A real-valued function f: A rightarrow B defi | vedclass

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(D) Given $f(x) = \frac{4-x^2}{4+x^2}$.To find the range of $f(x)$,let $y = \frac{4-x^2}{4+x^2}$.$y(4+x^2) = 4-x^2 \implies 4y + yx^2 = 4-x^2 \implies

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$(-1, 0]$

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(D) Given $f(x) = \frac{4-x^2}{4+x^2}$.To find the range of $f(x)$,let $y = \frac{4-x^2}{4+x^2}$.$y(4+x^2) = 4-x^2 \implies 4y + yx^2 = 4-x^2 \implies x^2(y+1) = 4-4y$.$x^2 = \frac{4(1-y)}{1+y}$.Since $x^2 \ge 0$,we have $\frac{1-y}{1+y} \ge 0$,which implies $y \in (-1, 1]$.Since $f$ is a bijection,$B$ must be the range of $f$,so $B = (-1, 1]$.Given $-4 \in A$,we check $f(-4) = \frac{4-(-4)^2}{4+(-4)^2} = \frac{4-16}{4+16} = \frac{-12}{20} = -0.6$.Since $f$ is a bijection,$A$ must be the domain such that the range is $B$. For $f(x)$ to be a bijection,$A$ must be a set such that each $y \in B$ has a unique $x \in A$. Since $x^2 = \frac{4(1-y)}{1+y}$,for each $y$,$x = \pm \sqrt{\frac{4(1-y)}{1+y}}$.If $A = [-4, 4]$,then $f(x)$ is not injective. However,the problem states $f$ is a bijection. Given $-4 \in A$,and $f(-4) = -0.6$,the set $A$ must be chosen such that the function is bijective. The range $B = (-1, 1]$.If $A = [-4, 0]$,then $f(x)$ is a bijection from $[-4, 0]$ to $(-1, 1]$.Thus $A = [-4, 0]$ and $B = (-1, 1]$.$A \cap B = [-4, 0] \cap (-1, 1] = (-1, 0]$.

$A$ real-valued function $f: A \rightarrow B$ defined by $f(x) = \frac{4-x^2}{4+x^2}$ for all $x \in A$ is a bijection. If $-4 \in A$,then $A \cap B =$

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