Very easy. We have to find some y = f(x) such that y² - y + 1 equals x^4-2x³ + x. It is intuitive that y should be a quadratic polynomial because the operation y² - y + 1 turns it in a quartic. So we guess that y should be of the form ax² +bx+c and equate y² - y + 1 = x^4 -2x³ + x. By comparing the coefficients of each power on the both sides of the equation, we get a, b, and c. a = 1, 3a+b =0 and a+b+c=0, therefore b = -3 and c = 2.

Ah?! IS THE PERSON FROM THUMBNAIL SYBER ( lol probably not) . BUT. WE. NEED. A FACE . REVEAL!!!

@Ron_DeForest

4 күн бұрын

Might be him. Besides the sunglasses, he fits the avatar.

Good exercise!) Thank you

Given that putting in a quadratic yields a quartic, f had to be a quadratic itself. I let f(t) = at^2 + bt + c, let t = x^2 - x + 1, and solved for a, b, and c. Even though it was mostly obvious that a = 1, I chose to stick to the safer option.

@jpolowin0

5 күн бұрын

I'm not sure that f _must_ be a quadratic, but it's worth testing to see if it gives self-consistent results. Ditto that a = 1. Making those assumptions gives a valid solution for f(y) = y² - 3y +2.

@robertlunderwood

5 күн бұрын

@@jpolowin0 Admittedly, it's not the most conventional of methods, but barring something completely off the wall, anything that could be reasonable is easily eliminated. I did the work to show that a = 1 just in case there may have been any strange cancellations.

@jpolowin0

5 күн бұрын

@@robertlunderwood Assuming a quadratic, nothing other than a=1 is possible in order to get a coefficient of 1 on the x^4 term in the result. I *think* that anything other than a=1 would mess up the -2 coefficient on the x^3 term. Obviously, there's no harm in doing the checking, and a=1 falls out immediately from that x^4 term.

Cool!

Let f(x) ::= g(x)•x³ + ax² + bx + c And expand for g, a, b, c

I came up with two alternative approaches, both based on the ansatz of f(x) being a polynomial, most likely quadratic: 1. Plug in several distinct values for x, until you get three or more distinct values for x^2 - x + 1. Determine the values of x^4 - 2*x^3 + x for each of these x values. Interpolate a polynomial to the pairs (x^2 - x + 1, x^4 + 2*x^3 + x). For this, I used the x values { 1, 2, 3 }, which gave the (t, f(t)) pairs { (1, 0), (3, 2), (7, 30) }. Interpolating in Lagrange form gives: f(x) = 2*(x-1)*(x-7)/(3-1)/(3-7) + 30*(x-1)*(x-3)/(7-1)/(7-3) = x^2 - 3*x + 2 2. Solve x^4 - 2*x^3 + x = 0 for x, giving the roots { 0, 1, (1 - sqrt(5))/2, (1 + sqrt(5))/2 }. Substitute these roots into x^2 - x + 1, giving the values 1, 1, 2, and 2. These are the roots of f(x) = 0; given that the coefficient of the x^4 term in the right hand side is one, this means that f(x) = (x-1)^a*(x-2)^b, where a, b are both from the set { 1, 2 }. Checking the cases shows that only a = 1, b = 1 works, meaning that f(x) = (x-1)*(x-2) = x^2 - 3*x + 2.//

We can write x^4-2x^3+x as (x-1)(x^3-x^2-x) since x=1 is a root of this polynom. Then, x^4-2x^3+x =x(x-1)(x^2-x-1) or (x^2-x)(x^2-x-1). Let y= x^2-x+1, so f(y)= (y-1)(y-2)=y^2-3y+2. Then, f(x)=x^2-3x+2.

@randompersonasdf

5 күн бұрын

neat

@Ayush-yj5qv

4 күн бұрын

Same

Just a simple long division

Similar to another solution in the comments, but too complicated in comparison, as I did not realize that substituting x² - x + 1 directly would have been easier 😂 x⁴ - 2x³ + x = (x² - x)(x² - x - 1) Substitute u = x² - x, then we have: f(u + 1) = u(u -1). Then let v = u + 1, so f(v) = (v - 1)(v - 2) = v² - 3v + 2 or f(x) = x² - 3x + 2.

@SyberMath

5 күн бұрын

Nice!

I want you to teach som first little theory

Goal: Find all functions f: ℝ→ℝ such that for all x in ℝ ; f(x^2 − x + 1) = x^4 − 2x^3 + x x^2 − x + 1 = (x − 1/2)^2 + 3/4 . let t = (x − 1/2)^2 + 3/4 ⇒ x = 1/2 + √(t − 3/4) or x = 1/2 − √(t − 3/4) . Now, when we substitute x, it is important to substitute with both expressions because we have to verify that we won't get a contradiction like f(t) = t and f(t) = t^2 . Of course, for the LHS, we will get just f(t) but for the RHS we may get different expressions depending on the expression we substitute with. Getting diffrenent expressions is a contradiction and it means that the equation doesn't have a solution. Substituting with 1/2 + √(t − 3/4) and 1/2 − √(t − 3/4) gives on the RHS the same expression t^2 − 3t + 2 which means there is no contradiction. But that not everything! t is not just any real number; t ranges from 3/4 to infinity so f(t) is not constrained for the values The complete solution is any function f: ℝ→ℝ such that for all t ≥ 3/4; f(t) = t^2 − 2t + 2 and for all t That would be all.

The answer for f(x)=0 are x = 1 or 2

f(x^2-x+1)=x(x-1)(x^2-x-1)...t=x^2-x+1=>x(x-1)=t-1,x^2-x-1=t-2...quindi..f(t)=(t-1)(t-2)...f(x)=(x-1)(x-2)

Let's say f(x) is a kuadratic Function

x^4 - 2 x^3 + x = ( x^2 - x) ^2 - x^2 + x = ( z - 1) ^2 - ( z - 1) Here in z = x^2 - x + 1 Hereby f ( z) = ( z - 1) ( z - 2) = z^2 - 3 z + 2

Is this Math or ya know

@Gezraf

5 күн бұрын

nah its quantum thermobiology

@browhat6935

5 күн бұрын

real ​@@Gezraf

@MateusMuila

5 күн бұрын

@@Gezraf I didn't know. Thanks for telling me.