Question

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

Answer 1

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}`