I Made Up A Functional Equation and Solved It

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[@ricardozabalayoe2672]

Well explained. Good examples.

[@sunny_25]

Problem was easy...

[@yoav613]

Nice. 7:08 x cubed x cubed where are you??..🤣

[@tharunsankar4926]

Here's my method:
from inspection, we can deduce that g(x) should be a linear function (or degree 1 polynomial). I chose a form x+u because the first coefficient is 1. Therefore:
f(x+u) = (x+u)^3 - 1
= x^3 + 3ux^2 + 3u^2x + u^3 - 1
But the last term is 0 so comparing the last constants to each other yields: u^3 -1 = 0 => u = 1, exp(2pi/3), exp(4pi/3).
Plugging them in again and comparing makes u = 1 the only answer. I am considering u to be complex as well, because there's no restriction that u can be real.

[@scottleung9587]

Nice - I used the second method.

[@SyberMath]

Excellent!

[@misterdubity3073]

@4:40 A new symbol I never saw before: a small circle which reads as "is composed with" .

[@Ahwke]

❤❤❤❤❤

[@SyberMath]

💗😍

[@ilyasb4792]

The first method isn't always guaranteed to work. This time it worked because the function was bijective on the whole domain, but it's not always the case.
Second method is better in a general case imho.

[@tamilselvanrascal5956]

🎉

[@georgesdermesropian1565]

I used the 2nd method

[@neuralwarp]

Are you assuming all functions are 1-to-1 ?

[@SyberMath]

Am I?

[@GreenMeansGOF]

Third Method: Substitute ax+b into f and compare coefficients.

[@greensword6555]

Can't wait 3bood 🥰

[@yakupbuyankara5903]

G(x)=X+1

[@emanuellandeholm5657]

I simply completed the cube to find one solution to g(x), but I didn't prove it's the only one.

[@rakenzarnsworld2]

g(x)=x+1

[@_ilsegugio_]

f(x)=x^3-1
f[g(x)]=(x+1)^3-1
let t=g(x), f(t)=f[g(x)]->t=x+1
t=g(x)=x+1
are there other solutions tho?

[@maxvangulik1988]

g(x)=x+1 by observation 💀

[@honestadministrator]

f( g ( x)) + 1 = x^3 + 3 x^2 + 3 x + 1
.= ( x + 1) ^3
Hereby [ g(x) ] ^ 3 = ( x + 1) ^3
g ( x) = x + 1, w (x +1), w^2 * (x +1)
Herein w is a root of z^2 + z + 1 = 0
Question of bijection / single valued function ends herein

[@alextang4688]

It is obviously that g(x)=ax+b.
Sub it into the equation and a, b can be found. 😋😋😋😋😋😋

[@trojanleo123]

g(x) = x+1

[@giuseppemalaguti435]

g(x)=x+1

[@mathswan1607]

g(x)=x+1