First three are bijections, but the last one it's not, can you see why?
Could you see if what i have done is right for the following questions:
Ive got to find whether they are injection surjection or bijection for any integer to any integer so z to z (sorry i cant use latex)
1. f(x) = x^3
bijection
2. g(x) = x-3
bijection
3. h(x) = 3x+1
bijection
4. i(x) = (x^2) -1
Surjection
No, the inverse function is NOT the reciprocal. I assume that by the time you are asked to find inverse functions, you have seen the definition and should know that. There is an unfortunate notation: we use which, with numbers, implies the reciprocal (the multiplicative inverse) but the definition for functions is " is the inverse function to f if and only if and for all x". That is, "undoes" whatever f does to x and vice versa.
A standard way of finding an inverse for a function, f:
1) Write the equation y= f(x).
2) Swap x and y: x= f(y).
3) Solve for y.
For example, your first function if so you would write that as and then swap x and y to get . To solve for y, you would use the cube root, the "opposite" of cubing, to get or, equivalently, .
However, these functions, we were told, were "from Z to Z". If that is still true, then this function does not have an inverse function because the cuberoot of an integer is not necessarily an integer.