Solving A Functional System of Equations in Two Ways

If you need to post a picture of your solution or idea:
#ChallengingMathProblems #FunctionalEquations

EXPLORE 😎:

PLAYLISTS 🎵 :

[MisterPenguin]

Thanks for fixing that errant g(x); it felt like I held in a sneeze. You're so conscientious.

[RyanLewis-Johnson-wqxs]

f(g(x))=4x^2-2x+1 g(x)=2x-7 f(x)=x^2+13x+43

[abshariadam]

The second is basically putting the inverse of g(x) in the f(g(x)) so it would reverse the process of the functions.

But, I really like the approach of the first method, which is kind of wild when you think of a function as an individual variables.

Thank you ❤

[matheusresendeguedes]

A third method would be to assume the answer is a polynomial and just interpolate.

[kxxnhpg]

The second method uses the inverse function of g 👍🏻

[barryzeeberg]

at 6:01, you forgot to replace "g(x)" with "x"

[judeugwu]

Aren’t there an infinite number of functions for f(x)?

[guyhoghton]

The second method relies on the fact that composition of functions is associative, i.e.
if f∘g denotes the function of x formed by f(g(x)) then
f∘(g∘h) = (f∘g)∘h for any functions f, g, h
and on the fact that g(g⁻¹(x)) = x where g⁻¹(x) denotes the inverse function of g(x).
Here g⁻¹(x) = ⁽ˣ ⁺ ⁷⁾/₂
Then f(x) = f( g(g⁻¹(x)) ) = f(g( ⁽ˣ ⁺ ⁷⁾/₂ ) = 4(⁽ˣ ⁺ ⁷⁾/₂)² - 2(⁽ˣ ⁺ ⁷⁾/₂) + 1 = x² + 13x + 43.

Another way is to consider that f(x) maps a linear function to a quadratic and so is a quadratic function itself, i.e.
f(x) = ax² + bx + c for some constants a, b, c.
To evaluate a, b, c we can simplify f(g(x)) = f(2x - 7) ≡ (2x - 7)²a - 2(2x - 7)b + c
and compare the coefficients of x², x, 1 with f(g(x)) ≡ 4x² - 2x + 1 and solve the resulting equations for a, b, c.
Alternatively substitute some values for x that give conveniently low values for g(x):
c = f(0) = f(g(⁷/₂)) = 4(⁴⁹/₄) - 2(⁷/₂) + 1 = 43 ... ①
a + b + 43 = f(1) = f(g(4)) = 4(16) - 2(4) + 1 = 57 ⇒ a + b = 14 ... ②
25a + 5b + 43 = f(5) = f(g(6)) = 4(36) - 2(6) + 1 = 133 ⇒ 25a + 5b = 90 ... ③
From ① ② ③ solve for a, b, c to get f(x) = x² + 13x + 43.

[TrebolManger]

2
a x + b x + c = F(G(X))
2 2
a x + b x + c = 4 x - 2 x + 1

f(2x-7) LEFT

2 2
a (2 x - 7) + b (2 x - 7) + c = 4 x - 2 x + 1
ALGEBRA LEFT

2 2
4 a x - 28 a x + 2 b x + 49 a - 7 b + c = 4 x - 2 x + 1
FACTORIZE BY X LEFT
2 2
4 a x + 2 x (b - 14 a) - 7 b + 49 a + c = 4 x - 2 x + 1
IGUALATION TERM TO TERM LEFT & RIGTH
EQUATIONS SYSTEM 3X3

2 2
4 a x = 4 x ∧ 2 x (b - 14 a) = - 2 x ∧ - 7 b + 49 a + c = 1
SOLVE
a = 1 ∧ b = 13 ∧ c = 43
ASSEMBLE F(X)
2
y = x + 13 x + 43

[nandakumarcheiro]

You are commtting a lot mistakes during solution,