Question

find range and domain​

Answer 1

Answer:

$f(x) = \sqrt{ {x}^{2} + 4 } \\ f(x) \: must \: be \: a \: real \: no. \\ therefore \: {x}^{2} + 4 \geqslant 0 \\ {x}^{2} \geqslant - 4$

which is true for all x element of real no.

D=R

For Range,

As square of a no. is always non negative therefore y is greater than 0

$y = \sqrt{ {x}^{2} + 4 } \\ on \: squaring \: both \: the \: sides \\ {y}^{2} = {x}^{2} + 4 \\ {y}^{2} - 4 = {x}^{2} \\ as \: {x}^{2} \: is \: \: real \: no. \\ {x}^{2} \geqslant 0 \\ {y}^{2} - 4 \geqslant 0 \\ (y + 2)(y - 2) \geqslant 0 \\ and \: y \geqslant 0$

Range=[2,♾️)