A function is defined on the closed interval by p(x)= cases x, & - x<0, 1, & x=0,…

MATHEMATICS · Multi-part · AP · Calculus AB · Analytical Applications of Differentiation · Analytical Applications of Differentiation

Problem

A function pp is defined on the closed interval [−π,2][-\pi,2] by p(x)={sin⁡x,−π≤x<0,1,x=0,1−x,0<x≤2.p(x)=\begin{cases} \sin x, & -\pi\le x<0,\\ 1, & x=0,\\ 1-x, & 0<x\le 2. \end{cases} You will investigate whether absolute extrema exist for pp on its domain. (a) Determine whether pp is continuous at x=0x=0. Provide a clear limit–based justification for your conclusion. (b) Decide whether it is mathematically valid to restrict attention only to interior points with p′(x)=0p'(x)=0 or p′(x)p'(x) undefined when looking for absolute extrema of pp on [−π,2][-\pi,2]. Explain how your answer to part (a) influences this decision. (c) For the portions of the domain where pp is differentiable, compute p′(x)p'(x) and find the interior point(s) where p′(x)=0p'(x)=0. (d) Evaluate pp at every point that could plausibly be an absolute maximum or minimum, including endpoints, interior critical points, and any other candidates justified by your analysis. (e) State whether pp attains an absolute maximum value and an absolute minimum value on [−π,2][-\pi,2]. If so, identify the values and where they occur; if not, give a rigorous explanation for the failure.

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