1. Graph the parabola $\displaystyle y=x-x^2$. Then find the domain and range of $\displaystyle f(x)=sqrt(x-x^2)$.
2. Graph the parabola $\displaystyle y=3-2x-x^2$. Then find the domain and range of $\displaystyle g(x)=sqrt(3-2x-x^2)$.
$\displaystyle y=x-x^2=x(1-x)$
This means the parabola equals 0 at x=1 and x=0 (this should help graph it). A graph shows that between x=1 and x=0 y is positive.
$\displaystyle \frac{d}{dx} x-x^2=1-2x$
Stationary point: $\displaystyle 1-2x=0 \Rightarrow \ x=\frac{1}{2}$.
The stationary point (it's a maximum, you can prove that part) is $\displaystyle \left( \frac{1}{2}, \frac{1}{4} \right)$. This is the second part of the graph and comes in handy later!
The transformation is $\displaystyle f(x)=\sqrt{x-x^2}$ but we know that $\displaystyle y=x-x^2$. Hence the transformation is $\displaystyle f(y) = \sqrt{y}$.
The domain therefore is $\displaystyle \{ x \in \mathbb{R}: \ 0 \leq x \leq 1 \}$ (otherwise we would have the square root of something negative).
The range is $\displaystyle \{ f(x) \in \mathbb{R}: \ f(x) \leq \frac{1}{2} \}$.