(1 point) The following functions all have {1, 2, 3, 4, 5} as both their domain and codomain. For each, determine whether it is (only) injective, (only) surjective, bijective, or neither injective nor surjective. 3 4 V 1. f = 1 2 3 41 2 3 4 5 ? v 2. f = 2 1 4 4-T if 4 c+2 ifs 4for 3. The function find is bijedive. To show that f is injective we assumes that flow = feb) for some aid in the domain of f Then there are two cases. . of a4 4 and be 4 them 4 -a = 4- 6 0 of a > = 4 and 6 1: 4 them 6 - a = 6-6 a = b An both the cases we have shown that f is injective. To show that of is swijechives and all we take an arbitrary y in the domain of f which is gl, 2, 3, 4, 5 3 land show that I an it into the domain of for such that! Since y is in the lodomain of flavor we know that y= 4-k for some K en 21, 2, 31 9, 5gor or . 7 7 6 - K for some Ki in 8 1 1 2, 3, 4, 5 3. - 7- 4 -K then we can take me,4 - k and we havef (n ) = 4 - ( 4 - K ) = K = y 2 i y 2 6 -k then we can take we have f(MJ = 6 - ( 6 - K ) = kry. In both the cases, we found an X in the domain of f such that tools f(x2 27 . which show that of is surjective, de Therefore fis bijective for 4 The function fixes is neither injective nor surjective didwo wo oot owl To Show that of is not injective. linda take an example of two distinct elements in the domain of f that map to the same element in the lodomain of f. we take as ar 2 and b= 3. both are less than 4 and faja a + 2 24 older3 There fore f is not injective. To show that of is not surjective take an example of an element in the lodomain of f that is not the image of any element in the domain of f. he can take yz1 Which is in the lodomain of f but there is no x in the domain of f such that if 27 6 4 . then fix) is al least 2 and if 2 7 4 then fix) is at most 2. Therefore if is not surjective. DO. thiis is neither injective nor surjective