The following defines a one-to-one function, find the invers

[ochocki]

y=(x+2)^3-8

I don't know how to use the square root sign via latex or any other means, so bear with me.

I switched the variables:

x=(y+2)^3-8

Then I couldn't really figure what to do from here, I thought about using a cube root to eliminate the cubed term, this resulted in:

x=y+2-2 (eliminated the cubed term and took the cube root of eight, 2.

this just results in x=y, and that seems wrong to me. What am I missing?

 

[pka]

\(\displaystyle \L \begin{array}{rcl}
y & = & \left( {x + 2} \right)^3 - 8 \\
x & = & \left( {y + 2} \right)^3 - 8 \\
x - 8 & = & \left( {y + 2} \right)^3 \\
\sqrt[3]{{x - 8}} & = & \left( {y + 2} \right) \\
\sqrt[3]{{x - 8}} - 2 & = &y \\
\end{array}\)

 

[stapel]

x=(y+2)^3-8

...I thought about using a cube root to eliminate the cubed term, this resulted in:

x=y+2-2 (eliminated the cubed term and took the cube root of eight, 2.

Does (a + b)³ equal a³ + b³?

Then why would cbrt[a + b] equal cbrt[a] + cbrt?

Eliz.

 

[ochocki]

That makes sense, thanks guys, I was somewhat close

I love this forum, you guys are so helpful.