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}

Related questions
See all questions in Domain and Range of a Function
Impact of this question
2619 views around the world
You can reuse this answer
Creative Commons License