a. Determine whether f is a bijection function. Briefly explain. f colon spaceX rightwards arrow R comma space f (x )space equals 3 space cube root of x. (R comma space Z set of real numbers and integers respectively) b. Redefine the domain and/or codomain of the function f colon spaceX rightwards arrow R comma space f (x )space equals space 2 square root of x , so that f is a bijection. (R comma space Z are sets of Real Numbers and Integers respectively) c. Determine the type of the function from R to R (R set of real numbers). f(x)

Answer and Step-by-step explanation:

Solution:

(a)f is bijection function such that:

given function :         f: x → R,  and  f(x) = cube root of x

a bijective function is one to one correspondence, each element of one set is paired with exactly one element of other set. And each element of other set paired with exactly one element of first set.

The domain of cube root function is the set of all real numbers.

A cube root can use all real numbers because it is possible for three negative to equal a negative.

(b)the domain of function:

f: x → R,  and  f(x) = 2 square root of x

a square root function contain a radical, since the radical cannot be negative, we can calculate for positive or zeroes value.

The domain of a function is its range.

Range of function is all valid outputs of that function.

square root is greater than or equal to zero so we use only nonnegative numbers in square root.

For square root function, range of function is values produced .x results in a radicand that is equal to or greater than zero.

your welcome haha, this is the best place to find answers

guess at question:

a vertical line intersects the graph of g(x,y) = 0 at two points. what conclusion follows about the function f(x) implicitly defined by g(x,f(x)) = 0?

the domain of f(x) cannot include the x coordinate of any such vertical line.

informally, f(x) is not a function, and

g(x,y)=0 cannot be solved for y uniquely.

the relation r = {(x,y) such that g(x,y) = 0} is a function if

for all (a,b) ∊ r and (c,d) ∊ r, a=c implies b=d.

step-by-step explanation: