Let f(x)=x^2-2x. Find all real numbers x such that f(x)=f(f(x)). Thank you.
if y = x^2 - 2 x
f(y) = (x^2-2x)^2 - 2(x^2-2x) = f(f(x))
so
x^2-2x = (x^2-2x)^2 - 2(x^2-2x)
of course x = zero will do :)
let z = x^2 - 2x
z = z^2 - 2 z
z^2 - 3 z = 0
z (z-3) = 0
so another solution is
x^2-2x = 0 = x (x-2
we already have x = 0 so now also have x = 2
and also
z = 3 so x^2-2x = 3 which leads to
x^2 - 2 x - 3 = 0
(x+1)(x-3) = 0
so
x = -1
and x = 3 also
so
x = 0
x = 2
x = -1
x = 3
that was fun :)
To find all real numbers x such that f(x) = f(f(x)), we need to solve the equation f(x) - f(f(x)) = 0.
First, let's find f(x) by substituting x^2 - 2x into f(x):
f(x) = (x^2 - 2x)^2 - 2(x^2 - 2x)
= x^4 - 4x^3 + 4x^2 - 2x^2 + 4x
= x^4 - 4x^3 + 2x^2 + 4x
Next, let's find f(f(x)) by substituting f(x) into f(x):
f(f(x)) = f(x^4 - 4x^3 + 2x^2 + 4x)
= ((x^4 - 4x^3 + 2x^2 + 4x)^2 - 2(x^4 - 4x^3 + 2x^2 + 4x))
= (x^8 - 8x^7 + 20x^6 - 16x^5 + 4x^4 + 8x^3 - 2x^4 + 16x^3 - 4x^2 - 8x + 4x^2 + 8x)
= x^8 - 8x^7 + 20x^6 - 18x^5 + 2x^4 + 24x^3
Now, we have the equation f(x) - f(f(x)) = 0:
x^4 - 4x^3 + 2x^2 + 4x - (x^8 - 8x^7 + 20x^6 - 18x^5 + 2x^4 + 24x^3) = 0
x^8 - 8x^7 + 20x^6 - 18x^5 + 2x^4 + 24x^3 - x^4 + 4x^3 - 2x^2 - 4x = 0
Combining like terms, we have:
x^8 - 8x^7 + 20x^6 - 19x^5 + x^4 + 28x^3 - 2x^2 - 4x = 0
Unfortunately, solving this polynomial equation for all real solutions is not straightforward and may require the use of numerical methods or advanced techniques like factoring, synthetic division, or the Rational Root Theorem.
In this case, it would be best to use numerical methods such as graphing the equation or using a calculator or software program to approximate the solutions.