Let a function be defined as
If is a function, and and are polynomials of degree ,
Find .
Details and assumptions
This problem is part of the set - Piecewise-defined Functions
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For f − 1 to be a function, f must be bijective. It must pass the horizontal line test, and its range must be equal to codomain. The range of f must be R + .
If x ∈ Z , f ( x ) = x ⌊ x ⌋ .
For any non-zero integer n , we can define this as
f ( x ) = n x , if n < x < ( n + 1 )
For any general n , range of f is
If n > 0 , f is an increasing function,
⟹ f ( n + ) < f ( x ) < f ( ( n + 1 ) − )
If n < 0 , f is a decreasing function,
⟹ f ( ( n + 1 ) − ) < f ( x ) < f ( n + )
The limiting values can be found by substitution.
x → n + lim f ( x ) = n 2 and x → ( n + 1 ) − lim f ( x ) = n ( n + 1 )
Since n + 1 is an integer as well, we can replace n + 1 by n
x → n − lim f ( x ) = ( n − 1 ) n = n 2 − n
On observation, for some positive integer k ,
n = − k n = k n = − ( k + 1 ) n = k + 1 k ( k − 1 ) < f ( x ) < k 2 k 2 < f ( x ) < k ( k + 1 ) k ( k + 1 ) < f ( x ) < ( k + 1 ) 2 ( k + 1 ) 2 < f ( x ) < ( k + 1 ) ( k + 2 )
We see that if we extend this table for all k ≥ 1 , then the range of f is all positive real numbers except the limiting values.
If g ( x ) and h ( x ) were x 2 and x 2 − x , or x 2 + x and x 2 , then range of f will become R + , and it will become bijective.
So the functions are g ( x ) = x 2 and h ( x ) = x 2 − x , or x 2 + x and x 2 .
In both cases, we get g ( x ) − h ( x ) = x .
We get the result 5 × ⌊ g ( 3 . 5 ) − h ( 3 . 5 ) ⌋ = 5 ⌊ 3 . 5 ⌋ = 5 × 3 = 1 5