Let f(x) = begin{cases} x^2 + k, & text{when | vedclass

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(A) For a function $f(x)$ to be continuous at $x = 0$,the left-hand limit,right-hand limit,and the value of the function at $x = 0$ must be equal.$1$.

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(A) For a function $f(x)$ to be continuous at $x = 0$,the left-hand limit,right-hand limit,and the value of the function at $x = 0$ must be equal.$1$. Right-hand limit: $\lim_{x 	o 0^+} f(x) = \lim_{x 	o 0^+} (x^2 + k) = 0^2 + k = k$.$2$. Left-hand limit: $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^-} (-x^2 - k) = -0^2 - k = -k$.$3$. Value of the function: $f(0) = 0^2 + k = k$.For continuity,$\lim_{x 	o 0^+} f(x) = \lim_{x 	o 0^-} f(x) = f(0)$.Therefore,$k = -k$.$2k = 0 \implies k = 0$.

Let $f(x) = \begin{cases} x^2 + k, & \text{when } x \ge 0 \\ -x^2 - k, & \text{when } x < 0 \end{cases}$. If the function $f(x)$ is continuous at $x = 0$,then $k =$

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