Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 2 - Limits - 2.8 Intermediate Value Theorem - Exercises - Page 86: 20

Answer

The function $f$ is not continuous at $x=0$. For $y=0$ in $(f(-2),f(2))$ there is no corresponding $x$ in $(-2,2)$

Work Step by Step

The Intermediate Value Theorem "asks" for a function $f$ to be continuous on an interval $[a,b]$ so that $f(a)\not=f(b)$. Let's check these conditions for the given function. The interval $[a,b]$ is $[-2,2]$, so $a=-2,b=2$. $f(a)=f(-2)=-(-2)^2=-4$ $f(b)=f(2)=2$ Therefore $f(-2)\not=f(2)$. Now let's check the continuity. The function $f$ is continuous on each of the intervals $[-2,0), (0,2]$. We have to check the continuity at $x=0$. Calculate the left and right limits at $x=0$: $$\begin{align*} \lim_{x\rightarrow0^-}f(x)&=\lim_{x\rightarrow0^-} (-x^2)=0\\ \lim_{x\rightarrow0^+}f(x)&=\lim_{x\rightarrow0^+} x=0\\ f(0)&=1 \end{align*}$$ As $\lim_{x\rightarrow0^-}f(x)=\lim_{x\rightarrow0^+}f(x)\not=f(0)$, it means the function is not continuous at $x=0$. We conclude that the Intermediate Value Theorem cannot be applied. A value between $f(-2)$ and $f(2)$ for which there is no $x$ in $(-2,2)$ is $0$.
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.