A function is defined on the closed interval by p(x)= cases 5- x^2 4 , & -3 x 0, 1+ x…

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

Problem

A function pp is defined on the closed interval [−3,4][-3,4] by p(x)={5−x24,−3≤x≤0,1+x,0<x≤4.p(x)=\begin{cases} 5-\dfrac{x^2}{4}, & -3\le x\le 0,\\ 1+\sqrt{x}, & 0<x\le 4. \end{cases} The graph of pp is not provided. (1) Part 1: (a) Verify that pp is continuous on [−3,4][-3,4]. Provide complete reasoning, including consideration of x=0x=0. (2) Part 2: (b) Explain why pp must attain both an absolute maximum value and an absolute minimum value on the interval [−3,4][-3,4]. Cite an appropriate theorem and discuss how its conditions are satisfied by pp. (3) Part 3: (c) Describe a systematic procedure that could be used to locate the absolute extrema of pp on [−3,4][-3,4] without actually carrying out any numerical calculations. Your description should cite every point or type of point that must be examined and explain why each is necessary.

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