How do you find the domain and range of $sqrt(25-x^2)$ ?

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Answer 1 · 2017-01-26T11:31:34

The domain is $= [-5,5]$
The range is $=[0,5]$

What is under the $sqrt$ sign is $>=0$

Therefore,

$25-x^2>=0$

$(5+x)(5-x)>=0$

Let $f(x)=(5+x)(5-x)$

We can now build a sign chart

$color(white)(aaaa)$ $x$ $color(white)(aaaa)$ $-oo$ $color(white)(aaaa)$ $-5$ $color(white)(aaaa)$ $5$ $color(white)(aaaa)$ $+oo$

$color(white)(aaaa)$ $5+x$ $color(white)(aaaaa)$ $-$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $5-x$ $color(white)(aaaaa)$ $+$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $-$

$color(white)(aaaa)$ $f(x)$ $color(white)(aaaaaa)$ $-$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $-$

Therefore,

$f(x)>=0$ when $x in [-5, 5]$

The domain of $f(x)$ is $D_f(x)= [-5,5]$

When $x=+-5$ , $sqrt(f(x))=0$

When $x=0$ , $sqrt(f(x))=5$ , this is the max. value

so,
The range is $R = [0,5]$

How do you find the domain and range of sqrt(25-x^2) sqrt(25-x^2) ?