If f(x) = begin{cases} frac{x^2 - 9}{x - 3}, | vedclass

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(B) For a function $f(x)$ to be continuous at $x = a$,the condition $\lim_{x 	o a} f(x) = f(a)$ must hold.First,calculate the limit as $x 	o 3$:$\li

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(B) For a function $f(x)$ to be continuous at $x = a$,the condition $\lim_{x 	o a} f(x) = f(a)$ must hold.First,calculate the limit as $x 	o 3$:$\lim_{x 	o 3} f(x) = \lim_{x 	o 3} \frac{x^2 - 9}{x - 3} = \lim_{x 	o 3} \frac{(x - 3)(x + 3)}{x - 3} = \lim_{x 	o 3} (x + 3) = 3 + 3 = 6$.Next,evaluate the function at $x = 3$:$f(3) = 2(3) + k = 6 + k$.Since the function is continuous at $x = 3$,we equate the limit to the function value:$6 + k = 6$.Solving for $k$,we get $k = 0$.

If $f(x) = \begin{cases} \frac{x^2 - 9}{x - 3}, & \text{if } x \neq 3 \\ 2x + k, & \text{otherwise} \end{cases}$ is continuous at $x = 3$,then $k = $

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