Question

Let $f(x) = x^2 - 2x$. Find all real numbers $x$ such that $f(x) = f(f(x))$. List your solutions separated by commas in any order​

Answer 1

Answers: -1, 0, 2, 3

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Step-by-step explanation:

The given function is f(x) = x^2 - 2x

Let's apply function composition like so

`f(x) = x^2 - 2x\\\\f(f(x)) = (f(x))^2 - 2(f(x))\\\\f(f(x)) = (x^2-2x)^2 - 2(x^2-2x)\\\\`

In the second step, I replaced every x with f(x). Then in the third step, I replaced f(x) with x^2-2x on the right side.

Setting this equal to f(x) gets us

`f(f(x)) = f(x)\\\\(x^2-2x)^2 - 2(x^2-2x) = x^2-2x\\\\`

which is a bit cluttered. However, we have "x^2-2x" show up three times. Let's say w = x^2-2x

If we replaced all those "x^2-2x" expressions with w, then we get,

`(x^2-2x)^2 - 2(x^2-2x) = x^2-2x\\\\w^2 - 2w = w\\\\`

Which looks more manageable

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Let's solve for w

w^2 - 2w = w

w^2 - 2w-w = 0

w^2 - 3w = 0

w(w - 3) = 0

w = 0 or w-3 = 0

w = 0 or w = 3

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If w = 0, then,

w = x^2 - 2x

x^2 - 2x = w

x^2 - 2x = 0

x(x - 2) = 0

x = 0 or x-2 = 0

x = 0 or x = 2 are two solutions (out of four total)

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Or, if w = 3, then,

w = x^2 - 2x

x^2 - 2x = w

x^2 - 2x = 3

x^2 - 2x - 3 = 0

(x - 3)(x + 1) = 0

x-3 = 0 or x+1 = 0

x = 3 or x = -1 are the other two solutions

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To summarize, the four solutions are: x = -1, x = 0, x = 2, x = 3