Question

Find the domain of: g(x)=sqrt of x(x-6)

Answer 1

\[x \left( x-6 \right)\ge 0\]
\[x ^{2}-6x \ge0 \]
\[add~ both~ sides~\left( \frac{ -6 }{ 2 } \right)^{2}~i.e.,9\]
\[x^{2}-6x+9\ge9\]
\[\left( x-3 \right)^{2}\ge 3^{2}\]
\[\left| x-3 \right|\ge3\]
\[x-3\le-3~or~x \le0\]
\[or~x-3\ge3,x \ge6\]
so domain is\[(-\infty,0]\cup[6,\infty)\]

Answer 2

I don't understand where to (-6/2)^2 comes from

Answer 3

the*

Answer 4

just to complete the squares.
\[\left( a-b \right)^{2}=a ^{2}-2ab+b ^{2},here \left( \frac{ -2b }{2 } \right)^{2}\]

Answer 5

\[we~ can~ also~ say~ if~ \left( x-a \right)\left( x-b \right)\ge0,then~x~does~\not~lie~\in \left( a,b \right)\]
assume a<b

Answer 6

ok, thank you!