# F : N -> N defined by f(x) = x^3 - 1

f : N --> N defined by f(x) = x^3 - 1

(i) f : N --> N defined by f(x) = x^3 - 1
(ii) g : Z --> Z defined by g(x) = 2x + 1
(iii) h : R --> R defined by h(x) = x(x + 3)(x - 3)

(***note that N,Z,R stands for natural #, Integer, and Real # respectively..)

(a) Which of the functions are one-to-one?

(b) Which of the functions are bijections?

(c) For those that are bijections find the inverse function.

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Here's the other one:

The function f : R --> R defined by f(x) = (3x - 1)/(x - 3) is not bijective however by suitably restricting the domain and codomain the function can be made to be bijective.

(a)State the domain and codomain that will make the function bijective.

What's a domain? codomain?

(b) Find the inverse of the bijective function.

(I can still remember a bit of inverse function.. i think.. well ill give it a try anyway)

f(x) = (3x - 1)/x - 3)
x = (3y - 1)/y - 3) "replace x with y"
x(y - 3) = 1(3y - 1)) "Cross multiplication"
xy - 3x[i/] = 3y - 1 "Will minus both sides with 3y"
xy - 3x - 3y = -1 "Will add both sides with 3x"
xy - 3y = -1 + 3x
y(x - 3) = -1 + 3x "Factor out y"
y = (-1 + 3x)/(x - 3) "Divide both sides with (x - 3)"

Is that correct??