Domain, Range & Inverse of a Function

[bllnsr]

Homework Statement

How to solve part (iv) & (v)

Homework Equations

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

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.

 

[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...

 

[HallsofIvy]

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 $x\ge 4$[itex]has inverse function $f^-1(x)= 4+ \sqrt{x- 16}$.<br /> <br /> By separating at the vertex, we cut the given function into to one-to-one that now have inverses.[/itex]