2013 MPFG Problem 17

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 $-2$ .

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 $f(f(x))$

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

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2013 MPFG Problem 17

Problem

Solution 1