<s>[INST] Domain and Range of f(x) Find the domain and range of Function?
$$F(x) = \frac{1}{\sqrt{25-x^2}}$$
Domanin f(x) has real value , if
$$25-x^2\geqslant0$$
$$-x^2\geqslant-25$$
$$x^2\le25$$
$$-5\le x\le5$$
$$D_f= [-5,5]$$
let y= f(x)
    $$y = \frac{1}{\sqrt{25-x^2}}$$
    then what should i do to find range?
 [/INST] As noted above, the domain is $(-5,5)$.
Another way to find the range is to find the values of $y$ for which the equation
$\displaystyle y=\frac{1}{\sqrt{25-x^2}}$ has a solution for x:
1) Since $\displaystyle\frac{1}{\sqrt{25-x^2}}>0,$ we have that $y>0$.
2) $\displaystyle y^2=\frac{1}{25-x^2} \implies 25-x^2=\frac{1}{y^2}\implies x^2=25-\frac{1}{y^2}=\frac{25y^2-1}{y^2}=\frac{(5y+1)(5y-1)}{y^2}$, 
so 
$\;\;\;\;\displaystyle x^2\ge0\implies \frac{(5y+1)(5y-1)}{y^2}\ge0 \implies y\le -1/5$ or $y\ge1/5$.
Since $y>0$ and $y\le-1/5$ or $y\ge1/5$, $\;\;\;\;y\ge1/5\;\;\;;$ so  the range is $[1/5,\infty)$.
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