Consider the piecewise defined function f(x) | vedclass

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(D) To determine the continuity of the function $f(x)$,we check the limits at the transition points $x = 0$ and $x = 4$.At $x = 0$:Left-hand limit: $\

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The function is continuous on the entire real line.

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(D) To determine the continuity of the function $f(x)$,we check the limits at the transition points $x = 0$ and $x = 4$.At $x = 0$:Left-hand limit: $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^-} \sqrt{-x} = 0$.Right-hand limit: $\lim_{x 	o 0^+} f(x) = \lim_{x 	o 0^+} 0 = 0$.Value of the function: $f(0) = 0$.Since $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+} f(x) = f(0) = 0$,the function is continuous at $x = 0$.At $x = 4$:Left-hand limit: $\lim_{x 	o 4^-} f(x) = \lim_{x 	o 4^-} 0 = 0$.Right-hand limit: $\lim_{x 	o 4^+} f(x) = \lim_{x 	o 4^+} (x - 4) = 4 - 4 = 0$.Value of the function: $f(4) = 0$.Since $\lim_{x 	o 4^-} f(x) = \lim_{x 	o 4^+} f(x) = f(4) = 0$,the function is continuous at $x = 4$.Since the function is continuous at all points in its domain,it is continuous on the entire real line.

Consider the piecewise defined function $f(x) = \begin{cases} \sqrt{-x} & \text{if } x < 0 \\ 0 & \text{if } 0 \leqslant x \leqslant 4 \\ x - 4 & \text{if } x > 4 \end{cases}$. Choose the answer which best describes the continuity of this function.

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