Question #aa804

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Answer 1 · 2017-10-17T06:52:31

a). #color(blue)({x in RR})#

b). #color(blue)({x in RR : 0 < x <=3}uu{x in RR : -3<= x < 0 })#

c). #color(blue)({x in RR : -4 <= x < sqrt(13)})#

#h(x) = (x-3)/(x^2+9)#

#x^2+9>0# for all real #x#, so domain is:

#color(blue)({x in RR})#

#g(x)= (sqrt(9-x^2))/x#

Since we have x in the denominator, #x != 0#

For real numbers:

#9-x^2>= 0#

#9-x^2>= 0=x^2<=9=> x<= sqrt(9)=+-3#

So we have:

#x<=3#,

# x <= -3# ,

#x > 0#

#x < 0#

#0 < x <=3# , #-3 <= x < 0#

So domain is:

#color(blue)({x in RR : 0 < x <=3}uu{x in RR : -3<= x < 0 })#

#h(x) = 1/((sqrt(x^2 -4) -3)#

#sqrt(x^2-4) >= 0# and #sqrt(x^2-4) -3 > 0# , #sqrt(x^2-4)-3 < 0#

#sqrt(x^2-4) >= 0 = x^2-4>=0=> x<= +- 4#

#sqrt(x^2-4) -3 > 0 = sqrt(x^2-4)> 3=x^2 -4>9#

#->x^2 > 13=> x > +-sqrt(13)#

#sqrt(x^2-4)-3 < 0= sqrt(x^2 -4)< 3= x^2 -4 < 9#

#-> x^2 < 13=> x < +-sqrt(13)#

#x <= 4# , #x<= -4# , #x> sqrt(13)# , #x> -sqrt(13)# , #x< sqrt(13)# , #x< -sqrt(13)#

#-4 <= x < sqrt(13)#

So domain is:

#color(blue)({x in RR : -4 <= x < sqrt(13)})#