University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Section 2.5 - Continuity - Exercises - Page 95: 59

Answer

$f(x)=\frac{x^2+4x-12}{x-2}$ for $x\ne2$ and $f(x)=4$ for $x=2$

Work Step by Step

$f(x)=\frac{x^2+4x-12}{x-2}$ for $x\ne2$ and $f(x)=4$ for $x=2$ This function is continuous for all $x$ except $x=2$, where it has a removable discontinuity. The reason is stated as follows: 1) For all $x\ne2$, we have $\lim_{x\to c}\frac{x^2+4x-12}{x-2}=\frac{c^2+4c-12}{c-2}=f(c)$, meaning that $f$ is continuous for all $x\ne2$. - For $x=2$, we have $f(2)=4$. Yet, $$\lim_{x\to2}f(x)=\lim_{x\to2}\frac{x^2+4x-12}{x-2}=\lim_{x\to2}\frac{(x-2)(x+6)}{x-2}=\lim_{x\to2}(x+6)$$ $$\lim_{x\to2}f(x)=2+6=8$$ As $\lim_{x\to2}f(x)\ne f(2)$, $f(x)$ is discontinuous at $x=2$. 2) However, this discontinuity is a removable one, because both $\lim_{x\to2}f(x)$ and $f(2)$ still exist; it is just that their values are not equal. If we extend $f(x)$ to include the value $f(2)=\lim_{x\to2}f(x)=8$, then the discontinuity will be removed.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.