Find the domain of definition and the range of the following function:
$f (x) = \sqrt{x - x^{2}}$

Open in App

Solution

$f (x) = \sqrt{x - x^{2}}$
$(x - x^{2}) \geq 0$
$⟹ (- x + x^{2}) \leq 0$
$⟹ x (x - 1) \leq 0$
$x ϵ [0, 1]$
$x - x^{2} ϵ [0, 1 / 4]$
so, domain is $x ϵ [0, 1]$
and range $ϵ [0, \frac{1}{2}]$

Suggest Corrections

Similar questions

f (x) = {log}_{\frac{x - 2}{x + 3}} 2

g (x) = \sqrt{x^{2} - 9} .

f (x) = \sqrt{x - 1} + 2 \sqrt{3 - x}

f (x) = \sqrt{2 - x} + \sqrt{1 + x}

f (x) = {log}_{2} \frac{sin x - cos x + 3 \sqrt{2}}{\sqrt{2}} .

f (x) = - | x |

f (x) = \sqrt{9 - x^{2}}

Join BYJU'S Learning Program