How do you find the domain and range of #f(x)= sqrt(4-x^2) /( x-3)#?

1 Answer
bp
Sep 16, 2015

Domain #{ x:RR, -2<=x<= 2}#

Range #{y:RR, -2/sqrt5<=y<=0}#

Explanation:

Domain is quite evident that #x !=3# and #x <= +2# and #-2<=x# which essentially boils down to #{ x:RR, -2<=x<= 2}#

For finding the range, consider the domain from which it can be seen that y would never have any positive value and at end points +2, -2 it is 0, hence #y<=0# on the upper side.

Next square both sides, so that it is a quadratic equation in x,

#x^2 (y^2+1) -6y^2 x + 9y^2-4=0#. Solve it for x using quadratic formula, x=# (3y^2 +- sqrt( -5y^2 +4) )/((y^2+1) #

For a function Real to Real #5y^2 <=4# or #+-y<=2/sqrt5#. Since it is already settled that #y<=0# on the upper side, reject #y <= 2/sqrt5#' Hence it should be #-y<= 2/sqrt5# or #y>= -2/sqrt5#.

The range of the function would this be# {y:RR, -2/sqrt5<=y<=0}#

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