If [tex]f : {x : -1 \le x \le 1} \rightarrow {x : -1 \le x \le 1}[/tex], then which is/are bijective?
[tex]f(x) = [x][/tex]
[tex]f(x) = sin(\pi x/2)[/tex]
[tex]f(x) = |x|[/tex]
[tex]f(x) = x|x|[/tex]
Answer (1)
To determine which of the given functions are bijective, we need to understand what it means for a function to be bijective. A function is bijective if it is both injective (one-to-one) and surjective (onto), which means:
Injective: Different inputs map to different outputs.
Surjective: Every possible output is covered by some input.
Let's evaluate each function for bijectiveness over the domain − 1 ≤ x ≤ 1 and the codomain − 1 ≤ x ≤ 1 :
f ( x ) = [ x ] :
This function represents the greatest integer function (floor function).
Since it maps all values in [ − 1 , 0 ) to − 1 , and all values in [ 0 , 1 ) to 0 , it is not one-to-one, thus not injective.
Hence, f ( x ) is not bijective.
f ( x ) = sin ( 2 π x ) :
This is a sine function transforming the interval [ − 1 , 1 ] to itself.
Since the sine function in this form is both increasing over [ − 1 , 1 ] and covers [ − 1 , 1 ] completely, it is both injective and surjective.
Thus, f ( x ) is bijective.
f ( x ) = ∣ x ∣ :
The absolute value function maps both positive and negative inputs to the same positive output.
Therefore, it is not injective since f ( − x ) = f ( x ) .
It cannot be onto [ − 1 , 1 ] because negative values are not covered in the codomain.
Thus, f ( x ) is not bijective.
f ( x ) = x ∣ x ∣ :
This function can be expressed as f ( x ) = x 2 when x ≥ 0 and f ( x ) = − x 2 when x < 0 .
For non-negative x , it only gives outputs in [ 0 , 1 ] (not negative values), and for x < 0 , it only gives negative outputs but no values greater than or equal to 0 .
It is not injective, as it fails to cover all of [ − 1 , 1 ] .
Thus, f ( x ) is not bijective.
In conclusion, among the given functions, f ( x ) = sin ( 2 π x ) is bijective over the specified domain and codomain.
