Question

If $f\left(x\right)=\frac{1}{\sqrt{1-x^2}}$, find the range of $x$ for which the function is defined.

• A. $x=-1$ or $x=1$
• B. $x<-1$ or $x>1$
• C. $x\le -1$ or $x\ge 1$
• D. $-1<x<1$
• E. $-1\le x\le 1$

Answer 1

The correct option is D $-1<x<1$

The domain of a function is the complete set of possible values of the independent variable.

The range of a function is the complete set of all possible resulting values of the dependent variable ($y$, usually), after we have substituted the domain.

Then $f\left(x\right)=\frac{1}{\sqrt{1-x^2}}$

The denominator (bottom) of a fraction cannot be zero.

The number under a square root sign must be positive.

So, $\sqrt{1-x^2}=0\Rightarrow x^2=1\Rightarrow x=\pm 1$

Then rang of $x$ is not $-1$ or $1$.

Then $-1\le x\ge 1$