Let f(x) be a polynomial such that
f(f(x))−x^2=xf(x).
Find f(−100).
101
rearranging,
f(f(x))=x^2 + x*f(x).
assume degree of f(x) to be m.
therefore,f(f(x)) must have degree m^2.
therefore,L.H.S has degree m^2.
let us assume m >2 .
therefore,R.H.S has degree m+1.
now,the polynomial on both sides must have same degree.
hence,m^2=m+1.
but,then m has no integer solution .
therefore our assumption was wrong and m<2.
possible values for m are 0 and 1.
therefore,f(x)=c or f(x)=b*x+c.
now,by trial and error we get f(x)=1-x .
To find the value of f(-100), we need to use the given equation and some algebraic manipulation.
Let's start by substituting x = -100 into the equation:
f(f(-100)) - (-100)^2 = (-100)f(-100)
Simplifying this, we have:
f(f(-100)) - 10000 = -100f(-100)
Next, we can replace f(-100) with a variable, let's say 'a', to make the equation easier to work with:
f(f(-100)) - 10000 = -100a
Now, we have a new equation:
f(f(-100)) = -100a + 10000
To find f(-100), we need to find f(f(-100)). This means we need to find the value of f(x) when x is f(-100).
Let's substitute x = f(-100) into the equation:
f(f(f(-100))) - ((f(-100))^2) = f(-100) * f(f(-100))
Now, we can replace f(f(-100)) with -100a + 10000 based on the equation we derived earlier:
f(-100a + 10000) - ((f(-100))^2) = f(-100) * (-100a + 10000)
To make further progress, we need more information about the polynomial f(x). Without additional information, it is not possible to determine the value of f(-100) based solely on the given equation.