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Difference between revisions of "Euc20197/Sub-Problem 2"

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== Problem ==
 
== Problem ==
  
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Consider the function <math>f(x) = x^2 - 2x</math>. Determine all real numbers <math>x</math> so that <math>x</math> satisfy <math>f(f(f(x))) = 3</math>
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Consider the function <math>f(x) = x^2 - 2x</math>. Determine all real numbers <math>x</math> so that <math>x</math> satisfy <math>f(f(f(x))) = 3</math>.
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==Solution 1==
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Let's say that <math>f(f(x))=a</math>. Then, if <math>f(a)=3</math>, then <math>a^2-2a-3=0</math>, so <math>a=-1</math> or <math>3</math>. Now, let's do <math>f(f(x))=a</math> and call <math>f(x)=b</math> (so basically we are doing <math>f(b)=a</math>). Here, <math>a=-1</math> or <math>3</math>. If <math>a=3</math>, then <math>b=3</math> or <math>-1</math>. If <math>a=-1</math>, then <math>b^2-2b=-1</math>, so <math>b^2-2b+1=0</math>. Here, <math>b=1</math> is the only solution. Now, let's do <math>f(x)=b</math>. Here, because from <math>f(b)=a</math> we have the possibilities of <math>b=1, -1,</math> or <math>3</math>, so we have <math>f(x)=1, -1,</math> or <math>3</math>. If <math>f(x)=3</math>, then <math>x=-1</math> or <math>3</math>. If <math>f(x)=-1</math>, then <math>x=1</math>. If <math>f(x)=1</math>, then we have <math>x^2-2x=1</math>, so <math>x^2-2x-1=0</math>. Here, after applying the quadratic formula, it will give us <math>x=1-\sqrt2</math> or <math>1+\sqrt2</math>, so our only possibilities are <math>\boxed{3, -1, 1, 1+\sqrt2,}</math> and <math>\boxed{1-\sqrt2}</math>.
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~Yuhao2012
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~minor changes by Baihly2024
  
 
== Video Solution ==
 
== Video Solution ==

Latest revision as of 16:30, 12 October 2025

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

~minor changes by Baihly2024

Video Solution

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

~North America Math Contest Go Go Go

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