What is the domain and range of $F (x) = \frac{1}{\sqrt{4 - x^{2}}}$ $F(x) = 1/ sqrt(4 - x^2)$ ?

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Answer 1 · 2018-07-08T09:40:18

The domain is $x \in (- 2, 2)$ $x in (-2,2)$ . The range is $[\frac{1}{2}, + \infty)$ $[1/2, +oo)$ .

The function is

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

What'under the $\sqrt{}$ $sqrt$ sign must be $\geq 0$ $>=0$ and we cannot divide by $0$ $0$

Therefore,

$4 - x^{2} > 0$ $4-x^2>0$

$\Rightarrow$ $=>$ , $(2 - x) (2 + x) > 0$ $(2-x)(2+x)>0$

$\Rightarrow$ $=>$ , ${(2-x>0),(2+x>0):}$

$=>$ , ${(x<2),(x> -2):}$

Therefore,

The domain is $x in (-2,2)$

Also,

$lim_(x->2^-)f(x)=lim_(x->2^-)1/sqrt(4-x^2)=1/O^+=+oo$

$lim_(x->-2^+)f(x)=lim_(x->-2^+)1/sqrt(4-x^2)=1/O^+=+oo$

When $x=0$

$f(0)=1/sqrt(4-0)=1/2$

The range is $[1/2, +oo)$

graph{1/sqrt(4-x^2) [-9.625, 10.375, -1.96, 8.04]}

What is the domain and range of F(x) = 1/ sqrt(4 - x^2)F(x)=1√4−x2F(x) = 1/ sqrt(4 - x^2)?