Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Diagnostic Tests - C Diagnostic Tests: Functions - Page xxix: 6

Answer

(a) $\color{blue}{f(-2)=-3},\ \color{green}{ f(1)=3}$ (b) See attachment.

Work Step by Step

(a) To find $f(-2)$, since $x=-2 \le 0$, we use the upper piece (upper branch) of the split function: $\color{blue}{f(x) = 1-x^2, x\le 0}$. $\begin{align*} f(-1) &= 1- (-2)^2 \\ &= 1 - 4 \\ \color{blue}{f(-1)}\ &\color{blue}{= -3.} \end{align*}$ To find $f(1)$, since $x=1 \gt 0$, we use the lower piece (or lower branch) of the function: $\color{green}{f(x) = 2x+1, x\gt 0}$. $\begin{align*} f(1) &= 2(1) + 1 \\ &= 2+1 \\ \color{green}{f(1)}\ &\color{green}{= 3.} \end{align*}$ (b) See attachment.
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