Graph I, locale all points of discontinuily, and discuss the behavior of f at these points. \[ I(x)=\left\{\begin{array}{ll} 2+x & \text { if } x<0 \\ 5-x & \text { if } x \geq 0 \end{array}\right. \] Find \( f(x) \) al any points of discontinuity. Select the corred choice heinw and, if necessary, fill in the answor box to complate your cheice. A. At the point(s) of disennimully, the value(s) of the function \( f(x) \) is/aie \( \square \) . (Use a comna to separalu arswers as needed.) B. There are no points of discontinuity. Find \( \lim _{x \rightarrow c} f(x) \), where cislam the poinl( 5 ) at which ' \( (x) \) is discontinuous. Select the corroct choice below and, if necessary, fill in the answer box to complete ycut choice. A. \( \lim _{x \rightarrow-} f(x)-\square \) \( \square \) (Use a comma to separate answers as needed) B. The limit does not exist.
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To locate the points of discontinuity for the piecewise function \( I(x) \), we need to look closely at the point where the definition of the function changes, which is at \( x = 0 \). Evaluating the left-hand limit as \( x \) approaches 0, we find: \[ \lim_{x \to 0^-} I(x) = 2 + 0 = 2 \] For the right-hand limit: \[ \lim_{x \to 0^+} I(x) = 5 - 0 = 5 \] Since the left limit (2) does not equal the right limit (5), \( I(x) \) is discontinuous at \( x = 0 \). At \( x = 0 \), the function is defined as: A. At the point(s) of discontinuity, the value(s) of the function \( I(x) \) is 5. Now, assessing the limits around the point of discontinuity: A. \( \lim_{x \rightarrow 0} I(x) = 2, 5 \) Thus, the final answer for the limit is: A. \( \lim_{x \rightarrow 0^-} I(x) = 2, \lim_{x \rightarrow 0^+} I(x) = 5 \).
