Question: (Recall that Z = {2, 1, 0, 1, 2, . . .} is the integers, R the real numbers, (0,) the positive reals, and [0,)

(Recall that Z = {2, 1, 0, 1, 2, . . .} is the integers,

R the real numbers, (0,) the positive reals, and [0,) the non-negative reals.)

Let f : [0, ) [0,); f(x) = x [ x/2]

Let g : Z Z; g(x) = 2x + 3

a) What is the domain and codomain of f?

b) What is the domain and codomain of g?

c) Is f one-to-one? If not, show two elements of the domain which map to the same element of the codomain.

d) Is f onto? If not, show an element of the codomain which isnt mapped onto by any element of the domain.

e) Is g one-to-one? If not, show two elements of the domain which map to the same element of the codomain.

f) Is g onto? If not, show an element of the codomain which isnt mapped onto by any element of the domain.

g) Is the inverse function f 1 defined? If so, what is f 1 (2)? If not, why not?

h) Is the inverse function g 1 defined? If so, what is g 1 (2)? If not, why not?

i) Is the composition f g defined? If so, what is (f g)(2)? If not, why not?

j) Is the composition g f defined? If so, what is (g f)(2)? If not, why not?

Let f : R R; f(x) = x ^3 + x

Let g : (0,) R; g(x) = log3 x

a) What is the domain and codomain of f?

b) What is the domain and codomain of g?

c) Is f one-to-one? If not, show two elements of the domain which map to the same element of the codomain.

d) Is f onto? If not, show an element of the codomain which isnt mapped onto by any element of the domain.

e) Is g one-to-one? If not, show two elements of the domain which map to the same element of the codomain.

f) Is g onto? If not, show an element of the codomain which isnt mapped onto by any element of the domain.

g) Is the inverse function f 1 defined? If so, what is f 1 (2)? If not, why not?

h) Is the inverse function g 1 defined? If so, what is g 1 (2)? If not, why not?

i) Is the composition f g defined? If so, what is (f g)(2)? If not, why not?

j) Is the composition g f defined? If so, what is (g f)(2)? If not, why not?

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