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- Feb 14, 2012

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

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- #1

- Feb 14, 2012

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

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- Feb 14, 2012

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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

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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:

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

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- Feb 14, 2012

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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?

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

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

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

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

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

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

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

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

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