Finding the Inverse of a One to One Function

[NobodysHero]

Could someone double check my work for the problem: Find the Inverse of the one to one function (y)= cbrt (3x-6). My first step was to change the cube root to an exponent and exchange the x and y so x= (3y-6)^1/3. I then multiplied the parenthesis by the exponent and got x=y-2. Finally I add 2 to both sides and got y=x+2. Is my work and answer correct?

 

[pka]

Could someone double check my work for the problem: Find the Inverse of the one to one function (y)= cbrt (3x-6). My first step was to change the cube root to an exponent and exchange the x and y so x= (3y-6)^1/3. I then multiplied the parenthesis by the exponent and got x=y-2. Finally I add 2 to both sides and got y=x+2. Is my work and answer correct?

No, that is incorrect. Look at \(\displaystyle y=\dfrac{x^3+6}{3}\).
Is that the inverse?

 

[HallsofIvy]

Could someone double check my work for the problem: Find the Inverse of the one to one function (y)= cbrt (3x-6). My first step was to change the cube root to an exponent and exchange the x and y so x= (3y-6)^1/3. I then multiplied the parenthesis by the exponent and got x=y-2. Finally I add 2 to both sides and got y=x+2. Is my work and answer correct?

Your first step, swapping x and y to get x=(3y- 6)^(1/3) is correct but pretty much everything after that is wrong. You cannot get rid of a 1/3 power by multiplying by 1/3! You need to take the third power (since x^3 is the opposite of x^(1/3)). And you NEVER do something like that to just one side of an equation. Taking the third power of BOTH sides gives you x^3= 3y- 6. Solve that for y. And you can check your answer: both f(f^(-1)(x)) and f^(-1)(f(x)) must be equal to x. Notice that f(x)+ 2= (3x- 6)^(1/3)+ 2 is NOT x!

 

[NobodysHero]

Thank you for the help I thought I had messed up somewhere. Now just to clarify Isn't cube root the same as 1/3 power?

 

[HallsofIvy]

Yes, it is. That was why I said "Your first step, swapping x and y to get x=(3y- 6)^(1/3) is correct".

 

[NobodysHero]

Ok. If it's the same thing why can't I multiply just (3y-6) by ^1/3? Not trying to be rude. I'm just trying to understand this for future problems like this.

 

[DrPhil]

Ok. If it's the same thing why can't I multiply just (3y-6) by ^1/3? Not trying to be rude. I'm just trying to understand this for future problems like this.

The cube root IS the 1/3 power, but that is NOT the same as multiplication by 1/3. It is not generally true that x^n = n*x Perhaps you should review the chapter that has exponents in it.

The inverse operation of taking the (1/3) power is to take the power 3. That is,
.......\(\displaystyle \displaystyle \left[x^{1/3}\right]^3 = x^{(1/3)(3)} = x\)

To eliminate a cube root on one side of an equation, you have to cube both sides.

 

[NobodysHero]

Thank you Dr. Phil. I thought that changing cube root to the ^1/3 was the same thing so I thought I didn't need to do it to the other side. Thanks for clarifying it for me. Thank you again HallsofIvy you seem to be my math savior lol.

 

[pka]

Thank you Dr. Phil. I thought that changing cube root to the ^1/3 was the same thing so I thought I didn't need to do it to the other side. Thanks for clarifying it for me. Thank you again HallsofIvy you seem to be my math savior lol.

I think that you are badly confused.

If \(\displaystyle y=\sqrt[3]{3x-6}\) then \(\displaystyle y^3=3x-6\). Swap \(\displaystyle x^3=3y-6\)

Solve for \(\displaystyle y\): \(\displaystyle y=\dfrac{x^3+6}{3}\).