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?
 A: You must also ensure that $\sqrt{25-x^2} \neq 0$ otherwise you would be dividing by zero. Therefore you must have $25 \neq x^2$, which means the domain is $]-5,5[$.
To find the range, notice that the image of an interval by a continuous function is always an interval by the Intermediate Value Theorem (you might not have seen that, but let's not worry about it). The point is you must find the maximum and the minimum value of $f$, and then see if they are attained. 
The minimum value of $f(x) = \frac 1{g(x)}$ is attained at the maximum value of $g(x)$ in general. Therefore, since the maximum value of $\sqrt{25-x^2}$ is attained at $x=0$, the minimum of $f$ is $\frac 15$. 
There is no maximum for $f$, because $\sqrt{25 - x^2}$ can get arbitrarily close to $0$, hence $f$ can get as big as you want. The range is therefore $[1/5,\infty[$.
Hope that helps,
A: 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 $[1/5,\infty)$.
