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)#
