Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Section 3.1 - Tangents and the Derivative at a Point - Exercises 3.1 - Page 109: 38

Answer

The graph of $U(x)$ does not consist of a vertical tangent at the origin.

Work Step by Step

1. The graph of $U(x)$ has a vertical tangent at $x$ when $U(x)$ is continuous at $x$ 2. The graph of $U(x)$ has a vertical tangent at $x$ when $U'(x)=\lim\limits_{h\to 0}\dfrac{U(x+h)-U(x)}{h}=\pm\infty$ Also, $\lim\limits_{x\to0^+}U(x)=\lim\limits_{x\to0^+}(1)=1$ and $\lim\limits_{x\to0^{-}}U(x)=\lim\limits_{x\to0^{-}}(0)=0$ Since $\lim\limits_{x\to0^+}U(x)\ne\lim\limits_{x\to0^{-}}U(x)$, thus, $\lim\limits_{x\to 0} U(x)$ does not exist. This implies that $U(x)$ is not continuous at the origin and the graph of $U(x)$ does not consist of a vertical tangent at the origin.
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