How do you find the domain and range of #f(x)= sqrt(4-x)/( (x+1)(x^2+1))#?

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Answer 1 · 2018-06-30T14:11:40

The domain is #x in (-oo, -1)uu(-1, 4]#. The range is #y in (-oo,+oo)#

As you cannot divide by #0#, the denominator must be #!=0#

Therefore,

#x+1!=0#

#=>#, #x!=-1#

Also, what's under the #sqrt# sign must be #>=0#

Therefore,

#4-x>=0#

#=>#, #x<=4#

Therefore,

The domain is #x in (-oo, -1)uu(-1, 4]#

To find the range,

#f(4)=0#

Let #y=sqrt(4-x)/((x+1)(x^2+1))#

The limits are as follows :

#lim_(x->-oo)y=O^-#

#lim_(x->-1^-)y=-oo#

#lim_(x->-1^+)y=+oo#

The range is #y in (-oo,+oo)#

graph{sqrt(4-x)/((x+1)(x^2+1)) [-10, 10.01, -5, 5]}