MHB Solve equation: $y+k^3=\sqrt[3]{k-y}$

  • Thread starter Thread starter anemone
  • Start date Start date
AI Thread Summary
The equation $y+k^3=\sqrt[3]{k-y}$ is discussed with a focus on the implications of the function $f(k)=y+k^3$ and its inverse. A participant questions why $k$ appears on the right-hand side of the equation when deriving the inverse function. The response clarifies that using the identity $f(f^{-1}(k))=k$ leads to the conclusion that $f^{-1}(k)=\sqrt[3]{k-y}$. The conversation emphasizes the elegance of the solution provided by a participant named Jacks, while also addressing the confusion regarding the function's properties. The discussion concludes with a reaffirmation of the mathematical reasoning behind the inverse function.
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Solve the equation $y+k^3=\sqrt[3]{k-y}$ where $k$ is a real parameter.
 
Mathematics news on Phys.org
Consider 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 \[\boxed{y = k - k^3}\]
 
Last edited by a moderator:
jacks said:
Consider 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 \[\boxed{y = k - k^3}\]

Hey jacks, thanks for participating and your solution is simple, elegant and nice! Well done, jacks!:)
 
Hi, anemone and jacks!

Thankyou for your answers. Jacks solution is very elegant!(Star)

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:
$$f(k)=y+k^3=\sqrt[3]{k-y}=\sqrt[3]{f(k)-y} \Rightarrow f(k) = k \Rightarrow y+k^3=k \Rightarrow y = k-k^3$$
 
lfdahl said:
Hi, anemone and jacks!

Thankyou for your answers. Jacks solution is very elegant!(Star)

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 lfdahl,

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?:)

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

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

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

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

$f^{-1}(k)=\sqrt[3]{k-y}$
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

Similar threads

Back
Top