Finite Math and Applied Calculus (6th Edition)

Published by Brooks Cole
ISBN 10: 1133607705
ISBN 13: 978-1-13360-770-0

Chapter 1 - Section 1.1 - Functions from the Numerical, Algebraic, and Graphical Viewpoints - Exercises - Page 53: 33a

Answer

$f(x)=\{-2 < $x$ <= 0: $ x^2$ ,\ \ 0 < $x$ < = 4: 1/x\}$ $f(-1)=1$ .
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Work Step by Step

$f(x)=\left\{\begin{array}{lll} -1 & if & -2 < x \leq 0\\ x & if & 0 < x \leq 4 \end{array}\right.$ We build two tables, one for $-2 < x \leq 0$, (part of a parabola $ y=x^{2})$ since $-2$ does not belong to this interval, we draw an empty circle for the point $(-2,4)$ $0$ is in the interval, solid circle for the point $(0,0)$ another for $0 < x \leq 4$ (part of the hyperbola graph for $ y=1/x)$ $\displaystyle \frac{1}{x}$ is not defined for x=0, it rises without bound when x approaches 0, $4$ is in the interval, solid circle for the point $(4,\displaystyle \frac{1}{4})$ $x=-1$ belongs to the interval $0 < x \leq 4$, so we use the rule for that interval $f(-1)=(-1)^{2}=1$ The technology formula for piece wise defined function has the form: $f(x)=\text{\{$ interval:expression, interval:expression $\}}$ Here, $f(x)=\{-2 < $x$ <= 0: $ x^2$ ,\ \ 0 < $x$ < = 4: 1/x\}$
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