Calculus Concepts: An Informal Approach to the Mathematics of Change 5th Edition

Published by Brooks Cole
ISBN 10: 1-43904-957-2
ISBN 13: 978-1-43904-957-0

Chapter 1 - Ingredients of Change: Functions and Limits - 1.3 Activities - Page 31: 34

Answer

(a) 0, (b) 0, (c) 0, (d) $f$ is continuous at $x=0$

Work Step by Step

Given $$f(x)=\left\{\begin{array}{ll}{2 x^{2}} & {\text { when } x<0} \\ {4 x} & {\text { when } x \geq 0}\end{array}\right.$$ (a) Since \begin{align*} \lim _{x \rightarrow0^{-}} f(x)&=\lim _{x \rightarrow0^{-}}2 x^2\\ &=\lim _{x \rightarrow0^{-}}2(0)\\ &=0 \end{align*} (b) Since \begin{align*} \lim _{x \rightarrow0^{+}} f(x)&=\lim _{x \rightarrow0^{+}} (4x)\\ &=4(0)\\ &=0 \end{align*} (c) $f(0)= 4(0)=0$ (d) Since $$\lim _{x \rightarrow0^{+}} f(x)=\lim _{x \rightarrow0^{-}} f(x)=f(0)=0$$ Thus, $f$ is continuous at $x=0$
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