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is that \(f(x)=(\sqrt[3]x/8) -4\) ?

\[f(x)=\sqrt[3]{\frac{ x }{ 8 }}-4\]

oh, ok
to get the inverse
let y= f(x)
and interchange 'x' and 'y'

just solve for x and at the end swap x and y.

\(y= \sqrt[3]{\dfrac{x}{8}}-4\\x= \sqrt[3]{\dfrac{y}{8}}-4\)
solve for 'y' !
can you ?

|dw:1372623159821:dw|

maybe start with cuping both sides?

ADD 4 and then cube both sides :)

same thing :P

\[x ^{3}=\frac{ y }{ 8 } -4\]

no, no....
we first would need to add 4 on both sides...
first do that before cubing.

Oh...

BUt adding 4 first will be easier :P

\[x+4=\sqrt[3]{\frac{ x }{ 8}}\]

\[x+4^{3}=\frac{ x }{ 8 } ?\]

looks fishy.

when you cube , you cube entirely,
so it'll be
(x+4)^3 = y/8
right ?

yea

now what you can do to isolate 'y' ?

multiply by 8

correct!
do that and tell me what u get ?

(x+40)^3=y?

no, sorry

\[f ^{-1}(x)=8(x+4)^{3}\]

correct! good :)

remember do entirely. WHen you + / * square or cupe you do it to whole side.