- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,753

Solve the equation $y+k^3=\sqrt[3]{k-y}$ where $k$ is a real parameter.

- Thread starter anemone
- Start date

- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,753

Solve the equation $y+k^3=\sqrt[3]{k-y}$ where $k$ is a real parameter.

- Thread starter
- Admin
- #3

- Feb 14, 2012

- 3,753

HeyConsider function $f(k) = y + k^3$, then $f^{-1}(k) = \sqrt[3]{k - y}$. Hence \[f^{-1}(k) = f(k)\] This can happen if and only if \[k = f(k) = f^{-1}(k)\] i.e. \[k = \sqrt[3]{k - y} = y + k^3\] So \[\boxed{y = k - k^3}\]

- Nov 26, 2013

- 731

Thankyou for your answers. Jacks solution is very elegant!

I have one question. I do not understand why the following implication is true:

$$ f(k)=y+k^3 \Rightarrow f^{-1}(k)=\sqrt[3]{\mathbf{k}-y} $$

Why is k appearing on the RHS?

I would deduce the following:

$$ f(k)=y+k^3 \Rightarrow f^{-1}(k)=\sqrt[3]{f(k)-y}$$

By definition:

$$f(k)=y+k^3=\sqrt[3]{k-y}=\sqrt[3]{f(k)-y} \Rightarrow f(k) = k \Rightarrow y+k^3=k \Rightarrow y = k-k^3$$

- Thread starter
- Admin
- #5

- Feb 14, 2012

- 3,753

Thankyou for your answers. Jacks solution is very elegant!

I have one question. I do not understand why the following implication is true:

$$ f(k)=y+k^3 \Rightarrow f^{-1}(k)=\sqrt[3]{\mathbf{k}-y} $$

Why is k appearing on the RHS?

I would deduce the following:

$$ f(k)=y+k^3 \Rightarrow f^{-1}(k)=\sqrt[3]{f(k)-y}$$

By definition:

$$f(k)=y+k^3=\sqrt[3]{k-y}=\sqrt[3]{f(k)-y} \Rightarrow f(k) = k \Rightarrow y+k^3=k \Rightarrow y = k-k^3$$

Hi

I am sorry for I only replied to you days after...I thought to myself to let jacks to handle it and I would only chime in if we didn't hear from jacks 24 hours later. But it somehow just slipped my mind.

Back to what you asked us...I believe if we use the identity

$f(f^{-1}(k))=k$,

and that for we have $f(k)=y+k^3$, we would end up with getting $f^{-1}(k)=\sqrt[3]{\mathbf{k}-y} $, does that answer your question, lfdahl?

$f(k)=y+k^3$

$f(f^{-1}(k))=k$

$y+(f^{-1}(k))^3=k$

$(f^{-1}(k))^3=k-y$

$f^{-1}(k)=\sqrt[3]{k-y}$

$f(f^{-1}(k))=k$

$y+(f^{-1}(k))^3=k$

$(f^{-1}(k))^3=k-y$

$f^{-1}(k)=\sqrt[3]{k-y}$