How do you find the inverse of #f(x)=x^3-2# and is it a function?

1 Answer
Mar 16, 2016

#f^-1(x)=color(white)(0)^3sqrtx+2#

Explanation:

We begin with #f(x)=x^3-2#. To find the inverse of any equation, just switch #x# and #y#. No, before we do thta, I'm going to change the equation. I'm going to rename #f(x)# to #y#, just so that I don't have to deal with the parentheses or anything. Now, all I'bve doe is change what #f(x)# is called, not it's actual value.

Anyways, we have #y=x^3-2#. Now we switch #x# and #y# and then solve for #y#.

Now we've got #x=y^3-2#. If we add #2# on both sides we have #x+2=y^3#. Now we just need to get #y# as simple as possible, which we'll do by cube rooting both sides of the equation. That leaves us with #y=color(white)(0)^3sqrtx+2#. Now we just change #y# back to #f(x)# and add a #color(white)(0)^-1# to write it in inverse notation, and we have #f^-1(x)=color(white)(0)^3sqrtx+2#.