given f(x)= { 1, if -2 </= x</= 0 x^2, if 0</=x</=2 -x+1, if x>2 a) what is the domain of f? b) what is the range of f? c) evaluate f(0), f(-1), f(-3), and f(|x+1|+3), if possible d) find all values of x such that f(x)=1 Solution a) Domain of f: [-2,) b)Maximum possible value of f is 4 and minimum is - Hence, range of f is (-,4] c)f(0) = 0 2 =0 or f(0) = 1 i.e f(x) has a jump discontinuity at x=0, -1 lies between -2 and 0 and hence f(-1) = 1 and -3 does not lie in the domain, so f(-3) does not exist f(|x + 1| +3) = {f(-x-1+3), if -2<=x<=-1; f(x+1+3), if -1<x<} = {f(-x+2), if -2<=x<=-1; f(x+4)(always greater than 2 for the given values), if -1<x<} ={x 2 , if -2<=x<=-1; -x+1, if -1<x< } d) f(x) = 1 on -2 <= x <= 0 and f(x) = 1 if x = 1 .