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: 35


(i) $x=-2, 1$ (ii) $g(x) = x+1$

Work Step by Step

(i) The holes of the function occur when the denominator of the function equals zero. The denominator is $(x+2)(x-1)$ and is $0$ when $x=-2, 1$. (ii) Note that the holes of the function are created because the denominator equals $0$ are the two points mentioned in part (i). We are able to remove these holes by cancelling the factors in the bottom with the factors in the top as such: $g(x) = \dfrac{(x+2)(x^2-1)}{(x+2)(x-1)} = \dfrac{(x+2)(x-1)(x+1)}{(x+2)(x-1)} = x+1$. Therefore, the desired function identical to $f(x)$ but without holes is $g(x) = x+1$.
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