#### 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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