What is the range of the function $$r(x) = \frac{1}{(1-x)^2}~?$$ Express your answer in interval notation.
\(r(x) = \dfrac{1}{(1-x)^2}\)
When x -> 1, r-> infinity. So the maximum value of r(x) is +infinity.
Now we just have to find the minimum value of r(x).
Horizontal asymptote of r(x) is y = 0, which means the r(x) is approaching 0 when x approachs +inf or -inf.
Therefore range of r(x) is (0, inf).