how do you find the range?

Question

how do you find the range of the problems f(x)=(x-3)/x and f(x)=square root of (3-x)

Answer 1

One way to do it is to find the domain of the inverse function, since that is the range of the original function.

To find the inverse of y=(x-3)/x, solve for x, and you get xy=x-3, or x(y-1)=-3, or x=3/(1-y). It follows that y=1 must be excluded, because otherwise a division by zero would occur. Therefore, there is no x for which f(x)=1. However, all other real numbers are included in the range.

For f(x)=sqrt(3-x), solve y=sqrt(3-x) for x, and you get y^2=3-x, or x=3-y^2. However, from the original function, we must have y >= 0, because of the square root. But we can see that for any y >= 0, we can find an x for which f(x)=y, so the range is all nonnegative real numbers.

Answer 2

(x-3)/x is defined for all x =/= 0 so x = {R \0}
√(3-x) is defined for all (3-x) ≥ 0 so x ≤ 3 or x = [-∞,3]


Doug