The second method is nicer because you don't have to assume a functional form for f(x). It also automatically settles the question of whether any other solutions are possible.
This one solves by inspection. Most functional equations that are soluble have solutions of the form ax + b or ax^2 + bx + c. As there is an iterated functional on the left and a square on the right the x^2 term must be zero. Now let x = 0. Both sides of the equation equal zero. This means that the constant term must be zero. Try f(x) = ax. This gives a = 5/2. This method doesnt always work but its often worth a try
Note that f(x) is not necessarily a nice function. F can be a function where for each value, you chose whether f(x) is 5x/2 or -5x/2-1. Even if you impose that the function is continues, you still get 2 extra solutions compared to the 2 trivial functions. In fact, at the point x=-2/5, the 2 fonctions cross. So you can chose that to use one of the formulas below -2/5 and the other one for above -2/5. Without the assumption of the function being continues, you can get infintely many non continues functions as solution.
I found a third method. Taking everything to LHS generates a quadratic equation of f(x) with a=1 b=1 and c=-(25x^2+10x)/4 f(x)=(-1+/- (5x+1))/2 f(x)=5/2 or -1 -5x/2
The second method is nicer because you don't have to assume a functional form for f(x). It also automatically settles the question of whether any other solutions are possible.
This one solves by inspection. Most functional equations that are soluble have solutions of the form ax + b or ax^2 + bx + c. As there is an iterated functional on the left and a square on the right the x^2 term must be zero. Now let x = 0. Both sides of the equation equal zero. This means that the constant term must be zero. Try f(x) = ax. This gives a = 5/2.
This method doesnt always work but its often worth a try
Note that f(x) is not necessarily a nice function. F can be a function where for each value, you chose whether f(x) is 5x/2 or -5x/2-1. Even if you impose that the function is continues, you still get 2 extra solutions compared to the 2 trivial functions. In fact, at the point x=-2/5, the 2 fonctions cross. So you can chose that to use one of the formulas below -2/5 and the other one for above -2/5. Without the assumption of the function being continues, you can get infintely many non continues functions as solution.
Interesting
It s pretty easy if you see (25x^2+ 10x)/4 can be written as (5x/2)^2 + 5x/2 so f(x) = 5x/2
I found a third method.
Taking everything to LHS generates a quadratic equation of f(x) with a=1 b=1 and c=-(25x^2+10x)/4
f(x)=(-1+/- (5x+1))/2
f(x)=5/2 or -1 -5x/2
Just use the quadratic solution. Ultimately this is all it is.
quadratic formula and realize that 25x² + 10x + 1 is equal to (5x + 1)² so the root and the square cancel out.
Spotted the two squares 👍
Long time that you started with the second method
Nice!
Quadratic?
Factorization gives
( f(x) - 5 x/2) ( f(x) + 5 x /2 + 1) = 0
f(x) = 5 x /2, - 5 x/2 -1
The 2nd one is bravo！❤
(f(x))² + f(x) = (25x² + 10x)/4
(f(x))² + f(x) + (-25x² - 10x)/4 = 0
Discriminant:
1² - 4 × 1 × (-25x² -10x)/4
1 + 25x² + 10x
Now apply in the quadratic formula:
First solution:
f(x) = [-1 + √(25x² + 10x + 1)]/2
f(x) = (-1 + |5x + 1|)/2
5x + 1 ≥ 0 ⇒ f(x) = 5x/2
5x + 1 < 0 ⇒ f(x) = (-5x -1 -1)/2 = (-5x -2)/2
Second solution:
f(x) = (-1 - |5x + 1|)/2
5x + 1 ≥ 0 ⇒ f(x) = (- 1 - 5x - 1)/2 = (-5x - 2)/2
5x + 1 < 0 ⇒ f(x) = (-1 + 5x + 1)/2 = 5x/2
So the solutions are *f(x) = 5x/2 ∧ f(x) = (-5x - 2)/2*
F(x) = 2.5x, and f(x) = -2.5x - 1😊
Great
5x/2...-5x/2-1
Happy New Year!
It is obviously that f(x) is a linear equation. We can say it f(x)=ax+b. Then we can solve it by comparing the terms. 😋😋😋😋😋😋
One obvious solution is f(x) = 5/2*x
easy
f(x)=(5/2)x