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Mathematics LibreTexts

1.6E Excercises

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  • Page ID
    10662
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    Exercise \(\PageIndex{1}\): Terms and Concepts


    1. In your own words, describe what it means for a function to be continuous.
    2.  In your own words, describe what the Intermediate Value Theorem states.
    3.  What is a “root” of a function?
    4.  T/F: If \(f\) is defined on an open interval containing c, and \(\lim\limits_{x\to c} f(x)\) exists, then \(f\) is continuous at c.
    5. T/F: If \(f\) is continuous at c, then \(\lim\limits_{x\to c} f(x)\) exist T/F: If \(f\) is continuous at c, then \(\lim\limits_{x\to c^+} f(x)=f(c)\).
    6.  T/F: If \(f\) is continuous on [a, b], then \(\lim\limits_{x\to a^-} f(x)=f(a)\).
    7. T/F: If f is continuous on [0, 1) and [1, 2), then \(f\) is continuous on [0, 2).
    8.  T/F: The sum of continuous functions is also continuous.
    Answer

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    Exercise \(\PageIndex{2}\): Continuity


    Determine on which interval(s)  the given function is continuous.

    1. \(f(x)=x^2-3x+9\)
    2.  \(g(x) = \sqrt{x^2-4}\)
    3. \(h(k) = \sqrt{1-k}+\sqrt{k+1}\)
    4.  \(f(t) = \sqrt{5t^2-30}\)
    5.  \(g(t) = \frac{1}{\sqrt{1-t^2}}\)
    6. \(g(x) = \frac{1}{1+x^2}\)
    7.  \(f(x) = e^x\)
    8.  \(g(s) = \ln s \)
    9.  \(h(t) = \cos t\)
    10.  \(f(k) = \sqrt{1-e^k}\)
    11.  \(f(x) = \sin (e^x+x^2)\)
    Answer

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    Exercise \(\PageIndex{3}\): Continuity


    For what values of \(x\), is \(f\) continuous:

    1.   Given $$f(x)= \left\{ \begin{array}{ccc}
    \displaystyle \frac{x^2-5x+6}{3-x} & \mbox{ if } x <3\\
    \\
    -1 & \mbox{ if } x =3 \\
    \\
    x-2 & \mbox{ if } x >3 \\
    \end{array}
    \right.$$
    2. Given $$f(x)= \left\{ \begin{array}{ccc}
    \displaystyle \frac{-6}{x} & \mbox{ if } x \leq -1\\
    \\
    3x^2 & \mbox{ if } x >-1 \\
    \end{array}
    \right.$$
    3. Given $$f(x)= \left\{ \begin{array}{ccc}
    \displaystyle \frac{|x-3|-2}{x(x-1)}& \mbox{ if } x \neq 0,x \neq 1\\
    \\
    1& \mbox{ if } x =1 \\
    \\
    1& \mbox{ if } x =0 \\
    \end{array}
    \right.$$

    Answer

    \(\mathbb{R}\setminus\{3\},\(\mathbb{R}\setminus\{-1\}, \(\mathbb{R}\setminus\{0,1\}\)

     

    Exercise \(\PageIndex{4}\):Continuity


    For what values of \(k\), each of the following function  is continuous every where:

    1.   $$f(x)= \left\{ \begin{array}{ccc}
    (x+k)^2 & \mbox{ if } x\geq 0\\
    \\
    (2-x)^2 & \mbox{ if } x <0 \\
    \end{array}
    \right.$$
    2.   $$f(x)= \left\{ \begin{array}{ccc}
    (x+k) & \mbox{ if } x\leq 2\\
    \\
    \sqrt{kx^2+20} & \mbox{ if } x >2 \\
    \end{array}
    \right.$$
    3. $$ f(x) = \left\{ \begin{array}{l} 2x+k, x < 3\\ \\ \displaystyle \frac{kx^2 + 1}{7}, x \geq 3\end{array}\right .$$

    4. $$ f(x) = \left\{ \begin{array}{cc} 
              x^2+kx+1 & \mbox{if $x \leq 1$,} \\
    \\
              2x+3 & \mbox{if $x > 1$.} \end{array} \right.
    $$


     

    Answer

    \(k=\pm 2, k=\pm 4\)

    Exercise \(\PageIndex{5}\):Continuity


     

    Given $$f(x)= \left\{ \begin{array}{ccc}
    \displaystyle \frac{\sin(2x)}{ax}& \mbox{ if } x < 0,\\
    \\
    x^2-5& \mbox{ if } 0 \leq x <2 \\
    \\
    bx-4& \mbox{ if } x \geq 2 \\
    \end{array}
    \right.$$

     

    1.  For what value of \(a\), \(f\) is continuous at \(x=0\).
    2.  For what value of \(b\), \(f\) is continuous at \(x=2\).
    Answer

    \(a=-10, b=3/2\)

    Exercise \(\PageIndex{6}\): Continuity


    Find  values of \(k\) and \(m,\) if possible, that will make the function \(f\) continuous everywhere.
    $$f(x) = \left\{ \begin{array}{ll} x^2+2,  x>1\\m(x+1)+k, & - 1< x \leq 1\\ 2x^2 + x + 5, & x \leq -1. \end{array}\right.$$

    Answer

    \(k=6 , m=-3/2\)

     

    Solution:

    when \(x=1\):

    \(\lim_{x \to 1^+} f(x)= \lim_{x \to 1} x^2+2=3\) and 

    \(\lim_{x \to 1^-} f(x)= \lim_{x \to 1} m(x+1)+k=2m+k=f(1)\),

    Since \(f\) is continuous at \(x=1\), \(2m+k=3\).

    when \(x=-1\):

    \(\lim_{x \to -1^+} f(x)= \lim_{x \to -1} m(x+1)+k=k \) and 

    \(\lim_{x \to -1^-} f(x)= \lim_{x \to -1} 2x^2+x+5=2-1+5=6=f(1)\),

    Since \(f\) is continuous at \(x=-1\), \(k=6\).

    Therefore,  \(2m+6=3\), implies \(m=-3/2\).

     

     

    Exercise \(\PageIndex{7}\): Intermediate Value Theorem


    Use the Intermediate Value Theorem to show that the equation \(x^3+x^2-2x-1=0\) has at least one solution in the interval \([1,3]\).

    Answer

    Let \(f(x)= x^3+x^2-2x-1\). Then \(f(x)\) is continuous and \(f(1)=-1\) and \(f(3)=29\).

    Since \(-1<0<29\),by the Intermediate Value Theorem there exist  at least one  real number \(k\) in the interval \([1,3]\) such that \(f(k)=0\).

    Contributors

    Gregory Hartman (Virginia Military Institute). Contributions were made by Troy Siemers and Dimplekumar Chalishajar of VMI and Brian Heinold of Mount Saint Mary's University. This content is copyrighted by a Creative Commons Attribution - Noncommercial (BY-NC) License. http://www.apexcalculus.com/

    Pamini Thangarajah (Mount Royal University, Calgary, Alberta, Canada)