Chapter 5.6 Practice

Questions

  1. Take \(f\) to be a function with the property that \[ \lim_{x\to 4^+}f(x)=1\quad\text{and}\quad \lim_{x\to 4^-}f(x)=1.\] Determine a value for \(f(4)\) so that \(f\) is continuous at \(4.\)

  2. For each of these choices of function \(f\) and real number \(x_0\), determine whether \(f\) is continuous at \(x_0\).

    1. \(f(x)=-3x+1\), \(x_0=1\)
    2. \(f(x)=\frac{x}{x+1}\), \(x_0=0\)
    3. \(f(x)=\frac{x}{x+1}\), \(x_0=-1\)
    4. \(f(x)=\sqrt{x-2}\), \(x_0=2\)
    5. \(f(x)=\sin(x)\), \(x_0=1\)
    6. \(f(x)=\exp(x)\), \(x_0=-3\)
    7. \(f(x)=\ln(x)\), \(x_0=1\)
    8. \(f(x)=\ln(x)\), \(x_0=0\)
    9. \(f(x)=\begin{cases}(x-3)\sin\left(\frac{1}{x-3}\right)&\text{if }x\not=3\\0&\text{if }x=3\\\end{cases}\), \(x_0=3\)
    10. \(f(x)=\begin{cases}x-4&\text{if }x<5\\4&\text{if }x=5\\\log_4(9-x) &\text{if }x>5\\\end{cases}\), \(x_0=5\)
    11. \(f(x)=\begin{cases}\cos\left(\frac{1}{x-2}\right)&\text{if }x\leq 2\\ 4x^2+3 &\text{if }x>2\\\end{cases}\), \(x_0=5\)

  3. For each function below, identify the maximal set \(S\) for which \(f\) is continuous. Then, determine if there is a continuous extension of \(f\) to \(\mathbb{R}\) or some set \(S'\) so that \(S\) is a proper subset of \(S'.\)

    1. \(f(x)=x^2\)
    2. \(f(x)=\log_2(x^2+1)\)
    3. \(f(x)=\ln((x-3)^2)\)
    4. \(f(x)=\frac{x-5}{x^2-25}\)
    5. \(f(x)=\frac{x^2-25}{x-5}\)
    6. \(f(x)=\frac{\sin(4(x-3))}{x-3}\)
    7. \(f(x)=\frac{\cos(4(x-3))}{x-3}\)
    8. \(f(x)=\frac{1-\cos(x-5)}{x-5}\)
    9. \(f(x)=\frac{1-\cos(x^2-25)}{x^2-25}\)
    10. \(f(x)=\sqrt{\frac{x-1}{x^2-1}}\)

  4. For each function \(f\) given below and domain \(S\), determine whether \(f\) attains its minimum value on \(S\) or its maximum value on \(S\).

    1. \(f(x)=4x+1\), \(S=[1,4]\)
    2. \(f(x)=x^2+5\), \(S=[0,4)\)
    3. \(f(x)=\frac{(x+2)}{(x-2)(x+5)}\), \(S=(-2,2)\)
    4. \(f(x)=\frac{(x+2)}{(x-2)(x+5)}\), \(S=[-2,2)\)

  5. For each function \(f\) and real number \(C\), determine whether the Intermediate Value Theorem guarantees that there is an \(x_0\) in the given set \(S\) so that \(f(x_0)=C.\)

    1. \(f(x)=x^2+1\), \(S=[-2,5]\), \(C=10\)
    2. \(f(x)=x^2+1\), \(S=[-2,5]\), \(C=65\)
    3. \(f(x)=\exp(x)\), \(S=[0,1]\), \(C=2\)
    4. \(f(x)=x+2\), \(S=[-1,3]\), \(C=2\)

  6. For each of these choices of function \(f\), decompose \(f\) into sums, products, and quotients of continuous functions to determine where \(f\) is continuous.

    1. \(f(x)=\frac{x(x-1)}{x^2+1}\)
    2. \(f(x)=\frac{x(x-1)}{x^2-1}\)
    3. \(f(x)=\frac{\sqrt{1-x}}{(x+3)(x-3)}\)
    4. \(f(x)=x^2\sin\left(x^2-5\right)\)
    5. \(f(x)=\cos(x)e^\frac{x-1}{x+2}\)
    6. \(f(x)=\sqrt{x^2+1}+5\ln(x-1)\)
    7. \(f(x)=\ln(-e^x+3)\)

  7. For each of these choices of function \(f\) and real value \(x_0\), decompose \(f\) as a compound function to determine \(\displaystyle \lim_{x\to x_0}f(x)\)

    1. \(f(x)=\exp\left(\frac{x^2+2x-8}{x-2}\right)\), \(x_0=-4\)
    2. \(f(x)=\exp\left(\frac{x^2+2x-8}{x-2}\right)\), \(x_0=2\)
    3. \(f(x)=\ln\left(\frac{\sin(3x)}{x}\right)\), \(x_0=0\)
    4. \(f(x)=\sqrt{\frac{1-\cos(x)}{x}}\), \(x_0=0\)
    5. \(f(x)=\sqrt{\frac{1-\cos(x)}{x^2}}\), \(x_0=0\)

  8. Construct a function \(f\) such that

    • \(f\) is defined on \(\mathbb{R}\)
    • \(f\) is continuous everywhere except at \(2\)
    • \(\displaystyle\lim_{x\to 2^-}f(x)=3\)
    • \(\displaystyle\lim_{x\to 2^+}f(x)=2\)
    • \(f\) is asymptotically equal to \(5x^3\) to the right
    • \(f\) has a horizontal asymptote of \(\frac{1}{2}\)

  9. For the function \(f\) given below, determine \(a\) and \(b\) so that is continuous on \(\mathbb{R}.\) \[f(x)=\begin{cases}2x+a&\text{if }x<-1\\ 5x&\text{if }-1\leq x<b\\ 2x+2&\text{if }b\leq x\end{cases}\]

  10. For each equation and interval \(I\) given below, explain the existence of a solution to the equation in the interval \(I\):

    1. \(x^2-\frac{1}{2}=0\), \(I=(0,1)\)
    2. \(x^3-15=0\), \(I=(2,3)\)
    3. \(\sqrt{x}-\cos(x)=0\), \(I=(0,\pi)\)

Answers

  1. \(f(4)=1\)

    1. Yes
    2. Yes
    3. No
    4. Yes
    5. Yes
    6. Yes
    7. Yes
    8. No
    9. Yes
    10. Yes
    11. No
    1. \(S=\mathbb{R}\). There is a continuous extension to \(\mathbb{R}\); it is the function itself: \(\tilde{f}=f\)
    2. \(S=\mathbb{R}\). There is a continuous extension to \(\mathbb{R}\); it is the function itself: \(\tilde{f}=f\)
    3. \(S=(-\infty,3)\cup(3,\infty)\). There is a no continuous extension to \(\mathbb{R}\) nor some set \(S'\) so that \(S\) is a proper subset of \(S'\)
    4. \(S=(-\infty,-5)\cup(-5,5)\cup(5,\infty)\). There is no continuous extension to \(\mathbb{R}\) but there is a continuous extension to \(S'=(-\infty,5)\cup(5,\infty):\) \[\tilde{f}(x)=\begin{cases}\frac{x-5}{x^2-25} &\text{if } x<-5\\ \frac{x-5}{x^2-25} &\text{if } -5<x<5\\ \frac{1}{10}&\text{if } x=5\\ \frac{x-5}{x^2-25} &\text{if } x>5\end{cases}\]
    5. \(S=(-\infty,5)\cup(5,\infty)\). There is a continuous extension to \(\mathbb{R}:\) \[\tilde{f}(x)=\begin{cases}\frac{x^2-25}{x-5} &\text{if } x\not=5\\ 10&\text{if } x=5\end{cases}\]
    6. \(S=(-\infty,3)\cup(3,\infty)\). There is a continuous extension to \(\mathbb{R}:\) \[\tilde{f}(x)=\begin{cases}\frac{\sin(4(x-3))}{x-3} &\text{if } x\not=3\\ 4&\text{if } x=3\end{cases}\]
    7. \(S=(-\infty,3)\cup(3,\infty)\). There is a no continuous extension to \(\mathbb{R}\) nor some set \(S'\) so that \(S\) is a proper subset of \(S'\)
    8. \(S=(-\infty,5)\cup(5,\infty)\). There is a continuous extension to \(\mathbb{R}:\) \[\tilde{f}(x)=\begin{cases}\frac{1-\cos(x-5)}{x-5} &\text{if } x\not=5\\ 0&\text{if } x=5\end{cases}\]
    9. \(S=(-\infty,-5)\cup(-5,5)\cup(5,\infty)\). There is a continuous extension to \(\mathbb{R}:\) \[\tilde{f}(x)=\begin{cases}\frac{1-\cos(x^2-25)}{x^2-25} &\text{if } x<-5\\ 0 &\text{if } x=-5\\\frac{1-\cos(x^2-25)}{x^2-25} &\text{if } -5<x<5\\ 0 &\text{if } x=5\\ \frac{1-\cos(x^2-25)}{x^2-25} &\text{if } x>5\end{cases}\]
    10. \(S=(-1,1)\cup(1,\infty)\). There is no continuous extension to \(\mathbb{R}\), but there is a continuous extension to \(S'=(-1,\infty).\) \[\tilde{f}(x)=\begin{cases}\sqrt{\frac{x-1}{x^2-1}} &\text{if }-1<x<1\\ \frac{\sqrt{2}}{2} &\text{if } x=1\\ \sqrt{\frac{x-1}{x^2-1}} &\text{if } x>1\end{cases}\]
    1. Yes, both maximum and minimum
    2. Yes to minimum, no to maximum
    3. No to both
    4. Yes to minimum, no to maximum
    1. Yes, since \(f\) is continuous on \(S\), \(f(-2)=5\), \(f(5)=26\), and \(C=10\) is in \([f(-2),f(26)],\) so the intermediate value theorem applies.
    2. No, because \(C=26\) is not in \([f(-2),f(26)]\) so there is no guarantee that the equality \(f(x)=C\) has a solution.
    3. Yes, since \(f\) is continuous on \(S\), \(f(0)=1\), \(f(1)=e\), and \(C=2\) is in \([f(0),f(1)],\) so the intermediate value theorem applies.
    4. Yes, since \(f\) is continuous on \(S\), \(f(-1)=1\), \(f(3)=5\), and \(C=2\) is in \([f(-1),f(3)],\) so the intermediate value theorem applies.
    1. Continuous on \(\mathbb{R}\)
    2. Continuous on \((-\infty,-1)\cup(-1,1)\cup(1,\infty)\)
    3. Continuous on \((-\infty,-3)\cup(-3,-1]\)
    4. Continuous on \(\mathbb{R}\)
    5. Continuous on \((-\infty,-2)\cup(-2,\infty)\)
    6. Continuous \((1,\infty)\)
    7. Continuous \((\ln(3),\infty)\)
    1. \(\displaystyle\lim_{x\to -4}\exp\left(\frac{x^2+2x-8}{x-2}\right)=1\)
    2. \(\displaystyle\lim_{x\to 2}\exp\left(\frac{x^2+2x-8}{x-2}\right)=e^6\)
    3. \(\displaystyle\lim_{x\to 0}\ln\left(\frac{\sin(3x)}{x}\right)=\ln(3)\)
    4. \(\displaystyle\lim_{x\to 0}\sqrt{\frac{1-\cos(x)}{x}}=0\)
    5. \(\displaystyle\lim_{x\to 0}\sqrt{\frac{1-\cos(x)}{x^2}}=\frac{\sqrt{2}}{2}\)

  8. There are many possible answers. Here is one possibility. \[f(x)=\begin{cases}\frac{x+3}{2x^2+1}&\text{if } x\leq -2\\ \frac{13}{18}(x-2)+3 &\text{if }-2<x\leq 3\\ 3(x-3)+2&\text{if }3<x<5\\ 5(x-5)^3+8&\text{if }x\geq 5\end{cases}\]

  9. \(a=-3\) and \(b=\frac{2}{3}\)

    1. \(f(x)=x^2-\frac{1}{2}\) is continuous \([0,1]\), \(f(0)=-\frac{1}{2}\) and \(f(1)=\frac{1}{2}\); since \(0\) is in \([f(0),f(1)]\), by the intermediate value theorem, \(f(x)=0\) has a solution in \((0,1).\)
    2. \(f(x)=x^3-15\) is continuous \([2,3]\), \(f(2)=-7\) and \(f(3)=12\); since \(0\) is in \([f(2),f(3)]\), by the intermediate value theorem, \(f(x)=0\) has a solution in \((2,3).\)
    3. \(f(x)=\sqrt{x}-\cos(x)\) is continuous \([0,\pi]\), \(f(0)=-1\) and \(f(\pi)=1+\sqrt{\pi}\); since \(0\) is in \([f(0),f(\pi)]\), by the intermediate value theorem, \(f(x)=0\) has a solution in \((0,\pi).\)

Return

© Copyright 2025 by the POC Writing Team: Bryan Carrillo, Yat Sun Poon, and David Weisbart. All rights reserved. No part of this publication may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopy, recording, or any information storage and retrieval system, without permission in writing from the POC Writing Team.