I was on part (v) when my tab crashed. Socratic really needs draft management a la Quora.
f(x)= 5-2 sin(2x) quad quad quad 0 le x le pi
graph{5-2 sin(2x) [-2.25, 7.75, -2, 7.12]}
(i) The 0 le x le pi means sin(2x) goes a full cycle, so hits its max at 1, giving f(x)=5-2(1)=3 and its min at -1 giving f(x)=5-2(-1)=7, so a range of 3 le f(x) le 7
(ii) We get a full cycle of a sine wave, compressed into x=0 to x=pi. It starts at the zero point and is upside down, amplitude two, due to the -2 factor. The five raises it five units.
Here's Socratic's grapher; I don't seem to be able to indicate the domain 0 le x le pi.
(iii) Solve f(x)=6
5 - 2 sin(2x) = 6
-1 = 2 sin(2x)
sin(2x) = -1/2 = sin(-pi/6)
There's the biggest cliche in trig, you knew it was coming. (I did anyway, because this is the second time I've gone through this.)
2x = -pi/6 + 2pi n or 2x = -{5pi}/6 + 2pi n quad integer n
x = -pi/12 + pi n or x = -{5pi}/12 + pi n
(iv) g(x)=5-2 sin (2x) for 0 le x le k.
We want the biggest k that gives an invertible piece of g which is the same as f so we can use our graph. We can go to the first minimum to the right of zero before we start getting duplicate g(x). That's where f(x)=3 or sin(2x)=1 i.e. 2x=pi/2 or x=pi/4.
So k=pi/4 and we can invert g(x) over 0 \le x le pi/4
Crashed again but I had saved it in my clipboard this time!
(v) Invert g over that domain.
y = 5-2 sin(2x)
2 sin (2x) = 5 - y
sin (2x) = {5-y}/2
Over our domain 2x is in the first quadrant so we need the principal value of the inverse sine:
2x = text{Arc}text{sin}({5-y}/2)
x = 1/ 2 text{Arc}text{sin}({5-y}/2)
g^{-1}(y) = 1/ 2 text{Arc}text{sin}({5-y}/2)
