Question

Find domain and range of √(x^2 - 3x) ?​

Answer 1

Answer:

For Domain,

${x}^{2} - 3x \geqslant 0$

$x(x - 3) \geqslant 0$

so domain is

$x \leqslant 0 \: \: or \: x \geqslant 3$

now Range,

${x}^{2} - 3x = {x}^{2} - 2 \times x \times \frac{3}{2} + ( \frac{3}{2} ) {}^{2} - ( \frac{3}{2} ) {}^{2} = (x - \frac{3}{2} ) {}^{2} - \frac{9}{4} = \sqrt{(x - \frac{3}{2} ) {}^{2} - \frac{9}{4} } \geqslant 0$

Square root is greater than equal to zero.

so Range is [ 0 , infinity )