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]}

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