A and B are subsets of R. Every element x of | vedclass

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(B) The function is defined as $y(x) = f(x) = \begin{cases} \frac{5x}{(x-3)(x+3)}, & x 
eq -1 \ -1, & x = -1 \end{cases}$.For $f(x)$ to be a well-de

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$R \setminus \{-3, 3\}$

Vedclass · Answer

(B) The function is defined as $y(x) = f(x) = \begin{cases} \frac{5x}{(x-3)(x+3)}, & x 
eq -1 \ -1, & x = -1 \end{cases}$.For $f(x)$ to be a well-defined function $f: A ightarrow B$,$A$ must be the domain of the function.The expression $\frac{5x}{(x-3)(x+3)}$ is defined for all real numbers except where the denominator is zero.Setting the denominator to zero: $(x-3)(x+3) = 0 \implies x = 3$ or $x = -3$.At $x = -1$,the function is explicitly defined as $y(-1) = -1$,which is a valid real number.Therefore,the domain $A$ includes all real numbers except $3$ and $-3$.Thus,$A = R \setminus \{-3, 3\}$.

$A$ and $B$ are subsets of $R$. Every element $x$ of $A$ is mapped to an element of $B$ by the rule,$y(x) = \begin{cases} \frac{5x}{(x-3)(x+3)} & \text{if } x \neq -1 \\ -1 & \text{if } x = -1 \end{cases}$,then $A =$

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