# Math Help - Range and Domain

1. ## Range and Domain

1. Graph the parabola $y=x-x^2$. Then find the domain and range of $f(x)=sqrt(x-x^2)$.

2. Graph the parabola $y=3-2x-x^2$. Then find the domain and range of $g(x)=sqrt(3-2x-x^2)$.

2. $y=x-x^2=x(1-x)$

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.

$\frac{d}{dx} x-x^2=1-2x$

Stationary point: $1-2x=0 \Rightarrow \ x=\frac{1}{2}$.

The stationary point (it's a maximum, you can prove that part) is $\left( \frac{1}{2}, \frac{1}{4} \right)$. This is the second part of the graph and comes in handy later!

The transformation is $f(x)=\sqrt{x-x^2}$ but we know that $y=x-x^2$. Hence the transformation is $f(y) = \sqrt{y}$.

The domain therefore is $\{ x \in \mathbb{R}: \ 0 \leq x \leq 1 \}$ (otherwise we would have the square root of something negative).

The range is $\{ f(x) \in \mathbb{R}: \ f(x) \leq \frac{1}{2} \}$.