Question

5. The domain of V3x - 7
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Answer 1

$as \: we \: know \: that \: \sqrt{3x - 7} \: can \: \\ never \: be \: a \: negative \: number \: so \\ \\ \sqrt{3x - 7} \geqslant 0 \\ \\ squaring \: \: on \: both \: sides \\ \\ 3x - 7 \geqslant 0 \\ \\ 3x \geqslant 7 \\ \\ x \geqslant \frac{7}{3} .... |answer| \\ \\ you \: can \: put \: all \: the \: positive \: real \: \\ number\: greater \: than \: \frac{7}{3} \\ \\ range.... \\ \\ let \: y = \sqrt{3x - 7} \\ \\ squaring \\ \\ {y}^{2} = 3x - 7 \\ \\ {y}^{2} + 7 = 3x \\ \\ \frac{ {y}^{2} }{3} + \frac{7}{3} = x \\ \\ so \: there \: is \: no \: restriction \: on \\ \: the \: value \: of \: y \: \\ so \: range \: is \: all \: real \: number$