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

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Answer 1 · 2018-06-30T14:11:40

The domain is $x \in (- \infty, - 1) \cup (- 1, 4]$ $x in (-oo, -1)uu(-1, 4]$ . The range is $y \in (- \infty, + \infty)$ $y in (-oo,+oo)$

As you cannot divide by $0$ $0$ , the denominator must be $\neq 0$ $!=0$

Therefore,

$x + 1 \neq 0$ $x+1!=0$

$\Rightarrow$ $=>$ , $x \neq - 1$ $x!=-1$

Also, what's under the $\sqrt{}$ $sqrt$ sign must be $\geq 0$ $>=0$

Therefore,

$4 - x \geq 0$ $4-x>=0$

$\Rightarrow$ $=>$ , $x \leq 4$ $x<=4$

Therefore,

The domain is $x \in (- \infty, - 1) \cup (- 1, 4]$ $x in (-oo, -1)uu(-1, 4]$

To find the range,

$f (4) = 0$ $f(4)=0$

Let $y = \frac{\sqrt{4 - x}}{(x + 1) (x^{2} + 1)}$ $y=sqrt(4-x)/((x+1)(x^2+1))$

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)=√4−x(x+1)(x2+1)f(x)= sqrt(4-x)/( (x+1)(x^2+1))?