Let the function be defined by g(x) = cases |x-3| + 1, & x 3, 2, & x = 3

MATHEMATICS · Multi-part · AP · Calculus AB · Limits and Continuity · Limits and Continuity

Problem

Let the function gg be defined by g(x)={∣x−3∣+1,x≠3,2,x=3.g(x) = \begin{cases} |x-3| + 1, & x \neq 3,\\ 2, & x = 3. \end{cases} Answer the following questions about gg. (1) Part 1: (a) Rewrite ∣x−3∣|x-3| as an explicit piecewise expression valid for all real xx. Clearly indicate the branch that applies on each side of x=3x=3. (2) Part 2: (b) Using your result from part (a), determine lim⁡x→3−g(x)\displaystyle \lim_{x\to 3^-} g(x) and lim⁡x→3+g(x)\displaystyle \lim_{x\to 3^+} g(x). Show algebraic justification for each one-sided limit. (3) Part 3: (c) Decide whether gg is continuous at x=3x=3. Provide a concise argument referencing the relationship among the one-sided limits, the two-sided limit, and the value g(3)g(3). Explain your reasoning in complete sentences.

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