how do you find the range of a fuction? 
i read the book but i dont understand it 
can someone help me with m/j 05 q7 (i) and (iii) please explain the answer in (iii) Thanks
(i) f : x -> 3 - 2 sinxAmplitude is -2. y= - 2 sinx will have maximum and minimum points at 2 and -2. And since, f : x -> 3 - 2 sinx => curve is translated 3 units upward, minimum and maximum points will become:
2+3=5
and -2+3=1.
The range in which x is found therefore is: 1 </= f(x) </= 5 (Because of the equal to sign here: 0
o </= x </= 360
o.)
(iii)You must have drawn the curve. For any curve to have an inverse, it has to be a one-one function, i.e. every horizontal line drawn across it should cut at only one point. From the graph, you'll notice that as you move from 90
o onwards, any horizontal line through that curve will cut twice. Therefore, the curve is one-one in the range: 0
o </= x </= 90
o.
=> A= 90
o.
N. B: Curve is also one-one in the ranges: 90
o </= x </= 270
o and 270
o </= x </= 360
o
