Find the domain and range of the following functions.
(i) `f(x)=sqrt(2x-3) " (ii) " f(x) =(1)/(x-2)`
(iii) `f(x) =x^(2) +3 " (iv) " f(x)=(1)/(x^(2)+2)`

Explanation

(i) `f(x)=sqrt(2x-3)` is defined if ` 2x-3 ge 0 " or " x ge 3//2`.
` :. x in [2//3, oo),` which is the domain of the function.
For range, `2x-3 ge 0`
` :. sqrt(2x-3) ge 0.`
Thus, range is `[0, oo)`
(ii) `f(x)=(1)/(x-2)`.
Clearly above function is defined for any real value of x but not for `x=2`. For `x=2`, denominator becomes zero and division with zero is undefined.
So, domain is `R-{2}`.
For range let `y=(1)/(x-2)`.
We have to find all possible values of y. So, let us get x in terms of y.
`x-2=(1)/(y)implies x=(1)/(y)+2=(1+2y)/(y)`
Clearly y cannot be zero. For any value of y other than zero there exists corresponding value of x.
Thus range is ` R -{0}`
(iii) ` f(x) = x^(2)+3`
Clearly x can take any real value. So, domain is R. For range ,`x^(2) ge 0` for real values of x.
` :. x^(2) +3 ge 3 " or " f(x) ge 3.`
So, range is `[3,oo).`
(iv) `f(x)=(1)/(x^(2)+2)`
Above function is not defined for those values of x for which `x^(2)+2=0`. But there are no real values of x for which `x^(2)= -2`. So, we can take any real value of x. Therefore, domain is set R.
For range let `y=(1)/(x^(2)+2)`.
So, `x^(2)+2=(1)/(y) " or " x^(2)=(1)/(y)-2=(1-2y)/(y)`.
Now `x^(2) ge 0` for all real values of x.
` :. (1-2y)/(y) ge 0`
` implies (2y-1)/(y) le 0 implies 0 lt y le 1//2.`
Therefore, range is `(0,(1)/(2)]`.