Calculus

Applications of the Derivative

Curve Sketching

Page 1
Problem 1
Page 2
Problems 2-8
Page 3
Problems 9-14

The general algorithm for sketching graph of a function \(y = f\left( x \right)\) looks as follows:

  1. Find the domain of the function, determine the points of discontinuity and vertical asymptotes (if they exist).
  2. Determine whether the function is even, odd, or neither, check the periodicity of the function.
  3. Determine oblique and horizontal asymptotes of the graph.
  4. Find the points of intersection of the graph with the coordinate axes and the intervals where the function has a constant sign \(\left(f\left( x \right) \gt 0\right.\) and \(\left.f\left( x \right) \lt 0\right)\).
  5. Calculate the first derivative \(f’\left( x \right)\), find the points of extrema and intervals of increasing /decreasing.
  6. Calculate the second derivative \(f^{\prime\prime}\left( x \right)\), find the points of inflection and intervals of concavity upward /downward.
  7. Draw a graph of the function.

We apply this algorithm for the investigation and plotting of different functions.

Solved Problems

Click on problem description to see solution.

Sketch graphs of the following functions (Examples \(1-15\)).

 Example 1

\[y = {x^3} – 3{x^2} + 2x.\]

 Example 2

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

 Example 3

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

 Example 4

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

 Example 5

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

 Example 6

\[y = \frac{{{x^2} – 1}}{{{x^3}}}.\]

 Example 7

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

 Example 8

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

 Example 9

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

 Example 10

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

 Example 11

\[y = x – \sqrt {{x^2} – 1} .\]

 Example 12

\[y = {x^4} – 2{x^2} – 1.\]

 Example 13

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

 Example 14

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:
\[
{x\left( {{x^2} – 3x + 2} \right) = 0,\;\;}\Rightarrow
{{x_1} = 0,\;{x_2} = 1,\;{x_3} = 2,}
\] i.e. 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 increase/decrease 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 derivatives of the cubic function x^3-3x^2+2x

Figure 1a.

A schematic view of the cubic function x^3-3x^2+2x

Figure 1b.

Page 1
Problem 1
Page 2
Problems 2-8
Page 3
Problems 9-14