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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