Calculus

Applications of the Derivative

Applications of Derivative Logo

Curve Sketching

  • The following steps are taken in the process of curve sketching:

    \(1.\) Domain

    Find the domain of the function and determine the points of discontinuity (if any).

    \(2.\) Intercepts

    Determine the \(x-\) and \(y-\)intercepts of the function, if possible. To find the \(x-\)intercept, we set \(y = 0\) and solve the equation for \(x.\) Similarly, we set \(x = 0\) to find the \(y-\)intercept. Find the intervals where the function has a constant sign \(\left({f\left( x \right) \gt 0}\right.\) and \(\left.{f\left( x \right) \lt 0}\right).\)

    \(3.\) Symmetry

    Determine whether the function is even, odd, or neither, and check the periodicity of the function. If \(f\left( { – x} \right) = f\left( x \right)\) for all \(x\) in the domain, then \(f\left( x \right)\) is even and symmetric about the \(y-\)axis. If \(f\left( { – x} \right) = -f\left( x \right)\) for all \(x\) in the domain, then \(f\left( x \right)\) is odd and symmetric about the origin.

    \(4.\) Asymptotes

    Find the vertical, horizontal and oblique (slant) asymptotes of the function.

    \(5.\) Intervals of Increase and Decrease

    Calculate the first derivative \(f^\prime\left( x \right)\) and find the critical points of the function. (Remember that critical points are the points where the first derivative is zero or does not exist.) Determine the intervals where the function is increasing and decreasing using the First Derivative Test.

    \(6.\) Local Maximum and Minimum

    Use the First or Second Derivative Test to classify the critical points as local maximum or local minimum. Calculate the \(y-\)values of the local extrema points.

    \(7.\) Concavity/Convexity and Points of Inflection

    Using the Second Derivative Test, find the points of inflection (at which \(f^{\prime\prime}\left( x \right) = 0\)). Determine the intervals where the function is convex upward \(\left({f^{\prime\prime}\left( x \right) \lt 0}\right)\) and convex downward \(\left({f^{\prime\prime}\left( x \right) \gt 0}\right).\)

    \(8.\) Graph of the Function

    Sketch a graph of \(f\left( x \right)\) using all the information obtained above.

    Further we use this algorithm for the investigation of functions.


  • Solved Problems

    Click a problem to see the solution.

    Sketch graphs of the following functions (Examples \({1-23}\)).

    Example 1

    \[y = {x^3} – 3{x^2} + 2x.\]

    Example 2

    \[f\left( x \right) = {x^2}\left( {x + 3} \right).\]

    Example 3

    \[y = {\left( {x + 2} \right)^2}\left( {x – 1} \right).\]

    Example 4

    \[y = \frac{1}{{1 + {x^2}}}.\]

    Example 5

    \[f\left( x \right) = \frac{{{x^2} + 1}}{{x – 1}}.\]

    Example 6

    \[f\left( x \right) = \frac{1}{{1 – {x^2}}}.\]

    Example 7

    \[f\left( x \right) = {x^2} + \frac{1}{x}.\]

    Example 8

    \[y = {x^3}{e^x}.\]

    Example 9

    \[f\left( x \right) = \frac{{{e^x}}}{x}.\]

    Example 10

    \[y = {x^2}{e^{\large\frac{1}{x}\normalsize}}.\]

    Example 11

    \[f\left( x \right) = x\ln x.\]

    Example 12

    \[y = \frac{{{x^2} – 1}}{{{x^3}}}.\]

    Example 13

    \[y = \sqrt[\large 3\normalsize]{{{x^2}\left( {x + 1} \right)}}.\]

    Example 14

    \[y = {x^2}\sqrt {x + 1} .\]

    Example 15

    \[f\left( x \right) = \frac{x}{{{x^2} – 1}}.\]

    Example 16

    \[y = \frac{{{{\left( {x – 1} \right)}^3}}}{{{x^2}}}.\]

    Example 17

    \[y = \sqrt[\large 3\normalsize]{{{x^3} – x}}.\]

    Example 18

    \[y = x – \sqrt {{x^2} – 1} .\]

    Example 19

    \[y = {x^4} – 2{x^2} – 1.\]

    Example 20

    \[y = \sin x\sin 2x.\]

    Example 21

    \[f\left( x \right) = 2\arctan x + x.\]

    Example 22

    \[f\left( x \right) = \ln \left( {{x^2} – 4x + 5} \right).\]

    Example 23

    Draw the curve given by the parametric equations \[{x = {t^3} + {t^2} – t,}\;\;\;\kern-0.3pt{y = {t^3} + 2{t^2} – 4t.}\]

    Example 1.

    \[y = {x^3} – 3{x^2} + 2x.\]

    Solution.

    The function is defined for all \(x \in \mathbb{R}.\) Consequently, this function has no vertical asymptotes. Check for oblique (slant) asymptotes. Calculate the slope of the asymptote:

    \[
    {k = \lim\limits_{x \to \pm \infty } \frac{{y\left( x \right)}}{x} }
    = {\lim\limits_{x \to \pm \infty } \frac{{{x^3} – 3{x^2} + 2x}}{x} }
    = {\lim\limits_{x \to \pm \infty } \left( {{x^2} – 3x + 2} \right) = + \infty .}
    \]

    This indicates that the function has also no oblique asymptotes.

    Determine the points of intersection of the graph with the coordinate axes:

    \[y\left( 0 \right) = 0.\]

    Next, solving the equation

    \[{x^3} – 3{x^2} + 2x = 0,\]

    we find:

    \[
    {x\left( {{x^2} – 3x + 2} \right) = 0,\;\;}\Rightarrow
    {{x_1} = 0,\;{x_2} = 1,\;{x_3} = 2,}
    \]

    that is the function has three real roots.

    The intervals where the function is positive or negative can be found solving the following inequality (Figure \(1a\)):

    \[
    {{x^3} – 3{x^2} + 2x > 0,\;\;}\Rightarrow
    {x\left( {x – 1} \right)\left( {x – 2} \right) > 0.}
    \]

    The first derivative of the function is

    \[
    {y’\left( x \right) = {\left( {{x^3} – 3{x^2} + 2x} \right)^\prime } }
    = {3{x^2} – 6x + 2.}
    \]

    We find the stationary points by setting the first derivative equal to zero:

    \[
    {y’\left( x \right) = 0,\;\;}\Rightarrow
    {3{x^2} – 6x + 2 = 0,\;\;}\Rightarrow
    {D = 36 – 4 \cdot 3 \cdot 2 = 12,\;\;}\Rightarrow
    {{x_{1,2}} = \frac{{6 \pm \sqrt {12} }}{6}} = {1 \pm \sqrt 3 \approx 0,42;\;1,58.}
    \]

    When passing through the point \(x = 1 – {\large\frac{{\sqrt 3 }}{3}\normalsize},\) the derivative changes sign from plus to minus (Figure \(1a\)). Hence, this point is the maximum point. Similarly, we establish that \(x = 1 + {\large\frac{{\sqrt 3 }}{3}\normalsize}\) is the minimum point. Calculate the approximate value of the function at the points of maximum and minimum:

    \[\require{cancel} {y\left( {1 – \frac{{\sqrt 3 }}{3}} \right) } = {{\left( {1 – \frac{{\sqrt 3 }}{3}} \right)^3} } – {3{\left( {1 – \frac{{\sqrt 3 }}{3}} \right)^2} } + {2\left( {1 – \frac{{\sqrt 3 }}{3}} \right) } = {1 – 3 \cdot \frac{{\sqrt 3 }}{3} } + {3 \cdot {\left( {\frac{{\sqrt 3 }}{3}} \right)^2} } – {{\left( {\frac{{\sqrt 3 }}{3}} \right)^3} } – {3\left[ {1 – \frac{{2\sqrt 3 }}{3} } + {{{\left( {\frac{{\sqrt 3 }}{3}} \right)}^2}} \right] } + {2 – \frac{{2\sqrt 3 }}{3} } = {\cancel{1} – \sqrt 3 + \cancel{1} } – {\frac{{\sqrt 3 }}{9} – \cancel{3} } + {2\sqrt 3 – \cancel{1} + \cancel{2} } – {\frac{{2\sqrt 3 }}{3} } = {\frac{{9\sqrt 3 – \sqrt 3 – 6\sqrt 3 }}{9} } = {\frac{{2\sqrt 3 }}{9} \approx 0,38;} \]

    Similarly, we find that

    \[
    {y\left( {1 + \frac{{\sqrt 3 }}{3}} \right) }
    = -{\frac{{2\sqrt 3 }}{9} \approx -0,38.}
    \]

    Thus, the function has a local maximum at the point

    \[\left( {1 – \frac{{\sqrt 3 }}{3}, \frac{{2\sqrt 3 }}{9}} \right) \approx \left( {0,42;\;0,38} \right).\]

    Respectively, the local minimum is reached at the point

    \[\left( {1 + \frac{{\sqrt 3 }}{3},-\frac{{2\sqrt 3 }}{9}} \right) \approx \left( {1,58;\;-0,38} \right)\]

    The intervals of increasing/decreasing follow from Figure \(1a.\)

    Consider the second derivative:

    \[
    {y^{\prime\prime}\left( x \right) = {\left( {3{x^2} – 6x + 2} \right)^\prime } }
    = {6x – 6;}
    \]

    \[
    {y^{\prime\prime}\left( x \right) = 0,\;\;}\Rightarrow
    {6x – 6 = 0,\;\;}\Rightarrow
    {x = 1.}
    \]

    If \(x \le 1\), the function is convex upward, and if \(x \ge 1\), it is convex downward. Hence, the point \(x = 1\) is a point of inflection. At this point we have:

    \[y\left( 1 \right) = {1^3} – 3 \cdot {1^2} + 2 \cdot 1 = 0.\]

    Given these results, we can draw a schematic graph of the function (see Figure \(1b\)).

    Signs of the derivative of the cubic function y=x^3-3x^2+2x
    Figure 1a.
    A schematic view of the function y=x^3-3x^2+2x
    Figure 1b.

    Example 2.

    \[f\left( x \right) = {x^2}\left( {x + 3} \right).\]

    Solution.

    The function is defined for all \(x \in \mathbb{R}.\) It has the following \(x-\)intercepts:

    \[{f\left( x \right) = 0,}\;\; \Rightarrow {{x^2}\left( {x + 3} \right) = 0,}\;\; \Rightarrow {{x_1} = 0,{x_2} = – 3.}\]

    The \(y-\)intercept is equal to

    \[f\left( 0 \right) = 0.\]

    The function is positive on the intervals \(\left( { – 3,0} \right)\) and \(\left( {0,+\infty} \right)\) and negative on \(\left( { -\infty,-3} \right).\)

    The function is neither even nor odd, and it has no asymptotes.

    Take the derivative:

    \[{f^\prime\left( x \right) }={ \left( {{x^2}\left( {x + 3} \right)} \right)^\prime }={ \left( {{x^3} + 3{x^2}} \right)^\prime }={ 3{x^2} + 6x.}\]

    Find the critical points:

    \[{f^\prime\left( x \right) = 0,}\;\; \Rightarrow {3{x^2} + 6x = 0,}\;\; \Rightarrow {3x\left( {x + 2} \right) = 0,}\;\; \Rightarrow {{x_1} = 0,{x_2} = – 2.}\]

    Sign chart of the derivative of the function y=x^2(x+3)
    Figure 2a.
    Graph of the function f(x)=x^2(x+3)
    Figure 2b.

    We can see from the sign chart that \(x = -2\) is a point of maximum, and \(x = 0\) is a point of minimum. The \(y-\)values of these points are

    \[{f\left( { – 2} \right) }={ {\left( { – 2} \right)^2}\left( { – 2 + 3} \right) }={ 4;}\]

    \[f\left( 0 \right) = 0.\]

    We differentiate once more to get the second derivative:

    \[{f^{\prime\prime}\left( x \right) }={ \left( {3{x^2} + 6x} \right)^\prime }={ 6x + 6.}\]

    \[{f^{\prime\prime}\left( x \right) = 0,}\;\; \Rightarrow {6x + 6 = 0,}\;\; \Rightarrow {x = – 1.}\]

    The graph of the function is concave downward on \(\left( { – \infty , – 1} \right)\) and concave upward on \(\left( { – 1, + \infty } \right).\) Therefore, \(x = -1\) is a point of inflection. The \(y-\)coordinate of this point is

    \[{f\left( { – 1} \right) }={ {\left( { – 1} \right)^2}\left( { – 1 + 3} \right) }={ 2.}\]

    Given these results, we can draw a schematic graph of the function (Figure \(2\)b).

    Example 3.

    \[y = {\left( {x + 2} \right)^2}\left( {x – 1} \right).\]

    Solution.

    The function is defined for all real \(x.\) Therefore, it has no vertical asymptotes. Check for oblique (slant) asymptotes:

    \[
    {k = \lim\limits_{x \to \pm \infty } \frac{{y\left( x \right)}}{x} }
    = {\lim\limits_{x \to \pm \infty } \frac{{{{\left( {x + 2} \right)}^2}\left( {x – 1} \right)}}{x} }
    = {\lim\limits_{x \to \pm \infty } \frac{{\left( {{x^2} + 4x + 4} \right)\left( {x – 1} \right)}}{x} }
    = {\lim\limits_{x \to \pm \infty } \frac{{{x^3} + {3{x^2}} – {4}}}{x} }
    = {\lim\limits_{x \to \pm \infty } \left( {{x^2} + 3x – \frac{4}{x}} \right) = + \infty .}
    \]

    Since the slope \(k\) is infinity, the function has no oblique asymptotes, too.

    Find the points of intersection of the graph with the coordinate axes:

    \[y\left( 0 \right) = {2^2} \cdot \left( { – 1} \right) = – 4;\]

    \[
    {y\left( x \right) = 0,\;\;}\Rightarrow
    {{\left( {x + 2} \right)^2}\left( {x – 1} \right) = 0,\;\;}\Rightarrow
    {{x_1} = – 2,\;{x_2} = 1.}
    \]

    The function is positive for \(x \gt 1\) and negative for \(x \in \left( { – \infty , – 2} \right) \cup \left( { – 2,1} \right)\) (Figure \(3a\)).

    Calculate the first derivative:

    \[\require{cancel}
    {y’\left( x \right) = {\left[ {{{\left( {x + 2} \right)}^2}\left( {x – 1} \right)} \right]^\prime } }
    = {2\left( {x + 2} \right)\left( {x – 1} \right) + {\left( {x + 2} \right)^2} }
    = {\left( {x + 2} \right)\left( {2x – \cancel{2} + x + \cancel{2}} \right) }
    = {3x\left( {x + 2} \right).}
    \]

    The stationary points are

    \[
    {y’\left( x \right) = 0,\;\;}\Rightarrow
    {3x\left( {x + 2} \right) = 0,\;\;}\Rightarrow
    {{x_1} = 0,\;{x_2} = – 2.}
    \]

    The derivative changes its sign as shown in Figure \(3a.\) Therefore, \(x = -2\) is the maximum point, and \(x = 0\) is the minimum point. The function has the following values at these extrema points:

    \[
    {y\left( { – 2} \right) = – 4,}\;\;\;\kern-0.3pt
    {y\left( 0 \right) = 0.}
    \]

    Find the second derivative:

    \[
    {y^{\prime\prime}\left( x \right) = {\left[ {3x\left( {x + 2} \right)} \right]^\prime } }
    = {3\left( {x + 2} \right) + 3x }
    = {6x + 6.}
    \]

    So the function is strictly convex upward for \(x \lt -1\) and strictly convex downward for \(x \gt -1.\) Hence, \(x = -1\) is the inflection point, and

    \[y\left( { – 1} \right) = {\left( { – 1 + 2} \right)^2}\left( { – 1 – 1} \right) = – 2.\]

    As a result, we can draw a graph of the function. The shape of the graph resembles a cubic parabola (Figure \(3b\)).

    Signs of the derivatives of the function y=(x+2)^2*(x-1)
    Figure 3a.
    A schematic view of the function y=(x+2)^2*(x-1)
    Figure 3b.

    Example 4.

    \[y = \frac{1}{{1 + {x^2}}}.\]

    Solution.

    The function is defined for all real values of \(x.\) Consequently, it has no vertical asymptotes. Since

    \[
    {\lim\limits_{x \to \pm \infty } y\left( x \right) }
    = {\lim\limits_{x \to \pm \infty } \frac{1}{{1 + {x^2}}} = 0,}
    \]

    then the graph of the function has horizontal asymptote \(y = 0,\) that is the \(x\)-axis is the horizontal asymptote.

    This function is even. Indeed,

    \[
    {y\left( { – x} \right) = \frac{1}{{1 + {{\left( { – 1} \right)}^2}}} }
    = {\frac{1}{{1 + {x^2}}} = y\left( x \right).}
    \]

    It is obvious that the function has no roots and positive for all \(x.\) At the point \(x = 0,\) its value is

    \[y\left( 0 \right) = \frac{1}{{1 + {0^2}}} = 1.\]

    Find the first derivative:

    \[
    {y’\left( x \right) = {\left( {\frac{1}{{1 + {x^2}}}} \right)^\prime } }
    = { – \frac{1}{{{{\left( {1 + {x^2}} \right)}^2}}} \cdot {\left( {1 + {x^2}} \right)^\prime } }
    = { – \frac{{2x}}{{{{\left( {1 + {x^2}} \right)}^2}}}.}
    \]

    This shows that \(x = 0\) is a stationary point. When passing through this point the derivative changes sign from plus to minus (Figure \(4a\)). Therefore, we have a maximum at \(x = 0.\) Its value is \(y\left( 0 \right) = 1.\)

    Calculate the second derivative:

    \[
    {y^{\prime\prime}\left( x \right) = {\left( { – \frac{{2x}}{{{{\left( {1 + {x^2}} \right)}^2}}}} \right)^\prime } }
    = { – \frac{{2{{\left( {1 + {x^2}} \right)}^2} – 2x \cdot 2\left( {1 + {x^2}} \right)}}{{{{\left( {1 + {x^2}} \right)}^4}}} }
    = {\frac{{8{x^2} – 2 – 2{x^2}}}{{{{\left( {1 + {x^2}} \right)}^3}}} }
    = {\frac{{6{x^2} – 2}}{{{{\left( {1 + {x^2}} \right)}^3}}}.}
    \]

    It is equal to zero at the following points:

    \[
    {y^{\prime\prime}\left( x \right) = 0,\;\;}\Rightarrow
    {\frac{{6{x^2} – 2}}{{{{\left( {1 + {x^2}} \right)}^3}}} = 0,\;\;}\Rightarrow
    {\frac{{2\left( {x – \sqrt 3 } \right)\left( {x + \sqrt 3 } \right)}}{{{{\left( {1 + {x^2}} \right)}^3}}} = 0,\;\;}\Rightarrow
    {{x_1} = – \sqrt 3 ,\;{x_2} = \sqrt 3 .}
    \]

    When passing through these points the second derivative changes its sign. Therefore, both points are inflection points. The function is strictly convex downward in the intervals \(\left( { – \infty , – \sqrt 3 } \right)\) and \(\left( {\sqrt 3 , + \infty } \right)\) and, accordingly, strictly convex upward in the interval \(\left( { – \sqrt 3 ,\sqrt 3 } \right).\) Since the function is even, the found inflection points have the same values of \(y:\)

    \[
    {y\left( { \pm \sqrt 3 } \right) = \frac{1}{{1 + {{\left( { \pm \sqrt 3 } \right)}^2}}} }
    = {\frac{1}{{1 + 3}} = \frac{1}{4}.}
    \]

    Figure \(4b\) presents a schematic graph of the function.

    Signs of the derivatives of the rational function y=1/(1+x^2)
    Figure 4a.
    Graph of the rational function y=1/(1+x^2)
    Figure 4b.

    Example 5.

    \[f\left( x \right) = \frac{{{x^2} + 1}}{{x – 1}}.\]

    Solution.

    The function is defined for all \(x\) except the point \(x = 1\) where it has a discontinuity.

    Find the \(y-\)intercept:

    \[{f\left( 0 \right) = \frac{{{0^2} + 1}}{{0 – 1}} }={ – 1.}\]

    The function is negative for \(x \lt 1\) and positive for \(x \gt 1,\) but it has no \(x-\)intercepts.

    Look for vertical asymptote near \(x = 1:\)

    \[{\mathop {\lim }\limits_{x \to 1 – 0} f\left( x \right) }={ \mathop {\lim }\limits_{x \to 1 – 0} \frac{{{x^2} + 1}}{{x – 1}} }={ – \infty ;}\]

    \[{\mathop {\lim }\limits_{x \to 1 + 0} f\left( x \right) }={ \mathop {\lim }\limits_{x \to 1 + 0} \frac{{{x^2} + 1}}{{x – 1}} }={ \infty .}\]

    There is a vertical asymptote at \(x = 1.\)

    Rewrite the function in the form

    \[{f\left( x \right) = \frac{{{x^2} + 1}}{{x – 1}} }={ \frac{{{x^2} – x + x – 1 + 2}}{{x – 1}} }={ \frac{{x\left( {x – 1} \right) + x – 1 + 2}}{{x – 1}} }={ x + 1 + \frac{2}{{x – 1}},}\]

    where \(\large{\frac{2}{{x – 1}}}\normalsize \to 0\) as \(x \to \infty.\) Hence the function has an oblique asymptote \(y = x + 1.\)

    Take the first derivative:

    \[{f^\prime\left( x \right) }= {\left( {\frac{{{x^2} + 1}}{{x – 1}}} \right)^\prime }={ \frac{{2x \cdot \left( {x – 1} \right) – \left( {{x^2} + 1} \right) \cdot \left( {x – 1} \right)}}{{{{\left( {x – 1} \right)}^2}}} }={ \frac{{2{x^2} – 2x – {x^2} – 1}}{{{{\left( {x – 1} \right)}^2}}} }={ \frac{{{x^2} – 2x – 1}}{{{{\left( {x – 1} \right)}^2}}}.}\]

    Determine the critical points:

    \[{f^\prime\left( x \right) = 0,}\;\; \Rightarrow {\frac{{{x^2} – 2x – 1}}{{{{\left( {x – 1} \right)}^2}}} = 0,}\;\; \Rightarrow {\left\{ {\begin{array}{*{20}{l}} {{x^2} – 2x – 1 = 0}\\ {x \ne 1} \end{array}} \right..}\]

    Solve the quadratic equation:

    \[{{x^2} – 2x – 1 = 0,}\;\; \Rightarrow {D = {\left( { – 2} \right)^2} – 4 \cdot \left( { – 1} \right) = 8,};\; \Rightarrow {{x_{1,2}} = \frac{{2 \pm \sqrt 8 }}{2} }={ 1 \pm \sqrt 2 .}\]

    Thus, the function has two critical points: \({x_1} = 1 – \sqrt 2 \approx – 0.41\) and \({x_2} = 1 + \sqrt 2 \approx 2.41\)

    The second derivative is written as

    \[{f^{\prime\prime}\left( x \right) }={ \left( {\frac{{{x^2} – 2x – 1}}{{{{\left( {x – 1} \right)}^2}}}} \right)^\prime }={ \frac{4}{{{{\left( {x – 1} \right)}^3}}}.}\]

    We see that the function is concave downward at \(x \lt 1\) and concave upward at \(x \gt 1,\) though it has no inflection points.

    Draw a sign chart for the function and its derivatives (Figure \(5a\)).

    Sign chart for the function f(x)=(x^2+1)/(x-1)
    Figure 5a.
    Graph of the function f(x)=(x^2+1)/(x-1)
    Figure 5b.

    The point \(x = 1 – \sqrt 2\) is a local maximum, and the point \(x = 1 + \sqrt 2\) is a local minimum. Calculate their \(y-\)coordinates:

    \[\require{cancel}{f\left( {1 – \sqrt 2 } \right) }={ \frac{{{{\left( {1 – \sqrt 2 } \right)}^2} + 1}}{{\cancel{1} – \sqrt 2 – \cancel{1}}} }={ \frac{{1 – 2\sqrt 2 + 2 + 1}}{{ – \sqrt 2 }} }={ \frac{{4 – 2\sqrt 2 }}{{ – \sqrt 2 }} }={ \frac{{4 – 4\sqrt 2 }}{2} }={ 2\left( {1 – \sqrt 2 } \right) \approx {- 0.83}}\] \[{f\left( {1 + \sqrt 2 } \right) }={ \frac{{{{\left( {1 + \sqrt 2 } \right)}^2} + 1}}{{\cancel{1} + \sqrt 2 – \cancel{1}}} }={ \frac{{1 + 2\sqrt 2 + 2 + 1}}{{ \sqrt 2 }} }={ \frac{{4 + 2\sqrt 2 }}{{\sqrt 2 }} }={ \frac{{4 + 4\sqrt 2 }}{2} }={ 2\left( {1 + \sqrt 2 } \right) \approx {4.83}}\]

    Now we can sketch a graph of the function (Figure \(5b\)).

    Page 1
    Problems 1-5
    Page 2
    Problems 6-14
    Page 3
    Problems 15-23