Difference between revisions of "2013 MPFG Problem 17"

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As such, <math>f(f(f(x))</math> will have <math>2\cdot4^{2} = 32</math> cycles in the domain <math>[-2,2]</math>. Each has <math>2</math> points that cross with <math>f(x)</math>. However, at <math>3</math> special locations, the crossing points of two consecutive <math>f(f(f(x)))</math> cycles overlapped (at <math>x=-1, 0, 1</math>).
 
As such, <math>f(f(f(x))</math> will have <math>2\cdot4^{2} = 32</math> cycles in the domain <math>[-2,2]</math>. Each has <math>2</math> points that cross with <math>f(x)</math>. However, at <math>3</math> special locations, the crossing points of two consecutive <math>f(f(f(x)))</math> cycles overlapped (at <math>x=-1, 0, 1</math>).
 
As such, number of total points is <math>2\cdot32-3 = \boxed{61}</math>
 
As such, number of total points is <math>2\cdot32-3 = \boxed{61}</math>
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~cassphe

Latest revision as of 09:10, 21 August 2025

Problem

Let f be the function defined by $f(x) = -2 sin(\pi x)$ . How many values of x such that $-2 \leq x \leq 2$ satisfy the equation $f(f(f(x))) = f(x)$?

Solution 1

We first create the graph of $f(x)$ by transforming a regular $sin$ graph by scaling down $x$ by $\pi$ and scaling up $y$ by $-2$.

Graph of f(x).jpg

We observe that the domain & range of $f(x)$ is restricted in $[-2,2]$ and the graph is restricted in a $2x2$ square. Trying out $f(f(x))$, we see that every $\frac{1}{4}$ of a cycle of $f(x)$ ranges over $[0,\pm 2]$, which means a cycle for $(f(x))$ will have $4$ cycles for $f(f(x))$. $f(x)$ and $f(f(x))$ have two crossing points in each small cycle of $f(f(x))$

Graph of f(f(x)).jpg

As such, $f(f(f(x))$ will have $2\cdot4^{2} = 32$ cycles in the domain $[-2,2]$. Each has $2$ points that cross with $f(x)$. However, at $3$ special locations, the crossing points of two consecutive $f(f(f(x)))$ cycles overlapped (at $x=-1, 0, 1$). As such, number of total points is $2\cdot32-3 = \boxed{61}$

~cassphe