How do you find the domain of #5/(x sqrt (x+2))#?

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Answer 1 · 2018-06-16T06:13:09

The domain is #x in (-2,0)uu(0,+oo)#

The denominator must be #!=0#

Therefore,

#x!=0#

and

#x+2!=0#, #=>#, #x!=-2#

and for the square root sign

#x+2>0#, #=>#, #x > -2#

Finally. the domain is #x in (-2,0)uu(0,+oo)#

graph{5/(xsqrt(x+2)) [-20.28, 20.27, -10.14, 10.14]}