#### Find the Domain and the Range of the function $y = \sqrt{1-x^2}$

Solution:

For real values of ${y}$, the quantity ${1-x^2}$ cannot be negative.

${1-x^2}\geq{0}$

Subtract ${1}$ from both sides of this inequality.

${-x^2}\geq{-1}$

Multiply this inequality with ${-1}$

$x^2\leq{1}$

(Note that multiplying by a negative number reverses the inequality.)

Since both sides of this inequality are non-negative, we can take the principal square root. Doing so, we get

$\sqrt{x^2}\leq{1}$

$\sqrt{x^2}$ is another definition of $|x|$

So, we can write this inequality as $|x|\leq{1}$

$\Rightarrow-1\leq{x}\leq1$

Therefore, ${x}$ is greater than or equal to ${-1}$ and less than or equal to ${1}$

Hence, Domain $=[-1,1]$

Now let’s find the range of this function.

As ${x}$ varies from ${-1}$ to ${1}$, ${x^2}$ varies from ${0}$ to ${1}$

Therefore, ${0}\leq{x^2}\leq{1}$

Multiply this inequality with $-1$

${0}\geq{-x^2}\geq{-1}$

Now add ${1}$ to this inequality.

${1}\geq{1-x^2}\geq{0}$

The quantity ${1-x^2}$ is non-negative on the domain of this function. So, taking the principal square root, we get

${1}\geq{\sqrt{1-x^2}}\geq{0}$

Therefore, Range $=[0,1]$

Hence, the Domain of the function $\boldsymbol{{y}=\sqrt{1-x^2}}$ is $\boldsymbol{[-1,1]}$ and its Range is $\boldsymbol{[0,1]}$.

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