A continuous function is defined by h(t)= cases t & for 0 t 2, 4 & for 2 < t 5

MATHEMATICS · Multi-part · AP · Calculus AB · Integration and Accumulation of Change · Integration and Accumulation of Change

Problem

A continuous function hh is defined by h(t)={tfor 0≤t≤2,4for 2<t≤5.h(t)=\begin{cases} t & \text{for } 0 \le t \le 2,\\ 4 & \text{for } 2 < t \le 5. \end{cases} Let HH be an antiderivative of hh with H(0)=−1H(0)=-1. (a) Compute the exact value of H(2)H(2). (b) Compute the exact value of H(5)H(5). (c) For 0≤x≤50\le x\le 5, at which xx-value(s) does HH increase most rapidly? Justify your answer.

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