# Curve Sketching

## Solved Problems

Click or tap a problem to see the solution.

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

### Example 15

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

### Example 16

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

### Example 17

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

### 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

$x = {t^3} + {t^2} - t, y = {t^3} + 2{t^2} - 4t.$

### 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.

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 $$16b$$.

### 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 $$17b$$).

### 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 $$18b$$.

### 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 $$19b$$.

### 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 $$20b$$).

### 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 $$21b$$.

### 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:

$f^\prime\left( x \right) = 0,\;\; \Rightarrow \frac{{2x - 4}}{{{x^2} - 4x + 5}} = 0,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {2x - 4 = 0}\\ {{x^2} - 4x + 5 \ne 0} \end{array}} \right.,\;\; \Rightarrow x = 2.$

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 $$23a.$$

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 $$23b.$$

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:

$y^{\prime\prime}\left( x \right) = {y^{\prime\prime}_{xx}} = \frac{{{{\left( {{y'_x}} \right)}'_t}}}{{{x'_t}}} = \frac{{{{\left( {\frac{{3{t^2} + 4t - 4}}{{3{t^2} + 2t - 1}}} \right)}^\prime }}}{{{{\left( {{t^3} + {t^2} - t} \right)}^\prime }}} = \frac{{ - {6{t^2}} + {18t} + {4}}}{{{{\left( {3{t^2} + 2t - 1} \right)}^3}}} = \frac{{ - 6\left( {t - \frac{{9 - \sqrt {105} }}{6}} \right)\left( {t - \frac{{9 + \sqrt {105} }}{6}} \right)}}{{{{\left( {t + 1} \right)}^3}{{\left( {3t - 1} \right)}^3}}}.$

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 $$23b$$.