Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 0 - Before Calculus - 0.1 Functions - Exercises Set 0.1 - Page 15: 36


(i) $x = 0$ (ii) $g(x) = |x| + 1$

Work Step by Step

(i) Holes in a function occur where the denominator is $0$. The given function is \[ f(x) = \frac{x^{2} + |x|}{|x|}, \] so the denominator is $|x|$. Now $|x|$ is $0$ when $x = 0$, and only when $x = 0$. So, $f(x)$ has a hole at $x = 0$ and only at $x = 0$. (ii) First, note that $x^{2}$ is always positive, so for any $x, \: x^{2} = |x| \cdot |x|$. Therefore, \[f(x) = \frac{x^{2} + |x|}{|x|} = \frac{|x| \cdot |x| + |x|}{|x|} = \frac{|x| \cdot |x|}{|x|} + \frac{|x|}{|x|} = |x| + 1, \mbox{for } x \ne 0. \] So now, if we take $g(x) = x + 1$, then $g(x)$ is identical to $f(x)$ everywhere except at $x=0$. That is, $f(x)$ is not defined at $x = 0$ so $f(x)$ has a hole at $x=0$. But $g(x)$ on the other hand is defined at $x=0$ and so $g(x)$ has no hole. So the graph of $g$ is identical to that of $f$ but without the hole, which is exactly what the question asks for.
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