Euc20197/Sub-Problem 2

Problem

Consider the function $f(x) = x^2 - 2x$ . Determine all real numbers $x$ so that $x$ satisfy $f(f(f(x))) = 3$ .

Solution 1

Let's say that $f(f(x))$ =a. Then, if $f(a)=3$ , then $a^2-2a-3=0$ , so $a=-1$ or $3$ . Now, let's do $f(f(x))=a$ and call $f(x)=b$ (so basically we are doing $f(b)=a$ ). Here, $a=-1$ or $3$ . If $a=3$ , then $b=3$ or $-1$ . If $a=-1$ , then $b^2-2b=-1$ , so $b^2-2b+1=0$ . Here, $b=1$ is the only solution. Now, let's do $f(x)=b$ . Here, because from $f(b)=a$ we have the possibilities of $b=1, -1,$ or $3$ , so we have $f(x)=1, -1,$ or $3$ . If $f(x)=3$ , then $x=-1$ or $3$ . If $f(x)=-1$ , then $x=1$ . If $f(x)=1$ , then we have $x^2-2x=1$ , so $x^2-2x-1=0$ . Here, after applying the quadratic formula, it will give us $x=1-\sqrt2$ or $1+\sqrt2$ , so our only possibilities are $\boxed{3, -1, 1, 1+\sqrt2,}$ and $\boxed{1-\sqrt2}$ .

~Yuhao2012

Video Solution

https://www.youtube.com/watch?v=M4gzTG8HnQ4

~North America Math Contest Go Go Go

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Euc20197/Sub-Problem 2

Problem

Solution 1

Video Solution