# Domain, Range & Inverse of a Function

1. Nov 6, 2012

### bllnsr

1. The problem statement, all variables and given/known data

How to solve part (iv) & (v)

2. Relevant equations
general form : $y = a(x-h)^2 + k$

3. The attempt at a solution
In part (iv) for finding domain and range I converted g(x) in general form and then compared it with general form.

2. Nov 7, 2012

### symbolipoint

The inverse relation of y=8x-x2 is x=8y-y2.

Also, the domain of the original relation is the range of the inverse relation in general. Specifics depend...

3. Nov 7, 2012

### HallsofIvy

Staff Emeritus
I don't believe that is quite what is being asked. Completing the square, 8x- x2= 16- 16+ 8x- x2= 16- (x-4)2. The graph of that is a parabola having vertex at (4, 16). The function f(x)= 16- (x- 4)2 with $x\le 4$ has inverse function $f^-1(x)= 4- \sqrt{x- 16}$ while the function f(x)= 16- (x- 4)2 with $x\ge 4 has inverse function [itex]f^-1(x)= 4+ \sqrt{x- 16}$.

By separating at the vertex, we cut the given function into to one-to-one that now have inverses.