If D is the domain and G is the range of the | vedclass

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(B) Given the function $f(x) = \sqrt{\frac{1-x^2}{1+x^2}}$.For the domain $D$, we require $\frac{1-x^2}{1+x^2} \geq 0$. Since $1+x^2 > 0$ for all real

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$[0, 1]$

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(B) Given the function $f(x) = \sqrt{\frac{1-x^2}{1+x^2}}$.For the domain $D$, we require $\frac{1-x^2}{1+x^2} \geq 0$. Since $1+x^2 > 0$ for all real $x$, we need $1-x^2 \geq 0$, which implies $x^2 \leq 1$, so $x \in [-1, 1]$. Thus, $D = [-1, 1]$.For the range $G$, let $y = \sqrt{\frac{1-x^2}{1+x^2}}$. Since $x \in [-1, 1]$, $1-x^2$ ranges from $0$ to $1$ and $1+x^2$ ranges from $1$ to $2$. Thus, $\frac{1-x^2}{1+x^2}$ ranges from $0$ to $1$. Taking the square root, $y \in [0, 1]$. Thus, $G = [0, 1]$.Finally, $D \cap G = [-1, 1] \cap [0, 1] = [0, 1]$.

If $D$ is the domain and $G$ is the range of the real-valued function $f(x)=\sqrt{\frac{1-x^2}{1+x^2}}$, then $D \cap G=$

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