you can reduce the 6 and 3. You end up with: `(12-y)^3/3 + 2`
Hope this helps.
• Is multiplying both sides by -1 to change the negative radical into a positive necessary? I did a question where there was a negative cubed root and to reverse it I raised both sides to the -3rd power. When I did the calculations this was sound but the question came back as wrong purely because I did not change it to a positive in the way it is done in this video.

In this video, it doesn't sound like something you have to do, rather something you can do if it's easier.

(1 vote)
• Raising something to a negative power will not change the sign of the base to be positive.
If you raised the radical to an exponent of "-3", you created the reciprocal raised to the power of 3.
[-∛(3x-6)]^(-3) = 1 / [-∛(3x-6)]^(3)
= 1 / [-1(3x-6)]
= 1 /[-3x+6]

Did you have to multiply by -1? No, you could have just raised both sides to the 3rd power. But, pay attention to the signs!
(y-12)^3 = [-∛(3x-6)]^(3)
(y-12)^3 = (-1)^3 [∛(3x-6)]^(3)
(y-12)^3 = -1 (3x-6)
(y-12)^3 = -3x+6
(y-12)^3 -6 = -3x
x = [(y-12)^3 -6]/(-3)
x = -[(y-12)^3 -6]/3

Hope this helps.
• what happens when the x is negative
(1 vote)
• Find both g^-1(x) given: g(x)= (x-3)^2+8 & h^-1(x) given: h(x)= (Radical x+1 ) -5 show your work and explain how they are related to one another.
(1 vote)
• g(x)=(x-3)^2+8; h(x)=√x+1 -5
y=(x-3)^2+8; y=√x+1 -5
y-8=(x-3)^2; y+5=√x+1
√y-8 =x-3; (y+5)^2=x+1
√y-8 +3=x; (y+5)^2 -1=x
(1 vote)

## Video transcript

- [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.