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: 30


$f(x)$ is continuous on $(-\infty,-2)\cup(-2,\infty)$

Work Step by Step

$f(x)=\frac{x^3-8}{x^2-4}$ for $x\ne2$ and $x\ne-2$ $f(x)=3$ for $x=2$ $f(x)=4$ for $x=-2$ - Domain: $f(x)$ is defined on $(-\infty,\infty)$ 1) Is $f(x)$ continuous at $x=2$ and $x=-2$? - $f(x)$ is continuous at $x=2$ if and only if $\lim_{x\to2}f(x)=f(2)=3$. Similarly, $f(x)$ is continuous at $x=-2$ if and only if $\lim_{x\to-2}f(x)=f(-2)=4$ - First, we would examine the function of $f(x)=\frac{x^3-8}{x^2-4}$: $$f(x)=\frac{x^3-8}{x^2-4}=\frac{(x-2)(x^2+2x+4)}{(x-2)(x+2)}=\frac{x^2+2x+4}{x+2}$$ Therefore, $$\lim_{x\to2}f(x)=\lim_{x\to2}\frac{x^2+2x+4}{x+2}$$ $$\lim_{x\to2}f(x)=\frac{2^2+2\times2+4}{2+2}=\frac{12}{4}=3$$ So $f(x)$ is continuous at $x=2$. $$\lim_{x\to-2}f(x)=\lim_{x\to-2}\frac{x^2+2x+4}{x+2}$$ As $x\to-2$, $(x+2)$ approaches $0$, which means $\frac{x^2+2x+4}{x+2}$, instead of approaching a definite value, would go for $\infty$. So $\lim_{x\to-2}f(x)$ does not exist. $f(x)$, hence, is not continuous at $x=-2$. 2) Is $f(x)$ continuous at other points? As $x\ne2$ and $x\ne-2$: - $\lim_{x\to c}(x^3-8)=c^3-8$ - $\lim_{x\to c}(x^2-4)=c^2-4$ So both $y=x^3-8$ and $y=x^2-4$ are continuous on $(-\infty,-2)\cup(-2,2)\cup(2,\infty)$ Applying Theorem 8, the quotient of 2 functions continuous at $x=c$ is also continuous at $x=c$ Therefore, $f(x)=\frac{x^3-8}{x^2-4}$ is continuous on $(-\infty,-2)\cup(-2,2)\cup(2,\infty)$. In conclusion, combining the results from 1) and 2), $f(x)$ is continuous on $(-\infty,-2)\cup(-2,\infty)$
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.