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> | ||
+ | |||
+ | ~cassphe |
Latest revision as of 09:10, 21 August 2025
Problem
Let f be the function defined by . How many values of x such that
satisfy the equation
?
Solution 1
We first create the graph of by transforming a regular
graph by scaling down
by
and scaling up
by
.
We observe that the domain & range of is restricted in
and the graph is restricted in a
square. Trying out
, we see that every
of a cycle of
ranges over
, which means a cycle for
will have
cycles for
.
and
have two crossing points in each small cycle of
As such, will have
cycles in the domain
. Each has
points that cross with
. However, at
special locations, the crossing points of two consecutive
cycles overlapped (at
).
As such, number of total points is
~cassphe