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Section 2-9 : Continuity

  1. The graph of \(f\left( x \right)\) is given below. Based on this graph determine where the function is discontinuous.
    This graph consists of four segments.  The first is in the domain \(x < -3\).  This segment is a decreasing function that starts at (-4,5) and ends at an open dot at (-3,2).  There is also a closed dot at (-3,5). The second segment is in the domain \(-3 < x \le -1\).  It starts with an open dot at (-3,2) and increases until approximately (-2.2, 4.8) and then decreases until it ends at a closed dot at (-1,4).  The third segment is in the domain \(-1 < x < \le 2\).  It starts with a closed dot at (-1,-3) has an open dot at (1,-1) and ends at a closed dot at (2,0).  The final segment is in the domain \(x > 2\).  The graph in the segment is an oscillating function that oscillates faster and faster as it approaches x = 2 from the right and the oscillation slows down as it moves away from x=2.
  2. The graph of \(f\left( x \right)\) is given below. Based on this graph determine where the function is discontinuous.
    This function has three parts to it.  In the domain \(-4 \le x <-1\) it is a parabola with a vertex at approximately (-2.2, 4.1) that opens down intersecting the x-axis at approximately (-3.8,0) and ends at (-1,1) in an open dot.  Also in this portion there is an open dot at (-2,4) and a closed dot at (-2,2).  In the domain \(-1 \le x \le 2\) it is a parabola with vertex at approximately (0.25, 5) that opens upwards.  The left end is a closed dot at (-1,-3) and the right end is a closed dot at (2,-1).  The final portion is in the domain \(2 < x \le 4\) and is a parabola with vertex at (3,5) which is indicated by a closed dot and opens downward.  The left end of the parabola is an open dot at (2,2).
  3. The graph of \(f\left( x \right)\) is given below. Based on this graph determine where the function is discontinuous.

    This function has three parts to it.  In the domain \(-5 \le x < 1\) it is a curve that looks pretty much like a parabola with vertex at approximately (-2.5, 3.2) and opens downward.  There is a closed dot at (-4,-2), and open dot at (-2,3), a closed dot at (-2,5) and the curve ends at an open dot at (1,-3).  The second portion of this graph is in the domain \(1 < x < 4\).  It starts with an open dot at (1,4), there is a closed dot at (1,2).  As the graph moves to the right it starts out fairly flat with a slight decrease but as it gets closer and closer to x=4 from the left the graph decreases faster and faster.  The final portion of the graph is in the domain \(4 < x < 6\).  As the graph approaches x=4 from the right it increases faster and faster and as it moves away from x=4 the graph decreases until it reaches the end of the domain at z=6.

For problems 4 – 13 using only Properties 1- 9 from the Limit Properties section, one-sided limit properties (if needed) and the definition of continuity determine if the given function is continuous or discontinuous at the indicated points.

  1. \(\displaystyle f\left( x \right) = \frac{{6 + 2x}}{{7x - 14}}\)
    1. \(x = - 3\)
    2. \(x = 0\)
    3. \(x = 2\)
  2. \(\displaystyle R\left( y \right) = \frac{{2y}}{{{y^2} - 25}}\)
    1. \(y = - 5\)
    2. \(y = - 1\)
    3. \(y = 3\)
  3. \(\displaystyle g\left( z \right) = \frac{{5z - 20}}{{{z^2} - 12z}}\)
    1. \(z = - 1\)
    2. \(z = 0\)
    3. \(z = 4\)
  4. \(\displaystyle W\left( x \right) = \frac{{2 + x}}{{{x^2} + 6x - 7}}\)
    1. \(x = - 7\)
    2. \(x = 0\)
    3. \(x = 1\)
  5. \(h\left( z \right) = \left\{ {\begin{array}{rl}{2{z^2}}&{z < - 1}\\{4z + 6}&{z \ge - 1}\end{array}} \right.\)
    1. \(z = - 6\)
    2. \(z = - 1\)
  6. \(g\left( x \right) = \left\{ {\begin{array}{rl}{x + {{\bf{e}}^x}}&{x < 0}\\{{x^2}}&{x \ge 0}\end{array}} \right.\)
    1. \(x = 0\)
    2. \(x = 4\)
  7. \(Z\left( t \right) = \left\{ {\begin{array}{rl}8&{t < 5}\\{1 - 6t}&{t \ge 5}\end{array}} \right.\)
    1. \(t = 0\)
    2. \(t = 5\)
  8. \(h\left( z \right) = \left\{ {\begin{array}{rl}{z + 2}&{z < - 4}\\0&{z = - 4}\\{18 - {z^2}}&{z > - 4}\end{array}} \right.\)
    1. \(z = - 4\)
    2. \(z = 2\)
  9. \(f\left( x \right) = \left\{ {\begin{array}{rc}{1 - {x^2}}&{x < 2}\\{ - 3}&{x = 2}\\{2x - 7}&{2 < x < 7}\\0&{x = 7}\\{{x^2}}&{x > 7}\end{array}} \right.\)
    1. \(x = 2\)
    2. \(x = 7\)
  10. \(g\left( w \right) = \left\{ {\begin{array}{rc}{3w}&{w < 0}\\0&{w = 0}\\{w + 6}&{0 < w < 8}\\{14}&{w = 8}\\{22 - w}&{w > 8}\end{array}} \right.\)
    1. \(w = 0\)
    2. \(w = 8\)

For problems 14 – 22 determine where the given function is discontinuous.

  1. \(\displaystyle f\left( x \right) = \frac{{11 - 2x}}{{2{x^2} - 13x - 7}}\)
  2. \(\displaystyle Q\left( z \right) = \frac{3}{{2{z^2} + 3z - 4}}\)
  3. \(\displaystyle h\left( t \right) = \frac{{{t^2} - 1}}{{{t^3} + 6{t^2} + t}}\)
  4. \(\displaystyle f\left( z \right) = \frac{{4z + 1}}{{5\cos \left( {{\textstyle{z \over 2}}} \right) + 1}}\)
  5. \(\displaystyle h\left( x \right) = \frac{{1 - x}}{{x\sin \left( {x - 1} \right)}}\)
  6. \(\displaystyle f\left( x \right) = \frac{3}{{4{{\bf{e}}^{x - 7}} - 1}}\)
  7. \(\displaystyle R\left( w \right) = \frac{{{{\bf{e}}^{{w^2} + 1}}}}{{{{\bf{e}}^w} - 2{{\bf{e}}^{1 - w}}}}\)
  8. \(g\left( x \right) = \cot \left( {4x} \right)\)
  9. \(f\left( t \right) = \sec \left( {\sqrt t } \right)\)

For problems 23 – 27 use the Intermediate Value Theorem to show that the given equation has at least one solution in the indicated interval. Note that you are NOT asked to find the solution only show that at least one must exist in the indicated interval.

  1. \(1 + 7{x^3} - {x^4} = 0\) on \(\left[ {4,8} \right]\)
  2. \({z^2} + 11z = 3\) on \(\left[ { - 15, - 5} \right]\)
  3. \(\displaystyle \frac{{{t^2} + t - 15}}{{t - 8}} = 0\) on \(\left[ { - 5,1} \right]\)
  4. \(\ln \left( {2{t^2} + 1} \right) - \ln \left( {{t^2} + 4} \right) = 0\) on \(\left[ { - 1,2} \right]\)
  5. \(10 = {w^3} + {w^2}{{\bf{e}}^{ - w}} - 5\) on \(\left[ {0,4} \right]\)

For problems 28 – 33 assume that \(f\left( x \right)\) is continuous everywhere unless otherwise indicated in some way. From the given information is it possible to determine if there is a root of \(f\left( x \right)\) in the given interval?

If it is possible to determine that there is a root in the given interval clearly explain how you know that a root must exist. If it is not possible to determine if there is a root in the interval sketch a graph of two functions each of which meets the given information and one will have a root in the given interval and the other will not have a root in the given interval.

  1. \(f\left( { - 5} \right) = 12\) and \(f\left( 0 \right) = - 3\) on the interval \(\left[ { - 5,0} \right]\).
  2. \(f\left( 1 \right) = 30\) and \(f\left( 9 \right) = 6\) on the interval \(\left[ {1,9} \right]\).
  3. \(f\left( {20} \right) = - 100\) and \(f\left( {40} \right) = - 100\) on the interval \(\left[ {20,40} \right]\).
  4. \(f\left( { - 4} \right) = - 10\), \(f\left( 5 \right) = 17\), \(\mathop {\lim }\limits_{x \to \,{1^ - }} f\left( x \right) = - 2\), and \(\mathop {\lim }\limits_{x \to \,{1^ + }} f\left( x \right) = 4\) on the interval \(\left[ { - 4,5} \right]\).
  5. \(f\left( { - 8} \right) = 2\), \(f\left( 1 \right) = 23\), \(\mathop {\lim }\limits_{x \to \, - {4^ - }} f\left( x \right) = 35\), and \(\mathop {\lim }\limits_{x \to \, - {4^ + }} f\left( x \right) = 1\) on the interval \(\left[ { - 8,1} \right]\).
  6. \(f\left( 0 \right) = - 1\), \(f\left( 9 \right) = 10\), \(\mathop {\lim }\limits_{x \to \,{2^ - }} f\left( x \right) = - 12\), and \(\mathop {\lim }\limits_{x \to \,{2^ + }} f\left( x \right) = - 3\) on the interval \(\left[ {0,9} \right]\).