Question #9738d

1 Answer
Jan 10, 2017

Answer:

The domain is #x in ] 10 ,12 [ #

Explanation:

What is under the #sqrt# is #>=0#

But the #sqrt# is in the denominator

So,

#sqrt((x-10)(12-x))>0#

Let #f(x)=sqrt((x-10)(12-x))#

Construct the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaaaaa)##10##color(white)(aaaaaaa)##12##color(white)(aaaa)##-oo#

#color(white)(aaaa)##x-10##color(white)(aaaa)##-##color(white)(aaaa)##color(red)(∣∣)##color(white)(aaa)##+##color(white)(aaa)##color(red)(∣∣)##color(white)(aa)##+#

#color(white)(aaaa)##12-x##color(white)(aaaa)##+##color(white)(aaaa)##color(red)(∣∣)##color(white)(aaa)##+##color(white)(aaa)##color(red)(∣∣)##color(white)(aa)##-#

#color(white)(aaaa)##f(x)##color(white)(aaaaaa)##-##color(white)(aaaa)##color(red)(∣∣)##color(white)(aaa)##+##color(white)(aaa)##color(red)(∣∣)##color(white)(aa)##-#

Therefore,

#f(x)>0#, when #x in ] 10 ,12 [ #