Let f(x) be a polynomial such that
f(f(x))−x^2=xf(x).
Find f(−100).
It is clear that f(x) is linear. If it were, say, quadratic,
f(x) = ax^2+bx+c with a≠0
f(f) = af^2+bf+c
= a(ax^2+bx+c)^2 + b(ax^2+bx+c) + c
is a 4th-degree polynomial.
xf(x)+x^2 = x(ax^2+bx+c) + x^2
is a 3rd-degree polynomial. That would mean a=0.
So, let
f(x) = ax+b
f(f) = af+b = a(ax+b)+b = a^2x+(ab+b)
xf(x)+x^2 = x(ax+b)+x^2 = (a+1)x^2 + bx
So, a+1=0, and a = -1
That makes b=1, and so
f(x) = 1-x
check:
f(f(x)) = 1-f = 1-(1-x) = x
xf(x)+x^2 = x(1-x)+x^2 = x
101
To find f(-100), we need to solve the given equation and determine the values of f(x).
Let's start by rewriting the given equation:
f(f(x)) - x^2 = x * f(x)
Since f(x) is a polynomial, let's assume that:
f(x) = a_n * x^n + a_(n-1) * x^(n-1) + ... + a_1 * x + a_0
Substituting this into our equation, we get:
f(f(x)) - x^2 = x * (a_n * x^n + a_(n-1) * x^(n-1) + ... + a_1 * x + a_0)
Expanding further, we have:
f(f(x)) - x^2 = a_n * x^(n+1) + a_(n-1) * x^n + ... + a_1 * x^2 + a_0 * x
Now, let's substitute x with f(x) in the equation:
f(f(f(x))) - f(x)^2 = f(x) * (a_n * f(x)^(n+1) + a_(n-1) * f(x)^n + ... + a_1 * f(x)^2 + a_0 * f(x))
Expanding this equation gives:
a_n * f(x)^(n+1) + a_(n-1) * f(x)^n + ... + a_1 * f(x)^2 + a_0 * f(x) - f(x)^2 = f(x) * (a_n * f(x)^(n+1) + a_(n-1) * f(x)^n + ... + a_1 * f(x)^2 + a_0 * f(x))
Canceling out like terms, we get:
a_n * f(x)^(n+1) + a_(n-1) * f(x)^n + ... + a_1 * f(x)^2 + a_0 * f(x) - f(x)^2 = 0
Now, let's equate the coefficients of like powers of f(x):
For the highest power, a_n * f(x)^(n+1) = 0, which implies either a_n = 0 or f(x)^(n+1) = 0.
Since f(x) is a polynomial, f(x)^(n+1) = 0 implies that f(x) = 0.
Therefore, either a_n = 0 or f(x) = 0.
Assuming f(x) = 0, substituting in the original equation, we have:
0 - x^2 = x * 0
0 - x^2 = 0
This implies that x = 0.
So, one possible solution is f(x) = 0.
Therefore, f(-100) = 0.
Please note that there may be other solutions for f(x), but we have found one possible value for f(-100) based on the given equation.