# What is the domain and range of f(x)= sqrt(4x-x^2)?

Jul 31, 2017

The domain is $x \in \left[0 , 4\right]$
The range is $f \left(x\right) \in \left[0 , 2\right]$

#### Explanation:

For the domain, what's under the square root sign is $\ge 0$

Therefore,

$4 x - {x}^{2} \ge 0$

$x \left(4 - x\right) \ge 0$

Let $g \left(x\right) = \sqrt{x \left(4 - x\right)}$

We can build a sign chart

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a a a}$$0$$\textcolor{w h i t e}{a a a a a a}$$4$$\textcolor{w h i t e}{a a a a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$0$$\textcolor{w h i t e}{a a}$$+$$\textcolor{w h i t e}{a a a a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$4 - x$$\textcolor{w h i t e}{a a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$\textcolor{w h i t e}{a a a}$$+$$\textcolor{w h i t e}{a a}$$0$$\textcolor{w h i t e}{a a a a}$$-$

$\textcolor{w h i t e}{a a a a}$$g \left(x\right)$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a}$$\textcolor{w h i t e}{a a a}$$0$$\textcolor{w h i t e}{a a}$$+$$\textcolor{w h i t e}{a a}$$0$$\textcolor{w h i t e}{a a a a}$$-$

Therefore

$g \left(x\right) \ge 0$ when $x \in \left[0 , 4\right]$

Let,

$y = \sqrt{4 x - {x}^{2}}$

hen,

${y}^{2} = 4 x - {x}^{2}$

${x}^{2} - 4 x + {y}^{2} = 0$

The solutions this quadratic equation is when the discriminant $\Delta \ge 0$

So,

$\Delta = {\left(- 4\right)}^{2} - 4 \cdot 1 \cdot {y}^{2}$

$16 - 4 {y}^{2} \ge 0$

$4 \left(4 - {y}^{2}\right) \ge 0$

$4 \left(2 + y\right) \left(2 - y\right) \ge 0$

Let $h \left(y\right) = \left(2 + y\right) \left(2 - y\right)$

We build the sign chart

$\textcolor{w h i t e}{a a a a}$$y$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a}$$- 2$$\textcolor{w h i t e}{a a a a}$color(white)(aaaaaa)2$\textcolor{w h i t e}{a a a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$2 + y$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$0$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$0$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$2 - y$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$0$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$0$$\textcolor{w h i t e}{a a a a}$$-$

$\textcolor{w h i t e}{a a a a}$$h \left(y\right)$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$0$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$0$$\textcolor{w h i t e}{a a a a}$$-$

Therefore,

$h \left(y\right) \ge 0$, when $y \in \left[- 2 , 2\right]$

This is not possible for the whole interval, so the range is $y \in \left[0 , 2\right]$

graph{sqrt(4x-x^2) [-10, 10, -5, 5]}