Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 11 - Review Exercises - Page 1179: 49


Continuous everywhere.

Work Step by Step

Recall that if $ f $ is a polynomial function, then we have $\lim_\limits{x\to a}f(x)=f(a)$. $\lim_\limits{x \to 5^{-}} f(x)= \lim_\limits{x \to 5^{-}} 4x=20$ and $\lim_\limits{x \to 5^{+}} f(x)= \lim_\limits{x \to 5^{+}}(x^2-5)=20$ So, $\lim_\limits{x \to 5^{-}} f(x)= \lim_\limits{x \to 5^{+}}f(x)$ exists. Now, $\lim_\limits{x \to 5} f(x)=20$ and $ f(5)=20$ so, $\lim_\limits{x \to 5} f(x) = f(5)$ Therefore, the function is continuous everywhere.
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