A piecewise function is defined on the closed interval by p(x)= cases 2x+5, & -2 x0$…

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

Problem

A piecewise function

p

is defined on the closed interval

[-2,6]

p(x)=\begin{cases} 2x+5, & -2\le x<1,\\ k\ln(x+2), & 1\le x\le6, \end{cases}

where

k

is a nonzero real constant. The natural logarithm is defined only for positive arguments, so

x+2>0

must hold wherever the logarithmic expression is used. (1) Part 1: (a) State the largest possible domain of

p

within

[-2,6]

without imposing any additional conditions on

k

. Briefly justify your answer referring to the restrictions of the component functions. (2) Part 2: (b) Determine the value of

k

that makes

p

continuous on the entire interval

[-2,6]

. Show all limit evaluations and algebraic manipulations. (3) Part 3: (c) Using the value of

k

from part (b), evaluate

\displaystyle \int_{-2}^{6} p(x)\,dx

. Explain why the integral is guaranteed to exist based on the continuity of

p