Calculus

Applications of the Derivative

Applications of Derivative Logo

Curve Sketching

Solved Problems

Sketch graphs of the following functions (Examples 15−23).

Example 15.

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

Solution.

Find the domain of the function:

\[{x^2} - 1 \ne 0,\;\; \Rightarrow {x^2} \ne 1,\;\; \Rightarrow x \ne \pm 1.\]

Calculate the roots:

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

Hence the function intersects the origin.

The function is odd:

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

Make sure that it has vertical asymptotes at \(x = \pm 1:\)

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

Check for horizontal asymptotes:

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

Consequently, \(y = 0\) (\(x-\)axis) is a horizontal asymptote. Clearly that the function has no oblique asymptotes (as it already has two-directional horizontal one).

Take the derivative:

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

We see that \(f^\prime\left( x \right) \lt 0\) for all \(x,\) that is the function is decreasing everywhere in its domain.

Determine the second derivative:

\[f^{\prime\prime}\left( x \right) = \left( { - \frac{{{x^2} + 1}}{{{{\left( {{x^2} - 1} \right)}^2}}}} \right)^\prime = \frac{{2{x^3} + 6x}}{{{{\left( {{x^2} - 1} \right)}^3}}}.\]

Find the points at which the second derivative is zero:

\[f^{\prime\prime}\left( x \right) = 0,\;\; \Rightarrow \frac{{2{x^3} + 6x}}{{{{\left( {{x^2} - 1} \right)}^3}}} = 0,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {2x\left( {{x^2} + 3} \right) = 0}\\ {x \ne \pm 1} \end{array}} \right.,\;\; \Rightarrow x = 0.\]

As the concavity changes sign around \(x = 0\) (see the sign chart), this point is a point of inflection.

Sign chart for the derivatives of function f(x) = x/(x^2-1).
Figure 15a.
Graph of the function f(x)=x/(x^2-1).
Figure 15b.

Now we can draw a schematic view of this rational function (Figure \(15b\)).

Example 16.

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

Solution.

The domain of the function is \(x \ne 0.\) The function has a discontinuity at \(x = 0.\) By calculating one-sided limits, we obtain:

\[\lim\limits_{x \to 0 - 0} y\left( x \right) = \lim\limits_{x \to 0 - 0} \frac{{{{\left( {x - 1} \right)}^3}}}{{{x^2}}} = - \infty ,\]
\[\lim\limits_{x \to 0 + 0} y\left( x \right) = \lim\limits_{x \to 0 + 0} \frac{{{{\left( {x - 1} \right)}^3}}}{{{x^2}}} = - \infty .\]

It follows that \(x = 0\) is a vertical asymptote.

Check for slant asymptotes:

\[\require{cancel} k = \lim\limits_{x \to \pm \infty } \frac{{y\left( x \right)}}{x} = \lim\limits_{x \to \pm \infty } \frac{{{{\left( {x - 1} \right)}^3}}}{{{x^3}}} = \lim\limits_{x \to \pm \infty } {\left( {1 - \frac{1}{x}} \right)^3} = 1;\]
\[b = \lim\limits_{x \to \pm \infty } \left[ {y\left( x \right) - kx} \right] = \lim\limits_{x \to \pm \infty } \left[ {\frac{{{{\left( {x - 1} \right)}^3}}}{{{x^2}}} - x} \right] = \lim\limits_{x \to \pm \infty } \frac{{\cancel{x^3} - 3{x^2} + 3x - 1 - \cancel{x^3}}}{{{x^2}}} = \lim\limits_{x \to \pm \infty } \left( { - 3 + \frac{3}{x} - \frac{1}{{{x^2}}}} \right) = - 3.\]

Consequently, there is an oblique asymptote given by the equation \(y = x - 3.\)

The graph of the function intersects the \(x\)-axis at \(x = 1.\) The function is positive for \(x \gt 1\) and negative for \(x \lt 1\) (except the point \(x = 0,\) where the function is not defined).

The first derivative is expressed by the following formula:

\[y'\left( x \right) = {\left( {\frac{{{{\left( {x - 1} \right)}^3}}}{{{x^2}}}} \right)^\prime } = \frac{{{{\left( {{{\left( {x - 1} \right)}^3}} \right)}^\prime }{x^2} - {{\left( {x - 1} \right)}^3}{{\left( {{x^2}} \right)}^\prime }}}{{{x^4}}} = \frac{{3{{\left( {x - 1} \right)}^2}{x^2} - 2x{{\left( {x - 1} \right)}^3}}}{{{x^4}}} = \frac{{{{\left( {x - 1} \right)}^2}\left( {x + 2} \right)}}{{{x^3}}}.\]

So the critical points are

\[{x_1} = 1,\;\;{x_2} = - 2,\;\;{x_3} = 0.\]

The function strictly decreases on the interval \( \left({- 2,0}\right)\) and strictly increases on the intervals \( \left({-\infty, -2}\right),\) \( \left({0, 1}\right)\) and \( \left({1, +\infty}\right)\) (Figure \(16a\)).

At \(x = -2,\) the function has a maximum equal

\[y\left( { - 2} \right) = \frac{{{{\left( { - 2 - 1} \right)}^3}}}{{{2^2}}} = - \frac{{27}}{4} = - 6,75.\]

There is no extreme point at \(x = 1\), because when passing through this point the sign of the derivative does not change.

Compute the second derivative:

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

It is seen that \(y^{\prime\prime} = 0\) at \(x = 1,\) and \(y^{\prime\prime} \lt 0\) to the left of this point and \(y^{\prime\prime} \gt 0\) to the right. Thus, the function is convex upward on the intervals \(\left( { - \infty ,0} \right),\) \(\left( { 0, 1} \right)\) and is convex downward on the interval \(\left( {1, +\infty} \right).\) The point \(x = 1\) is an inflection point and \(y\left( 1 \right) = 0.\)

A schematic graph of the function is given in Figure \(16b\).

Signs of the derivatives of the rational function y=(x-1)^3/x^2.
Figure 16a.
Graph of the rational function y=(x-1)^3/x^2.
Figure 16b.

Example 17.

\[y = \sqrt[3]{{{x^3} - x}}.\]

Solution.

The function is defined for all real \(x.\) Therefore it has no vertical asymptotes. It is easy to show that this function is odd. Indeed:

\[y\left( { - x} \right) = \sqrt[3]{{{{\left( { - x} \right)}^3} - \left( { - x} \right)}} = \sqrt[3]{{ - \left( {{x^3} - x} \right)}} = - \sqrt[3]{{{x^3} - x}} = - y\left( x \right).\]

Check for oblique (slant) asymptotes by calculating the following limits:

\[k = \lim\limits_{x \to + \infty } \frac{{y\left( x \right)}}{x} = \lim\limits_{x \to + \infty } \frac{{\sqrt[3]{{{x^3} - x}}}}{x} = \lim\limits_{x \to + \infty } \sqrt[3]{{\frac{{{x^3} - x}}{{{x^3}}}}} = \lim\limits_{x \to + \infty } \sqrt[3]{{1 - \frac{1}{{{x^2}}}}} = 1;\]
\[b = \lim\limits_{x \to + \infty } \left[ {y\left( x \right) - kx} \right] = \lim\limits_{x \to + \infty } \left( {\sqrt[3]{{{x^3} - x}} - x} \right) = \lim\limits_{x \to + \infty } \left( {\sqrt[3]{{{x^3} - x}} - \sqrt[3]{{{x^3}}}} \right) = 0.\]

Hence, the graph of the function has an oblique asymptote as \(x \to +\infty\) given by the equation \(y = x.\) By virtue of the oddness of the function, this asymptote also exists as \(x \to -\infty.\)

Determine the points of intersection of the graph with the coordinate axes and intervals, in which the function has a constant sign.

\[y\left( 0 \right) = \sqrt[3]{{{0^3} - 0}} = 0;\]
\[y\left( x \right) = 0,\;\; \Rightarrow \sqrt[3]{{{x^3} - x}} = 0,\;\; \Rightarrow {x^3} - x = 0,\;\; \Rightarrow x\left( {{x^2} - 1} \right) = 0,\;\; \Rightarrow {x_1} = 0,\;{x_2} = - 1,\;{x_3} = 1.\]

The inequality \(y\left( x \right) \gt 0\) can be written as

\[\sqrt[3]{{{x^3} - x}} \gt 0,\;\; \Rightarrow {x^3} - x \gt 0,\;\; \Rightarrow x\left( {x - 1} \right)\left( {x + 1} \right) \gt 0.\]

We can solve it using the method of intervals (Figure \(17a\)). It can be seen that the function is positive on the intervals \(\left( { - 1,0} \right)\) and \(\left( { 1, +\infty} \right)\) and, respectively, negative on the intervals \(\left( { -\infty, -1} \right)\) and \(\left( {0,1} \right).\)

Find the derivative:

\[y'\left( x \right) = \left( {\sqrt[3]{{{x^3} - x}}} \right)^\prime = \left( {{{\left( {{x^3} - x} \right)}^{\frac{1}{3}}}} \right)^\prime = \frac{1}{3}{\left( {{x^3} - x} \right)^{ - \frac{2}{3}}} \cdot \left( {3{x^2} - 1} \right) = \frac{{3{x^2} - 1}}{{3\sqrt[3]{{{{\left( {{x^3} - x} \right)}^2}}}}}.\]

The derivative does not exist at \(x = 0\) and \(x = \pm 1\) and is zero at the following points:

\[ y'\left( x \right) = 0,\;\; \Rightarrow \frac{{3{x^2} - 1}}{{3\sqrt[3]{{{{\left( {{x^3} - x} \right)}^2}}}}} = 0,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {3{x^2} - 1 = 0}\\ {3\sqrt[3]{{{{\left( {{x^3} - x} \right)}^2}}} \ne 0} \end{array}} \right.,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {{x^2} = \frac{1}{3}}\\ {{x^3} - x \ne 0} \end{array}} \right.,\;\; \Rightarrow x = \pm \frac{1}{{\sqrt 3 }} \approx \pm 0,58.\]

Thus, the function has \(5\) critical points. We explore the sign of the derivative when passing through these points (Figure \(17a\)). As a result, we conclude that \(x = - {\frac{1}{{\sqrt 3 }}}\) is a maximum point and \(x = {\frac{1}{{\sqrt 3 }}}\) is a minimum point (which is quite natural because the function is odd). The maximum and minimum values are

\[y\left( { - \frac{1}{{\sqrt 3 }}} \right) = \sqrt[3]{{{{\left( { - \frac{1}{{\sqrt 3 }}} \right)}^3} - \left( { - \frac{1}{{\sqrt 3 }}} \right)}} = - \sqrt[[3]{{\frac{1}{{3\sqrt 3 }} - \frac{1}{{\sqrt 3 }}}} = - \sqrt[3]{{\frac{{1 - 3}}{{3\sqrt 3 }}}} = - \sqrt[3]{{\frac{2}{{3\sqrt 3 }}}} \approx -0,73;\;\; \Rightarrow y\left( {\frac{1}{{\sqrt 3 }}} \right) = 0,73.\]

Calculate the second derivative:

\[y^{\prime\prime}\left( x \right) = \left[ {\frac{1}{3}{{\left( {{x^3} - x} \right)}^{ - \frac{2}{3}}}\left( {3{x^2} - 1} \right)} \right]^\prime = \frac{2}{3}{\left( {{x^3} - x} \right)^{ - \frac{5}{3}}} \cdot \left[ {9x\left( {{x^3} - x} \right) - {{\left( {3{x^2} - 1} \right)}^2}} \right] = \frac{2}{{9\sqrt[3]{{{{\left( {{x^3} - x} \right)}^5}}}}} \cdot \left[ {\cancel{9{x^4}} - 9{x^2} - \cancel{9{x^4}} + 6{x^2} - 1} \right] = - \frac{{6{x^2} + 2}}{{3\sqrt[3]{{{{\left( {{x^3} - x} \right)}^5}}}}}.\]

The second derivative is nowhere equal to zero and (as well as the first derivative) does not exist at \(x = 0\) and \(x = \pm 1.\) When passing through these points the second derivative changes its sign (Figure \(17a\)). Therefore, these points are points of inflection. Their coordinates have been determined to be as follows: \(\left( {0,0} \right),\) \(\left( {-1,0} \right)\) and \(\left( {1,0} \right).\)

According to the analysis we plot a graph of the given function (Figure \(17b\)).

Signs of the derivatives of a cubic root function.
Figure 17a.
A schematic view of a cubic root function.
Figure 17b.

Example 18.

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

Solution.

First we determine the domain of the function:

\[{x^2} - 1 \ge 0,\;\; \Rightarrow \left( {x - 1} \right)\left( {x + 1} \right) \ge 0,\;\; \Rightarrow x \in \left( { - \infty , - 1} \right] \cup \left[ {1, + \infty } \right).\]

The function has the following values at the boundary points \(x = \pm 1:\)

\[y\left( { - 1} \right) = - 1 - \sqrt {{{\left( { - 1} \right)}^2} - 1} = - 1,\;\;y\left( 1 \right) = 1 - \sqrt {{1^2} - 1} = 1.\]

Consider the inequality \(y\left( x \right) \gt 0:\)

\[ x - \sqrt {{x^2} - 1} \gt 0,\;\; \Rightarrow x \gt \sqrt {{x^2} - 1} ,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {{x^2} \gt {x^2} - 1}\\ {x \gt 0}\\ {{x^2} - 1 \ge 0} \end{array}} \right.,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} { - 1 \lt 0}\\ {x \gt 0}\\ {{x^2} - 1 \ge 0} \end{array}} \right.,\;\; \Rightarrow x \ge 1.\]

Hence, the function is positive in the interval \(\left[ {1, + \infty } \right).\) Similarly, we can show that the function is negative in the interval \(\left( {-\infty, -1 } \right].\)

There are no vertical asymptotes for this function. Explore slant asymptotes (including horizontal ones):

\[{k_1} = \lim\limits_{x \to + \infty } \frac{{y\left( x \right)}}{x} = \lim\limits_{x \to + \infty } \frac{{x - \sqrt {{x^2} - 1} }}{x} = \lim\limits_{x \to + \infty } \left( {\frac{x}{x} - \sqrt {\frac{{{x^2} - 1}}{{{x^2}}}} } \right) = \lim\limits_{x \to + \infty } \left( {1 - \sqrt {1 - \frac{1}{{{x^2}}}} } \right) = 0;\]
\[{b_1} = \lim\limits_{x \to + \infty } \left[ {y\left( x \right) - {k_1}x} \right] = \lim\limits_{x \to + \infty } \left( {x - \sqrt {{x^2} - 1} } \right) = \lim\limits_{x \to + \infty } \frac{{\cancel{x^2} - \cancel{x^2} + 1}}{{x + \sqrt {{x^2} - 1} }} = 0;\]
\[ {k_2} = \lim\limits_{x \to - \infty } \frac{{y\left( x \right)}}{x} = \lim\limits_{x \to - \infty } \frac{{x - \sqrt {{x^2} - 1} }}{x} = \lim\limits_{x \to - \infty } \frac{{ - x + \sqrt {{{\left( { - x} \right)}^2} - 1} }}{{ - x}} = \left[ {\begin{array}{*{20}{l}} { - x = z,}\\ {x \to - \infty ,}\\ {z \to + \infty } \end{array}} \right] = \lim\limits_{z \to + \infty } \frac{{z + \sqrt {{z^2} - 1} }}{z} = \lim\limits_{z \to + \infty } \left( {\frac{z}{z} + \sqrt {\frac{{{z^2} - 1}}{{{z^2}}}} } \right) = \lim\limits_{z \to + \infty } \left( {1 + \sqrt {1 - \frac{1}{{{z^2}}}} } \right) = 2;\]
\[ {b_2} = \lim\limits_{x \to - \infty } \left[ {y\left( x \right) - {k_2}x} \right] = \lim\limits_{x \to - \infty } \left( {x - \sqrt {{x^2} - 1} - 2x} \right) = \lim\limits_{x \to - \infty } \left( { - x - \sqrt {{{\left( { - x} \right)}^2} - 1} } \right) = \left[ {\begin{array}{*{20}{l}} { - x = z,}\\ {x \to - \infty ,}\\ {z \to + \infty } \end{array}} \right] = \lim\limits_{z \to + \infty } \left( {z - \sqrt {{z^2} - 1} } \right) = \lim\limits_{z \to + \infty } \frac{{\cancel{z^2} - \cancel{z^2} + 1}}{{\left( {z + \sqrt {{z^2} - 1} } \right)}} = 0.\]

Thus, the graph of the function has two asymptotes: the horizontal asymptote \(y = 0\) (\(x\)-axis) as \(x \to +\infty\) and the slant asymptote \(y = 2x\) as \(x \to -\infty.\)

Calculate the derivative:

\[y'\left( x \right) = \left( {x - \sqrt {{x^2} - 1} } \right)^\prime = 1 - \frac{{\cancel{2}x}}{{\cancel{2}\sqrt {{x^2} - 1} }} = \frac{{\sqrt {{x^2} - 1} - x}}{{\sqrt {{x^2} - 1} }}.\]

The first derivative does not exist at \(x = \pm 1.\) The numerator of the derivative is nowhere equal to zero:

\[ x - \sqrt {{x^2} - 1} = 0,\;\; \Rightarrow \sqrt {{x^2} - 1} = x,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {{x^2} - 1 = {x^2}}\\ {x \ge 0}\\ {{x^2} - 1 \ge 0} \end{array}} \right.,\;\; \Rightarrow - 1 \ne 0,\;\; \Rightarrow x \in \emptyset .\]

Therefore, the function has no other critical points. The signs of the derivative are indicated in Figure \(18a,\) i.e. the function decreases on the interval \(\left( { - \infty , - 1} \right)\) and increases on the interval \(\left( {1, +\infty} \right).\)

We compute the second derivative and determine the concavity of the graph:

\[y^{\prime\prime}\left( x \right) = {\left( {1 - \frac{x}{{\sqrt {{x^2} - 1} }}} \right)^\prime } = - \frac{{1 \cdot \sqrt {{x^2} - 1} - x \cdot \frac{{\cancel{2}x}}{{\cancel{2}\sqrt {{x^2} - 1} }}}}{{{{\left( {\sqrt {{x^2} - 1} } \right)}^2}}} = \frac{{{x^2} - \left( {{x^2} - 1} \right)}}{{\sqrt {{{\left( {{x^2} - 1} \right)}^3}} }} = \frac{1}{{\sqrt {{{\left( {{x^2} - 1} \right)}^3}} }}.\]

As you can see, the second derivative is positive in the whole domain. Consequently, both (left and right) branches of the graph are convex downward. A view of the function is shown in Figure \(18b\).

Signs of the derivatives of the function y=x-sqrt(x^2-1).
Figure 18a.
Graph of the function y=x-sqrt(x^2-1).
Figure 18b.

Example 19.

\[y = {x^4} - 2{x^2} - 1.\]

Solution.

The function is defined on the entire real axis. It has no vertical asymptotes. Make sure that it has no oblique/horizontal asymptotes, too. Indeed,

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

The function is even since

\[y\left( { - x} \right) = {\left( { - x} \right)^4} - 2{\left( { - x} \right)^2} - 1 = {x^4} - 2{x^2} - 1 = y\left( x \right).\]

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

\[y\left( 0 \right) = {0^4} - 2 \cdot {0^2} - 1 = - 1;\]
\[y\left( x \right) = 0,\;\; \Rightarrow {x^4} - 2{x^2} - 1 = 0,\;\; \Rightarrow \left[ {{x^2} = z} \right],\;\; \Rightarrow {z^2} - 2z - 1 = 0,\;\; \Rightarrow D = 4 - 4 \cdot \left( { - 1} \right) = 8,\;\; \Rightarrow {z_{1,2}} = \frac{{2 \pm \sqrt 8 }}{2} = 1 \pm \sqrt 2 \approx 2,41;\; - 0,41.\]

The real solutions for \(x\) exist only at \(z = 1 + \sqrt 2.\) Then we get:

\[{x_{1,2}} = \pm \sqrt {1 + \sqrt 2 } \approx \pm 1,55.\]

It is clear that the function is positive on the intervals \(\left( { - \infty ; - 1,55} \right),\) \(\left( {1,55; +\infty} \right),\) and respectively, negative in the interval \(\left( { -1,55; + 1,55} \right)\) (Figure \(19a\)).

Find the first derivative and local extrema:

\[y'\left( x \right) = \left( {{x^4} - 2{x^2} - 1} \right)^\prime = 4{x^3} - 4x;\]
\[y'\left( x \right) = 0,\;\; \Rightarrow 4{x^3} - 4x = 0,\;\; \Rightarrow 4x\left( {x - 1} \right)\left( {x + 1} \right) = 0,\;\; \Rightarrow {x_1} = 0,\;{x_2} = - 1,\;{x_3} = 1.\]

Using the sign change rule in the first derivative (Figure \(19a\)), we establish that \(x = -1\) and \(x = 1\) are points of minimum, and \(x = 0\) is a point of maximum. The values of the function at these points are

\[y\left( { - 1} \right) = y\left( 1 \right) = - 2,\;\;y\left( 0 \right) = - 1.\]

Investigate the concavity of the function. The second derivative is given by

\[y^{\prime\prime}\left( x \right) = \left( {4{x^3} - 4x} \right)^\prime = 12{x^2} - 4.\]

It is equal to zero at the following points:

\[y^{\prime\prime}\left( x \right) = 0,\;\; \Rightarrow 12{x^2} - 4 = 0,\;\; \Rightarrow {x^2} = \frac{1}{3},\;\; \Rightarrow {x_{1,2}} = \pm \frac{1}{{\sqrt 3 }} \approx \pm 0,58.\]

When passing through any of the points the sign of the second derivative is reversed (Figure \(19a\)). Therefore, these points are points of inflection. Their \(y\)-coordinates are the same because the function is even and are equal

\[y\left( { \pm \frac{1}{{\sqrt 3 }}} \right) = {\left( { \pm \frac{1}{{\sqrt 3 }}} \right)^4} - 2{\left( { \pm \frac{1}{{\sqrt 3 }}} \right)^2} - 1 = \frac{1}{9} - \frac{2}{3} - 1 = - \frac{{14}}{9} \approx - 1,56.\]

The function is strictly convex downward on the intervals \(\left( { - \infty , - {\frac{1}{{\sqrt 3 }}}} \right)\) and \(\left( {{\frac{1}{{\sqrt 3 }}}, +\infty} \right),\) and is strictly convex upward on the interval \(\left( {- {\frac{1}{{\sqrt 3 }}}, {\frac{1}{{\sqrt 3 }}}} \right).\) A graph of the function is shown in Figure \(19b\).

Signs of the derivatives of the quadric function y=x^4-2x^2-1.
Figure 19a.
A schematic view of the quadric function y=x^4-2x^2-1.
Figure 19b.

Example 20.

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

Solution.

The function is defined for all \(x \in \mathbb{R}.\) Using a product-to-sum identity, we can express the function in the form

\[y\left( x \right) = \sin x\sin 2x = \frac{{\cos \left( {x - 2x} \right) - \cos \left( {x + 2x} \right)}}{2} = \frac{1}{2}\left( {\cos x - \cos 3x} \right).\]

The function is periodic with period \(2\pi\) and even as

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

There are no asymptotes for the function. When \(x = 0,\) the function becomes zero: \(y\left( 0 \right) = 0.\)

Compute the roots of the function:

\[y\left( x \right) = 0,\;\; \Rightarrow \sin x\sin 2x = 0;\]
\[1)\;\sin x = 0,\;\Rightarrow {x_1} = \pi n, n \in Z;\]
\[2)\;\sin 2x = 0,\;\Rightarrow {x_2} = \frac{{\pi k}}{2}, k \in Z.\]

The general solution are the values \(x = {\frac{{\pi k}}{2}},\) \(k \in Z.\) On the interval \(\left[ {0,2\pi } \right],\) the function has zeros at the points \(0,\;{\frac{\pi }{2}},\) \(\pi,\) \({\frac{{3\pi }}{2}},\) \(2\pi .\)

To determine the intervals of constant sign, we solve the inequality

\[y\left( x \right) \gt 0,\;\; \Rightarrow \sin x\sin 2x \gt 0.\]

Here, there are two solutions:

\[1)\;{\left\{ {\begin{array}{*{20}{l}} {\sin x \gt 0}\\ {\sin 2x \gt 0} \end{array},} \right.}\;\;\Rightarrow \left\{ {\begin{array}{*{20}{l}} {2\pi n \lt x \lt \pi + 2\pi n}\\ {2\pi k \lt 2x \lt \pi + 2\pi k} \end{array}} \right., \text{ where } n,k \in Z,\;\;\Rightarrow \left\{ {\begin{array}{*{20}{l}} {2\pi n \lt x \lt \pi + 2\pi n}\\ {\pi k \lt x \lt {\frac{\pi }{2}} + \pi k} \end{array}} \right.,\;\;\Rightarrow 2\pi k \lt x \lt {\frac{\pi }{2}} + 2\pi k,k \in Z;\]
\[2)\;\left\{ {\begin{array}{*{20}{l}} {\sin x \lt 0}\\ {\sin 2x \lt 0} \end{array},} \right.\;\;\Rightarrow \left\{ {\begin{array}{*{20}{l}} {\pi + 2\pi n \lt x \lt 2\pi + 2\pi n}\\ {\pi + 2\pi k \lt 2x \lt 2\pi + 2\pi k} \end{array}} \right., \text{ where } n,k \in Z,\;\;\Rightarrow \left\{ {\begin{array}{*{20}{l}} {\pi + 2\pi n \lt x \lt 2\pi + 2\pi n}\\ {{\frac{\pi }{2}} + \pi k \lt x \lt \pi + \pi k} \end{array}} \right.,\;\;\Rightarrow {\frac{{3\pi }}{2}} + 2\pi k \lt x \lt 2\pi + 2\pi k,k \in Z.\]

As it can be seen, the solution of the inequality \(y\left( x \right) \gt 0\) are angles in the first and fourth quadrants. Combining the two branches of solutions, we can write:

\[y\left( x \right) \ge 0\;\;\text{at}\;\; - \frac{\pi }{2} + 2\pi n \le x \le \frac{\pi }{2} + 2\pi n,\;n \in Z.\]

Accordingly,

\[y\left( x \right) \le 0\;\;\text{at}\;\;\frac{\pi }{2} + 2\pi k \le x \le \frac{3\pi }{2} + 2\pi k,\;k \in Z.\]

Here we used non-strict inequalities to include zero values of the function (Figure \(20a\)).

Let us turn to the study of monotonicity and extreme points. The derivative can be written as

\[y'\left( x \right) = \left( {\sin x\sin 2x} \right)^\prime = \left[ {\frac{1}{2}\left( {\cos x - \cos 3x} \right)} \right]^\prime = \frac{1}{2}\left( {3\sin 3x - \sin x} \right).\]

Find the roots of the derivative. The problem is reduced to solving the trigonometric equation

\[3\sin 3x - \sin x = 0.\]

Using the triple angle formula, we obtain:

\[3\sin 3x - \sin x = 0,\;\; \Rightarrow 3\left( {3\sin x - 4{{\sin }^3}x} \right) - \sin x = 0,\;\; \Rightarrow 9\sin x - 12{\sin ^3}x - \sin x = 0,\;\; \Rightarrow \sin x\left( {2 - 3{{\sin }^2}x} \right) = 0.\]
\[1)\;\sin x = 0,\;\; \Rightarrow {x_1} = \pi n,\;n \in Z;\]
\[2)\;2 - {\sin ^2}x = 0,\;\;\Rightarrow {\sin ^2}x = \frac{2}{3},\;\;\Rightarrow \sin x = \pm \sqrt {\frac{2}{3}} ,\;\;\Rightarrow x = \arcsin \left( { \pm \sqrt {\frac{2}{3}} } \right) = \pm \arcsin \sqrt {\frac{2}{3}} ;\;\;\Rightarrow {x_2} = {\left( { - 1} \right)^k}\arcsin \sqrt {\frac{2}{3}} + \pi k, k \in Z;\;\; {x_3} = {\left( { - 1} \right)^{m + 1}}\arcsin \sqrt {\frac{2}{3}} + \pi m,m \in Z.\]

Here the angle \(\arcsin \sqrt {\frac{2}{3}} \) is approximately equal to \(0,3\pi\;\text{rad}\) or \(55^{\circ}.\)

Thus, the segment \(\left[ {0,2\pi } \right]\) contains the following stationary points:

\[0,\;\;\arcsin \sqrt {\frac{2}{3}} ,\;\;\pi - \arcsin \sqrt {\frac{2}{3}} ,\;\;\pi ,\;\;\pi + \arcsin \sqrt {\frac{2}{3}} ,\;\;2\pi - \arcsin \sqrt {\frac{2}{3}} ,\;\;2\pi .\]

When passing through any of these points the sign of the first derivative is reversed. Hence, all these points are local extrema, and the minimum points among them (Figure \(20a\)) are

\[x = 0,\;\;\;x = \pi - \arcsin \sqrt {\frac{2}{3}} ,\;\;\;x = \pi + \arcsin \sqrt {\frac{2}{3}} ,\;\;\;x = 2\pi .\]

The other points are the points of maximum:

\[x = \arcsin \sqrt {\frac{2}{3}} ,\;\;\;x = \pi ,\;\;\;x = 2\pi - \arcsin \sqrt {\frac{2}{3}} .\]

We calculate the values of the function at the points of maximum and minimum:

\[y\left( 0 \right) = 0;\;\;y\left( {\pi - \arcsin \sqrt {\frac{2}{3}} } \right) = - \frac{4}{{3\sqrt 3 }} \approx - 0,77;\;\;y\left( {\pi + \arcsin \sqrt {\frac{2}{3}} } \right) = - \frac{4}{{3\sqrt 3 }} \approx - 0,77;\;\;y\left( {\arcsin \sqrt {\frac{2}{3}} } \right) =\frac{4}{{3\sqrt 3 }} \approx 0,77;\;\;y\left( \pi \right) = 0;\;\;y\left( {2\pi - \arcsin \sqrt {\frac{2}{3}} } \right) = \frac{4}{{3\sqrt 3 }} \approx 0,77.\]

Now we investigate the concavity of the graph and points of inflection. Differentiating the first derivative again yields:

\[y^{\prime\prime}\left( x \right) = \left[ {\frac{1}{2}\left( {3\sin 3x - \sin x} \right)} \right]^\prime = \frac{9}{2}\cos 3x - \frac{1}{2}\cos x.\]

The inflection points satisfy the equation

\[y^{\prime\prime}\left( x \right) = 0,\;\; \Rightarrow \frac{9}{2}\cos 3x - \frac{1}{2}\cos x = 0,\;\; \Rightarrow 9\cos 3x - \cos x = 0.\]

Using the triple angle formula for cosine, we have:

\[9\cos 3x - \cos x = 0,\;\; \Rightarrow 9\left( {4{{\cos }^3}x - 3\cos x} \right) - \cos x = 0,\;\; \Rightarrow 36{\cos ^3}x - 27\cos x - \cos x = 0,\;\; \Rightarrow 9{\cos ^3}x - 7\cos x = 0,\;\; \Rightarrow \cos x\left( {9{{\cos }^2}x - 7} \right) = 0.\]
\[1)\;\cos x = 0,\;\;\Rightarrow {x_1} = \frac{\pi }{2} + \pi n, n \in Z;\]
\[2)\;9{\cos ^2}x - 7 = 0,\;\;\Rightarrow {\cos ^2}x = \frac{7}{9},\;\;\Rightarrow \cos x = \pm \sqrt {\frac{7}{9}} ,\;\;\Rightarrow x = \arccos \left( { \pm \sqrt {\frac{7}{9}} } \right),\;\;\Rightarrow {x_2} = \pm \arccos \sqrt {\frac{7}{9}} + 2\pi k, k \in Z,\;\; {x_3} = \pm \arccos \left( { - \sqrt {\frac{7}{9}} } \right) + 2\pi m = \pm \left( {\pi - \arccos \sqrt {\frac{7}{9}} } \right) + 2\pi m, m \in Z.\]

Since the angle \(\arccos \sqrt {\frac{7}{9}} \) is approximately equal to \(0,16\pi\;\text{rad}\) or \(28^{\circ},\) then the segment \(\left[ {0,2\pi } \right]\) contains the following points where the second derivative is zero:

\[x = \arccos \sqrt {\frac{7}{9}} ,\;\;\frac{\pi }{2},\;\;\pi - \arccos \sqrt {\frac{7}{9}} ,\;\;\pi + \arccos \sqrt {\frac{7}{9}} ,\;\;\frac{{3\pi }}{2},\;\;2\pi - \arccos \sqrt {\frac{7}{9}}.\]

One can show that when passing through these points, the second derivative changes its sign to the opposite (Figure \(20a\)). Therefore, these points are points of inflection. Calculate the corresponding values of the function:

\[y\left( {\arccos \sqrt {\frac{7}{9}} } \right) = \frac{{4\sqrt 7 }}{{27}} \approx 0,39;\;\;y\left( {\frac{\pi }{2}} \right) = 0;\;\;y\left( {\pi - \arccos \sqrt {\frac{7}{9}} } \right) = - \frac{{4\sqrt 7 }}{{27}} \approx - 0,39;\;\;y\left( {\pi + \arccos \sqrt {\frac{7}{9}} } \right) = - \frac{{4\sqrt 7 }}{{27}} \approx - 0,39;\;\;y\left( {\frac{{3\pi }}{2}} \right) = 0;\;\;y\left( {2\pi - \arccos \sqrt {\frac{7}{9}} } \right) = \frac{{4\sqrt 7 }}{{27}} \approx 0,39.\;\;\]

By bringing together all the data we can draw a schematic graph of the function (Figure \(20b\)).

Signs of the derivatives of the trig function y=sin(x)sin(2x).
Figure 20a.
A schematic view of the trig function y=sin(x)sin(2x).
Figure 20b.

Example 21.

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

Solution.

The function is defined for all \(x \in \mathbb{R}.\) Find the intercepts:

\[f\left( 0 \right) = 2\arctan 0 + 0 = 0,\]

that is the graph intersects the origin.

Check for oblique asymptotes:

\[k = \lim\limits_{x \to \pm \infty } \frac{{f\left( x \right)}}{x} = \lim\limits_{x \to \pm \infty } \frac{{2\arctan x + x}}{x} = \lim\limits_{x \to \pm \infty } \left( {\frac{{2\arctan x}}{x} + 1} \right) = 1.\]
\[{b_1} = \lim\limits_{x \to + \infty } \left[ {f\left( x \right) - kx} \right] = \lim\limits_{x \to + \infty } \left( {2\arctan x + \cancel{x} - \cancel{x}} \right) = \lim\limits_{x \to + \infty } \left( {2\arctan x} \right) = 2 \cdot \frac{\pi }{2} = \pi ;\]
\[{b_2} = \lim\limits_{x \to - \infty } \left[ {f\left( x \right) - kx} \right] = \lim\limits_{x \to - \infty } \left( {2\arctan x + \cancel{x} - \cancel{x}} \right) = \lim\limits_{x \to - \infty } \left( {2\arctan x} \right) = 2 \cdot \left({-\frac{\pi }{2}}\right) = -\pi .\]

Thus, there are two oblique asymptotes - in the positive and negative direction:

\[x \to \infty :\;y = kx + {b_1} = x + \pi ;\]
\[x \to -\infty :\;y = kx + {b_2} = x - \pi .\]

Compute the derivative:

\[f^\prime\left( x \right) = \left( {2\arctan x + x} \right)^\prime = \frac{2}{{1 + {x^2}}} + 1 = \frac{{3 + {x^2}}}{{1 + {x^2}}} \gt 0,\]

that is the function is increasing everywhere.

Take the second derivative:

\[f^{\prime\prime}\left( x \right) = \left( {\frac{{3 + {x^2}}}{{1 + {x^2}}}} \right)^\prime = \frac{{2x \cdot \left( {1 + {x^2}} \right) - \left( {3 + {x^2}} \right) \cdot 2x}}{{{{\left( {1 + {x^2}} \right)}^2}}} = \frac{{2x + \cancel{2{x^3}} - 6x - \cancel{2{x^3}}}}{{{{\left( {1 + {x^2}} \right)}^2}}} = - \frac{{4x}}{{{{\left( {1 + {x^2}} \right)}^2}}}.\]
\[f^{\prime\prime}\left( x \right) = 0,\;\; \Rightarrow - \frac{{4x}}{{{{\left( {1 + {x^2}} \right)}^2}}} = 0,\;\; \Rightarrow x = 0.\]

The function is concave upward at \(x \lt 0\) and concave downward at \(x \gt 0.\) The point \(x = 0\) is a point of inflection.

A graph of the function is shown in Figure \(21b\).

Sign chart of the function f(x)=2arctan(x)+x.
Figure 21a.
Graph of the function f(x)=2arctan(x)+x.
Figure 21b.

Example 22.

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

Solution.

First we find the domain of the function:

\[{x^2} - 4x + 5 \gt 0,\;\; \Rightarrow D = 16 - 4 \cdot 5 = - 4 \lt 0.\]

The quadratic function has no roots and is positive everywhere. Hence, the given function is defined for all \(x \in \mathbb{R}.\)

Calculate the \(x-\) and \(y-\)intercepts:

\[f\left( x \right) = 0,\;\; \Rightarrow \ln \left( {{x^2} - 4x + 5} \right) = 0,\;\; \Rightarrow {x^2} - 4x + 5 = 1,\;\; \Rightarrow {x^2} - 4x + 4 = 0,\;\; \Rightarrow {\left( {x - 2} \right)^2} = 0,\;\; \Rightarrow x = 2.\]
\[f\left( 0 \right) = \ln \left( {{0^2} - 4 \cdot 0 + 5} \right) = \ln 5 \approx 1.61\]

To check for horizontal asymptote, we need to compute the following limit:

\[\lim\limits_{x \to \pm \infty } f\left( x \right) = \lim\limits_{x \to \pm \infty } \ln \left( {{x^2} - 4x + 5} \right) = \ln \left[ {\mathop {\lim }\limits_{x \to \pm \infty } \left( {{x^2} - 4x + 5} \right)} \right] = + \infty .\]

Hence, the function has no horizontal asymptotes. Similarly, we can make sure that there are no oblique asymptotes. Using L'Hopital's rule, we have

\[k = \lim\limits_{x \to \pm \infty } \frac{{f\left( x \right)}}{x} = \lim\limits_{x \to \pm \infty } \frac{{\ln \left( {{x^2} - 4x + 5} \right)}}{x} = \left[ {\frac{\infty }{\infty }} \right] = \lim\limits_{x \to \pm \infty } \frac{{\left( {\ln \left( {{x^2} - 4x + 5} \right)} \right)^\prime}}{{x^\prime}} = \lim\limits_{x \to \pm \infty } \frac{{\frac{{2x - 4}}{{{x^2} - 4x + 5}}}}{1} = \lim\limits_{x \to \pm \infty } \frac{{2x - 4}}{{{x^2} - 4x + 5}} = 0.\]
\[b = \lim\limits_{x \to \pm \infty } \left[ {f\left( x \right) - kx} \right] = \lim\limits_{x \to \pm \infty } \left[ {\ln \left( {{x^2} - 4x + 5} \right)} \right] = + \infty .\]

The first derivative is given by

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

Determine the critical points:

\[f^\prime\left( x \right) = 0,\;\; \Rightarrow \frac{{2x - 4}}{{{x^2} - 4x + 5}} = 0,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {2x - 4 = 0}\\ {{x^2} - 4x + 5 \ne 0} \end{array}} \right.,\;\; \Rightarrow x = 2.\]
Sign chart of the function f(x)=ln(x^2-4x+5).
Figure 22a.
Graph of the function f(x)=ln(x^2-4x+5).
Figure 22b.

As you can see from the sign chart, \(x = 2\) is a point of local minimum. Its \(y-\)value is

\[f\left( 2 \right) = \ln \left( {{2^2} - 4 \cdot 2 + 5} \right) = \ln 1 = 0.\]

Find the second derivative and determine the points where it is equal to zero:

\[f^{\prime\prime}\left( x \right) = \left( {\frac{{2x - 4}}{{{x^2} - 4x + 5}}} \right)^\prime = \frac{{ - 2{x^2} + 8x - 6}}{{{{\left( {{x^2} - 4x + 5} \right)}^2}}}.\]
\[f^{\prime\prime}\left( x \right) = 0,\;\; \Rightarrow \frac{{ - 2{x^2} + 8x - 6}}{{{{\left( {{x^2} - 4x + 5} \right)}^2}}} = 0,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} { - 2{x^2} + 8x - 6 = 0}\\ {{{\left( {{x^2} - 4x + 5} \right)}^2} \ne 0} \end{array}} \right.,\;\; \Rightarrow \left( {x - 1} \right)\left( {x - 3} \right) = 0,\;\; \Rightarrow {x_1} = 1,{x_2} = 3.\]

It follows from the sign chart that both these points are points of inflection.

\[f\left( 1 \right) = \ln \left( {{1^2} - 4 \cdot 1 + 5} \right) = \ln 2;\]
\[f\left( 3 \right) = \ln \left( {{3^2} - 4 \cdot 3 + 5} \right) = \ln 2.\]

Thus, the function has the following inflection points: \(\left( {1,\ln 2} \right)\) and \(\left( {3,\ln 2} \right).\)

Now we can draw the graph of the function (Figure \(22b\)).

Example 23.

Draw the curve given by the parametric equations

\[x = {t^3} + {t^2} - t, y = {t^3} + 2{t^2} - 4t.\]

Solution.

First we investigate the graphs of the functions \(x\left( t \right)\) and \(x\left( t \right)\). Both functions are cubic polynomials, which are defined for all \(x \in \mathbb{R}.\) Find the derivative \(x'\left( t \right):\)

\[x'\left( t \right) = \left( {{t^3} + {t^2} - t} \right)^\prime = 3{t^2} + 2t - 1.\]

Solving the equation \(x'\left( t \right) = 0,\) we determine the stationary points of the function \(x\left( t \right):\)

\[x'\left( t \right) = 0,\;\; \Rightarrow 3{t^2} + 2t - 1 = 0,\;\; \Rightarrow {t_{1,2}} = \frac{{ - 2 \pm \sqrt {16} }}{6} = - 1;\;\frac{1}{3}.\]

At \(t = 1,\) the function \(x\left( t \right)\) reaches a maximum equal to

\[x\left( { - 1} \right) = \left( { - 1} \right)^3 + \left( { - 1} \right)^2 - \left( { - 1} \right) = 1,\]

and at the point \(t = {\frac{1}{3}},\) it has a minimum equal to

\[x\left( {\frac{1}{3}} \right) = {\left( {\frac{1}{3}} \right)^3} + {\left( {\frac{1}{3}} \right)^2} - \left( {\frac{1}{3}} \right) = \frac{1}{{27}} + \frac{1}{9} - \frac{1}{3} = - \frac{5}{{27}}.\]

Consider the derivative \(y'\left( t \right):\)

\[y'\left( t \right) = \left( {{t^3} + 2{t^2} - 4t} \right)^\prime = 3{t^2} + 4t - 4.\]

Similarly, we find the stationary points of \(y\left( t \right):\)

\[y'\left( t \right) = 0,\;\; \Rightarrow 3{t^2} + 4t - 4 = 0,\;\; \Rightarrow {t_{1,2}} = \frac{{ - 4 \pm \sqrt {64} }}{6} = - 2;\;\frac{2}{3}.\]

The function \(y\left( t \right)\) reaches a maximum at the point \(t = -2:\)

\[y\left( { - 2} \right) = {\left( { - 2} \right)^3} + 2{\left( { - 2} \right)^2} - 4\left( { - 2} \right) = 8\]

and a minimum at \(t = {\frac{2}{3}}:\)

\[y\left( {\frac{2}{3}} \right) = {\left( {\frac{2}{3}} \right)^3} + 2{\left( {\frac{2}{3}} \right)^2} - 4 \cdot \frac{2}{3} = \frac{8}{{27}} + \frac{8}{9} - \frac{8}{3} = - \frac{{40}}{{27}}.\]

The graphs of the functions \(x\left( t \right)\), \(y\left( t \right)\) are shown schematically in Figure \(23a.\)

The variables x and y as functions of the parameter t.
Figure 23a.

Note that since

\[\lim\limits_{t \to \pm \infty } x\left( t \right) = \pm \infty ,\;\;\lim\limits_{t \to \pm \infty } y\left( t \right) = \pm \infty ,\]

the curve \(y\left( x \right)\) has neither vertical nor horizontal asymptotes. Moreover, since

\[k = \lim\limits_{t \to \pm \infty } \frac{{y\left( t \right)}}{{x\left( t \right)}} = \lim\limits_{t \to \pm \infty } \frac{{{t^3} + 2{t^2} - 4t}}{{{t^3} + {t^2} - t}} = \lim\limits_{t \to \pm \infty } \frac{{1 + \frac{2}{t} - \frac{4}{{{t^2}}}}}{{1 + \frac{1}{t} - \frac{1}{{{t^2}}}}} = 1,\]
\[b = \lim\limits_{t \to \pm \infty } \left[ {y\left( t \right) - kx\left( t \right)} \right] = \lim\limits_{t \to \pm \infty } \left({ {t^2} - {3t}} \right) = + \infty ,\]

the curve \(y\left( x \right)\) has no oblique asymptotes, too.

Determine the points of intersection of the graph \(y\left( x \right)\) with the coordinate axes. The intersection with the \(x\)-axis occurs at the following points:

\[y\left( t \right) = {t^3} + 2{t^2} - 4t = 0,\;\; \Rightarrow t\left( {{t^2} + 2t - 4} \right) = 0;\]
\[1)\;{t_1} = 0;\]
\[2)\;{t^2} + 2t - 4 = 0,\;\;\Rightarrow D = 4 - 4 \cdot \left( { - 4} \right) = 20,\;\; \Rightarrow {t_{2,3}} = \frac{{ - 2 \pm \sqrt {20} }}{2} = - 1 \pm \sqrt 5 .\]
\[x\left( {{t_1}} \right) = x\left( 0 \right) = 0;\]
\[x\left( {{t_2}} \right) = x\left( { - 1 - \sqrt 5 } \right) = {\left( { - 1 - \sqrt 5 } \right)^3} + {\left( { - 1 - \sqrt 5 } \right)^2} - \left( { - 1 - \sqrt 5 } \right) = - \left( {1 + 3\sqrt 5 + 15 + 5\sqrt 5 } \right) + \left( {1 + 2\sqrt 5 + 5} \right) + 1 + \sqrt 5 = - 16 - 8\sqrt 5 + 6 + 2\sqrt 5 + 1 + \sqrt 5 = - 9 - 5\sqrt 5 \approx 20,18;\]
\[x\left( {{t_3}} \right) = x\left( { - 1 + \sqrt 5 } \right) = {\left( { - 1 + \sqrt 5 } \right)^3 + \left( { - 1 + \sqrt 5 } \right)^2} - \left( { - 1 + \sqrt 5 } \right) = - \left( {1 - 3\sqrt 5 + 15 - 5\sqrt 5 } \right) + \left( {1 - 2\sqrt 5 + 5} \right) + 1 - \sqrt 5 = - 16 + 8\sqrt 5 + 6 - 2\sqrt 5 + 1 - \sqrt 5 = - 9 + 5\sqrt 5 \approx 2,18.\]

In the same way we find the points of intersection of the graph with the \(y\)-axis:

\[x\left( t \right) = {t^3} + {t^2} - t = 0,\;\; \Rightarrow t\left( {{t^2} + t - 1} \right) = 0;\]
\[1)\;{t_1} = 0;\]
\[2)\;{t^2} + t - 1 = 0,\;\;\Rightarrow D = 1 - 4 \cdot \left( { - 1} \right) = 5,\;\;\Rightarrow {t_{2,3}} = \frac{{ - 1 \pm \sqrt {5} }}{2}.\]
\[y\left( {{t_1}} \right) = y\left( 0 \right) = 0;\]
\[y\left( {{t_2}} \right) = y\left( {\frac{{ - 1 - \sqrt 5 }}{2}} \right) = 3 + 2\sqrt 5 \approx 7,47;\]
\[y\left( {{t_3}} \right) = y\left( {\frac{{ - 1 + \sqrt 5 }}{2}} \right) = 3 - 2\sqrt 5 \approx - 1,47.\]

Divide the \(t\)-axis into \(5\) intervals:

\[\left( { - \infty , - 2} \right),\;\;\left( { - 2, - 1} \right),\;\;\left( { - 1,\frac{1}{3}} \right),\;\;\left( {\frac{1}{3},\frac{2}{3}} \right),\;\;\left( {\frac{2}{3}, + \infty } \right).\]

On the first interval \(\left( { - \infty , - 2} \right),\) the values of \(x\) and \(y\) increase from \(-\infty\) to \(x\left( { - 2} \right) = - 2\) and \(y\left( { - 2} \right) = 8.\) This is shown schematically in Figure \(23b.\)

Five intervals of the parametric function x=t^3+t^2-t, y=t^3+2t^2-4t.
Figure 23b.

On the second interval \(\left( { - 2, - 1} \right),\) the variable \(x\) increases from \(x\left( { - 2} \right) = - 2\) to \(x\left( { - 1} \right) = 1,\) and the variable \(y\) decreases from \(y\left( { - 2} \right) = 8\) to \(y\left( { - 1} \right) = 5.\)

Here we have a decreasing portion of the curve \(y\left( x \right).\) It intersects the vertical axis at the point \(\left( {0,3 + 2\sqrt 5 } \right).\)

On the third interval \(\left( { - 1,{\frac{1}{3}}} \right),\) both variables decrease. The value of \(x\) ranges from \(x\left( { - 1} \right) = 1\) to \(x\left( {\frac{1}{3}} \right) = - {\frac{5}{{27}}}.\) Accordingly, the value of \(y\) decreases from \(y\left( { - 1} \right) = 5\) to \(y\left( {\frac{1}{3}} \right) = - {\frac{29}{{27}}}.\) The curve \(y\left( x \right)\) intersects the origin.

On the fourth interval \(\left( {\frac{1}{3},\frac{2}{3}} \right),\) the variable \(x\) increases from \(x\left( {\frac{1}{3}} \right) = - {\frac{5}{{27}}}\) to \(x\left( {\frac{2}{3}} \right) = {\frac{2}{{27}}},\) and the variable \(y\) decreases from \(y\left( {\frac{1}{3}} \right) = - {\frac{29}{{27}}}\) to \(y\left( {\frac{2}{3}} \right) = - {\frac{40}{{27}}}.\) On this interval the curve \(y\left( x \right)\) intersects the \(y\)-axis at the point \(\left( {0,3 - 2\sqrt 5 } \right).\)

Finally, on the last interval \(\left( {{\frac{2}{3}}, + \infty } \right),\) both functions \(x\left( t \right)\), \(y\left( t \right)\) increase. The curve \(y\left( x \right)\) intersects the \(x\)-axis at the point \(x = - 9 + 5\sqrt 5 \approx 2,18.\)

To clarify the shape of the curve \(y\left( x \right),\) we determine the maximum and minimum points. The derivative \(y'\left( x \right)\) is expressed in the form

\[y'\left( x \right) = {y'_x} = \frac{{{y'_t}}}{{{x'_t}}} = \frac{{{{\left( {{t^3} + 2{t^2} - 4t} \right)}^\prime }}}{{{{\left( {{t^3} + {t^2} - t} \right)}^\prime }}} = \frac{{3{t^2} + 4t - 4}}{{3{t^2} + 2t - 1}} = \frac{{\cancel{3}\left( {t + 2} \right)\left( {t - \frac{2}{3}} \right)}}{{\cancel{3}\left( {t + 1} \right)\left( {t - \frac{1}{3}} \right)}} = \frac{{\left( {t + 2} \right)\left( {t - \frac{2}{3}} \right)}}{{\left( {t + 1} \right)\left( {t - \frac{1}{3}} \right)}}.\]

One can show that the curve has a maximum at the point \(t = - 2\) (that is at the boundary of the \(I\)st and \(II\)nd intervals). There is also a minimum at \(t = \frac{2}{3}\) (i.e. at the boundary of the \(IV\text{th}\) and \(V\text{th}\) intervals).

When passing through the point \(t = {\frac{1}{3}}\) the derivative also changes sign from plus to minus, but the curve \(y\left( x \right)\) is not a single-valued function in this area. Therefore, this point is not an extremum.

We also investigate the concavity/convexity of the given curve. The second derivative \(y^{\prime\prime}\left( x \right)\) has the form:

\[y^{\prime\prime}\left( x \right) = {y^{\prime\prime}_{xx}} = \frac{{{{\left( {{y'_x}} \right)}'_t}}}{{{x'_t}}} = \frac{{{{\left( {\frac{{3{t^2} + 4t - 4}}{{3{t^2} + 2t - 1}}} \right)}^\prime }}}{{{{\left( {{t^3} + {t^2} - t} \right)}^\prime }}} = \frac{{ - {6{t^2}} + {18t} + {4}}}{{{{\left( {3{t^2} + 2t - 1} \right)}^3}}} = \frac{{ - 6\left( {t - \frac{{9 - \sqrt {105} }}{6}} \right)\left( {t - \frac{{9 + \sqrt {105} }}{6}} \right)}}{{{{\left( {t + 1} \right)}^3}{{\left( {3t - 1} \right)}^3}}}.\]
Sign of the second derivative of the parametric function x=t^3+t^2-t, y=t^3+2t^2-4t.
Figure 23c.

Hence, the second derivative changes its sign when passing through the following points (see Figure \(23c\)):

\[{t_1} = - 1:\;\;x\left( { - 1} \right) = 1,\;\;y\left( { - 1} \right) = 5;\]
\[{t_2} = \frac{{9 - \sqrt {105} }}{6}:\;\;x\left( {\frac{{9 - \sqrt {105} }}{6}} \right) \approx 0,24;\;\;y\left( {\frac{{9 - \sqrt {105} }}{6}} \right) \approx 0,91;\]
\[{t_3} = \frac{1}{3}:\;\;x\left( {\frac{1}{3}} \right) = - \frac{5}{{27}},\;\;y\left( {\frac{1}{3}} \right) = - \frac{{29}}{{27}};\]
\[{t_4} = \frac{{9 + \sqrt {105} }}{6}:\;\;x\left( {\frac{{9 + \sqrt {105} }}{6}} \right) \approx 40,1;\;\;y\left( {\frac{{9 + \sqrt {105} }}{6}} \right) \approx 40,8.\]

Therefore, these points are points of inflection of the curve \(y\left( x \right).\)

A schematic graph of the curve \(y\left( x \right)\) is shown above in Figure \(23b\).

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