Question

Find the domain of definition of the following function.
$y=\sqrt{\frac{7-x}{\sqrt{4x^2-19x+12}}}.$

Answer 1

$y=\sqrt{\frac{7-x}{\sqrt{4x^2-19x+12}}}$

$(7-x)\ge 0$

$4x^2-19x+12>0$

$-x\ge -7$

$x\le 7$

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$=\frac{19\pm \sqrt{{\left(19\right)}^2-4\times 4\times 12}}{2\times 4}$

$=\frac{19\pm \sqrt{361-192}}{8}$

$=\frac{19\pm \sqrt{169}}{8}$

$=\frac{19\pm 13}{8}$

$=(\frac{19+13}{8},\frac{19-13}{8})$

$=\frac{32}{8},\frac{6}{8}$

$=4,\frac{3}{4}$

When $x>4$, $4x^2-19x+12>0$

When, $0<x<4$, $4x^2-19x+12<0$

So, the domain of $y=(-\infty ,0]\bigcup [4,\infty )$