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Graph t, locate all points of discontinuty, end discuss the behavior of t at these points. \[ I(x)=\left\{\begin{array}{ll} 1+x & \text { if } x<1 \\ 6-x & \text { if } x \geq 1 \end{array}\right. \] Locate all points of discontinuity. Solect the conect choice below and, if necessary, fill in the answor box to complete your choice. A. The poirl(s) of d scontinuity is/are \( \mathrm{x}= \) \( \square \) (Use a comma to separate answers as needed.) B. There are no points of discontinuity Discuss the behavior of t at its point(s) of discontinuity. Find \( \mathrm{f}(\mathrm{x}) \) at any points of discontinuity. Select the corred choice below and, if necessary, fill in the answer hox to complete your choice A. At the poinl(s) of discontinuity, the value(s) of the function \( f(x) \) is/are 5 . (Use a comma to separate answers as needed.) B. There are no points of discontinuity.

Ask by Phillips Kelley. in the United States
Feb 03,2025

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**Points of Discontinuity:** \( x = 1 \) **Function Value at Discontinuity:** 5

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Beyond the Answer

To analyze the function \( I(x) \), we find that it is piecewise defined with a potential point of discontinuity at \( x = 1 \). Evaluating the left-hand limit as \( x \) approaches 1 gives \( I(1^-) = 1 + 1 = 2 \), whereas the value of the function at \( x = 1 \) is \( I(1) = 6 - 1 = 5 \). Therefore, there is a discontinuity at \( x = 1 \). A. The point(s) of discontinuity is/are \( x = 1 \). At the point of discontinuity, we can see that the left limit does not equal the function value, indicating the jump. So, the left-hand limit approaches 2 while the function value at that point is 5, resulting in different behaviors based on the direction of approach. A. At the point(s) of discontinuity, the value(s) of the function \( I(x) \) is/are 5.

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