finding inverse function f^-1

[kpx001]

Am i doing this correctly ?

1. y = 4 + cuberoot x

x = 4 + cuberoot y
x^3 = 64 + y
x^3 - 64 = y


2. f(x) = (2 - x^3)^5

x = (2 - y^3 )^5

What do i do next?

 

[pka]

Here is the first one done correctly.

\(\displaystyle \L \begin{array}{l}
y = 4 + \sqrt[3]{x}\quad \Rightarrow \quad x = 4 + \sqrt[3]{y} \\
x - 4 = \sqrt[3]{y}\quad \Rightarrow \quad \left( {x - 4} \right)^3 = y \\
\end{array}\)

 

[kpx001]

thanks. So am I isolating the new Y?
if so for the second problem would i do

(2 - y^3 ) (2 - y^3 ) (2 - y^3 ) (2 - y^3 ) (2 - y^3 ) ?

 

[pka]

First change: \(\displaystyle \L y = \left( {2 - x^3 } \right)^5 \quad \Rightarrow \quad x = \left( {2 - y^3 } \right)^5\).

Now solve for y.
\(\displaystyle \L \begin{array}{l}
\sqrt[5]{x} = \left( {2 - y^3 } \right) \\
y^3 = 2 - \sqrt[5]{x} \\
y = \sqrt[3]{{2 - \sqrt[5]{x}}} \\
\end{array}\)

 

[stapel]

So am I isolating the new Y?

It sounds as though they haven't covered this topic in class (since otherwise you'd have known the answer to that question). Try this methodology:

. . . . .i) Rename "f(x)" as "y".

. . . . .ii) Switch "x" and "y".

. . . . .iii) Solve to get "y=" again.

. . . . .iv) If (and that's a big "if") the new relation "y=" is itself a function,
. . . . .then rename "y" as "f^-1(x)".

for the second problem would i do

(2 - y^3 ) (2 - y^3 ) (2 - y^3 ) (2 - y^3 ) (2 - y^3 ) ?

Why expand the thing? Why not just take roots? :shock:

Eliz.