We choose a rectangular coordinate system on the plane and plot the values of the argument on the abscissa axis X, and on the y-axis - the values of the function y = f(x).
Function Graph y = f(x) the set of all points is called, for which the abscissas belong to the domain of the function, and the ordinates are equal to the corresponding values of the function.
In other words, the graph of the function y \u003d f (x) is the set of all points in the plane, the coordinates X, at which satisfy the relation y = f(x).
On fig. 45 and 46 are graphs of functions y = 2x + 1 and y \u003d x 2 - 2x.
Strictly speaking, one should distinguish between the graph of a function (the exact mathematical definition of which was given above) and the drawn curve, which always gives only a more or less accurate sketch of the graph (and even then, as a rule, not of the entire graph, but only of its part located in the final parts of the plane). In what follows, however, we will usually refer to "chart" rather than "chart sketch".
Using a graph, you can find the value of a function at a point. Namely, if the point x = a belongs to the scope of the function y = f(x), then to find the number f(a)(i.e. the function values at the point x = a) should do so. Need through a dot with an abscissa x = a draw a straight line parallel to the y-axis; this line will intersect the graph of the function y = f(x) at one point; the ordinate of this point will be, by virtue of the definition of the graph, equal to f(a)(Fig. 47).
For example, for the function f(x) = x 2 - 2x using the graph (Fig. 46) we find f(-1) = 3, f(0) = 0, f(1) = -l, f(2) = 0, etc.
A function graph visually illustrates the behavior and properties of a function. For example, from a consideration of Fig. 46 it is clear that the function y \u003d x 2 - 2x takes positive values when X< 0 and at x > 2, negative - at 0< x < 2; наименьшее значение функция y \u003d x 2 - 2x accepts at x = 1.
To plot a function f(x) you need to find all points of the plane, coordinates X,at which satisfy the equation y = f(x). In most cases, this is impossible, since there are infinitely many such points. Therefore, the graph of the function is depicted approximately - with greater or lesser accuracy. The simplest is the multi-point plotting method. It consists in the fact that the argument X give a finite number of values - say, x 1 , x 2 , x 3 ,..., x k and make a table that includes the selected values of the function.
The table looks like this:
Having compiled such a table, we can outline several points on the graph of the function y = f(x). Then, connecting these points with a smooth line, we get an approximate view of the graph of the function y = f(x).
However, it should be noted that the multi-point plotting method is very unreliable. In fact, the behavior of the graph between the marked points and its behavior outside the segment between the extreme points taken remains unknown.
Example 1. To plot a function y = f(x) someone compiled a table of argument and function values:
The corresponding five points are shown in Fig. 48.
Based on the location of these points, he concluded that the graph of the function is a straight line (shown in Fig. 48 by a dotted line). Can this conclusion be considered reliable? Unless there are additional considerations to support this conclusion, it can hardly be considered reliable. reliable.
To substantiate our assertion, consider the function
.
Calculations show that the values of this function at points -2, -1, 0, 1, 2 are just described by the above table. However, the graph of this function is not at all a straight line (it is shown in Fig. 49). Another example is the function y = x + l + sinx; its meanings are also described in the table above.
These examples show that in its "pure" form, the multi-point plotting method is unreliable. Therefore, to plot a given function, as a rule, proceed as follows. First, the properties of this function are studied, with the help of which it is possible to construct a sketch of the graph. Then, by calculating the values of the function at several points (the choice of which depends on the set properties of the function), the corresponding points of the graph are found. And, finally, a curve is drawn through the constructed points using the properties of this function.
We will consider some (the most simple and frequently used) properties of functions used to find a sketch of a graph later, but now we will analyze some commonly used methods for plotting graphs.
Graph of the function y = |f(x)|.
It is often necessary to plot a function y = |f(x)|, where f(x) - given function. Recall how this is done. By definition of the absolute value of a number, one can write
This means that the graph of the function y=|f(x)| can be obtained from the graph, functions y = f(x) as follows: all points of the graph of the function y = f(x), whose ordinates are non-negative, should be left unchanged; further, instead of the points of the graph of the function y = f(x), having negative coordinates, one should construct the corresponding points of the graph of the function y = -f(x)(i.e. part of the function graph
y = f(x), which lies below the axis X, should be reflected symmetrically about the axis X).
Example 2 Plot a function y = |x|.
We take the graph of the function y = x(Fig. 50, a) and part of this graph with X< 0 (lying under the axis X) is symmetrically reflected about the axis X. As a result, we get the graph of the function y = |x|(Fig. 50, b).
Example 3. Plot a function y = |x 2 - 2x|.
First we plot the function y = x 2 - 2x. The graph of this function is a parabola, the branches of which are directed upwards, the top of the parabola has coordinates (1; -1), its graph intersects the abscissa axis at points 0 and 2. On the interval (0; 2) the function takes negative values, therefore this part of the graph reflect symmetrically about the x-axis. Figure 51 shows a graph of the function y \u003d |x 2 -2x |, based on the graph of the function y = x 2 - 2x
Graph of the function y = f(x) + g(x)
Consider the problem of plotting the function y = f(x) + g(x). if graphs of functions are given y = f(x) and y = g(x).
Note that the domain of the function y = |f(x) + g(х)| is the set of all those values of x for which both functions y = f(x) and y = g(x) are defined, i.e. this domain of definition is the intersection of the domains of definition, the functions f(x) and g(x).
Let the points (x 0, y 1) and (x 0, y 2) respectively belong to the function graphs y = f(x) and y = g(x), i.e. y 1 \u003d f (x 0), y 2 \u003d g (x 0). Then the point (x0;. y1 + y2) belongs to the graph of the function y = f(x) + g(x)(for f(x 0) + g(x 0) = y 1+y2),. and any point of the graph of the function y = f(x) + g(x) can be obtained in this way. Therefore, the graph of the function y = f(x) + g(x) can be obtained from function graphs y = f(x). and y = g(x) by replacing each point ( x n, y 1) function graphics y = f(x) dot (x n, y 1 + y 2), where y 2 = g(x n), i.e., by shifting each point ( x n, y 1) function graph y = f(x) along the axis at by the amount y 1 \u003d g (x n). In this case, only such points are considered. X n for which both functions are defined y = f(x) and y = g(x).
This method of plotting a function graph y = f(x) + g(x) is called the addition of graphs of functions y = f(x) and y = g(x)
Example 4. In the figure, by the method of adding graphs, a graph of the function is constructed
y = x + sinx.
When plotting a function y = x + sinx we assumed that f(x) = x, a g(x) = sinx. To build a function graph, we select points with abscissas -1.5π, -, -0.5, 0, 0.5,, 1.5, 2. Values f(x) = x, g(x) = sinx, y = x + sinx we will calculate at the selected points and place the results in the table.
The construction of graphs of functions containing modules usually causes considerable difficulties for schoolchildren. However, everything is not so bad. It is enough to remember several algorithms for solving such problems, and you can easily plot even the most seemingly complex function. Let's see what these algorithms are.
1. Plotting the function y = |f(x)|
Note that the set of function values y = |f(x)| : y ≥ 0. Thus, the graphs of such functions are always located completely in the upper half-plane.
Plotting the function y = |f(x)| consists of the following simple four steps.
1) Construct carefully and carefully the graph of the function y = f(x).
2) Leave unchanged all points of the graph that are above or on the 0x axis.
3) The part of the graph that lies below the 0x axis, display symmetrically about the 0x axis.
Example 1. Draw a graph of the function y = |x 2 - 4x + 3|
1) We build a graph of the function y \u003d x 2 - 4x + 3. It is obvious that the graph of this function is a parabola. Let's find the coordinates of all points of intersection of the parabola with the coordinate axes and the coordinates of the vertex of the parabola.
x 2 - 4x + 3 = 0.
x 1 = 3, x 2 = 1.
Therefore, the parabola intersects the 0x axis at points (3, 0) and (1, 0).
y \u003d 0 2 - 4 0 + 3 \u003d 3.
Therefore, the parabola intersects the 0y axis at the point (0, 3).
Parabola vertex coordinates:
x in \u003d - (-4/2) \u003d 2, y in \u003d 2 2 - 4 2 + 3 \u003d -1.
Therefore, the point (2, -1) is the vertex of this parabola.
Draw a parabola using the received data (Fig. 1)
2) The part of the graph lying below the 0x axis is displayed symmetrically with respect to the 0x axis.
3) We get the graph of the original function ( rice. 2, shown by dotted line).
2. Plotting the function y = f(|x|)
Note that functions of the form y = f(|x|) are even:
y(-x) = f(|-x|) = f(|x|) = y(x). This means that the graphs of such functions are symmetrical about the 0y axis.
Plotting the function y = f(|x|) consists of the following simple chain of actions.
1) Plot the function y = f(x).
2) Leave that part of the graph for which x ≥ 0, that is, the part of the graph located in the right half-plane.
3) Display the part of the graph specified in paragraph (2) symmetrically to the 0y axis.
4) As the final graph, select the union of the curves obtained in paragraphs (2) and (3).
Example 2. Draw a graph of the function y = x 2 – 4 · |x| + 3
Since x 2 = |x| 2 , then the original function can be rewritten as follows: y = |x| 2 – 4 · |x| + 3. And now we can apply the algorithm proposed above.
1) We build carefully and carefully the graph of the function y \u003d x 2 - 4 x + 3 (see also rice. one).
2) We leave that part of the graph for which x ≥ 0, that is, the part of the graph located in the right half-plane.
3) Display the right side of the graph symmetrically to the 0y axis.
(Fig. 3).
Example 3. Draw a graph of the function y = log 2 |x|
We apply the scheme given above.
1) We plot the function y = log 2 x (Fig. 4).
3. Plotting the function y = |f(|x|)|
Note that functions of the form y = |f(|x|)| are also even. Indeed, y(-x) = y = |f(|-x|)| = y = |f(|x|)| = y(x), and therefore, their graphs are symmetrical about the 0y axis. The set of values of such functions: y ≥ 0. Hence, the graphs of such functions are located completely in the upper half-plane.
To plot the function y = |f(|x|)|, you need to:
1) Construct a neat graph of the function y = f(|x|).
2) Leave unchanged the part of the graph that is above or on the 0x axis.
3) The part of the graph located below the 0x axis should be displayed symmetrically with respect to the 0x axis.
4) As the final graph, select the union of the curves obtained in paragraphs (2) and (3).
Example 4. Draw a graph of the function y = |-x 2 + 2|x| – 1|.
1) Note that x 2 = |x| 2. Hence, instead of the original function y = -x 2 + 2|x| - one
you can use the function y = -|x| 2 + 2|x| – 1, since their graphs are the same.
We build a graph y = -|x| 2 + 2|x| – 1. For this, we use algorithm 2.
a) We plot the function y \u003d -x 2 + 2x - 1 (Fig. 6).
b) We leave that part of the graph, which is located in the right half-plane.
c) Display the resulting part of the graph symmetrically to the 0y axis.
d) The resulting graph is shown in the figure with a dotted line (Fig. 7).
2) There are no points above the 0x axis, we leave the points on the 0x axis unchanged.
3) The part of the graph located below the 0x axis is displayed symmetrically with respect to 0x.
4) The resulting graph is shown in the figure by a dotted line (Fig. 8).
Example 5. Plot the function y = |(2|x| – 4) / (|x| + 3)|
1) First you need to plot the function y = (2|x| – 4) / (|x| + 3). To do this, we return to algorithm 2.
a) Carefully plot the function y = (2x – 4) / (x + 3) (Fig. 9).
Note that this function is linear-fractional and its graph is a hyperbola. To build a curve, you first need to find the asymptotes of the graph. Horizontal - y \u003d 2/1 (the ratio of the coefficients at x in the numerator and denominator of a fraction), vertical - x \u003d -3.
2) The part of the chart that is above or on the 0x axis will be left unchanged.
3) The part of the chart located below the 0x axis will be displayed symmetrically with respect to 0x.
4) The final graph is shown in the figure (Fig. 11).
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Build a function
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Benefits of online charting
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A function graph is a visual representation of the behavior of some function on the coordinate plane. Plots help to understand various aspects of a function that cannot be determined from the function itself. You can build graphs of many functions, and each of them will be given by a specific formula. The graph of any function is built according to a certain algorithm (if you forgot the exact process of plotting a graph of a particular function).
Steps
Plotting a Linear Function
- If the slope is negative, the function is decreasing.
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From the point where the line intersects with the Y axis, draw a second point using the vertical and horizontal distances. A linear function can be plotted using two points. In our example, the point of intersection with the Y-axis has coordinates (0.5); from this point move 2 spaces up and then 1 space to the right. Mark a point; it will have coordinates (1,7). Now you can draw a straight line.
Use a ruler to draw a straight line through two points. To avoid mistakes, find the third point, but in most cases the graph can be built using two points. Thus, you have plotted a linear function.
Drawing points on the coordinate plane
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Define a function. The function is denoted as f(x). All possible values of the variable "y" are called the range of the function, and all possible values of the variable "x" are called the domain of the function. For example, consider the function y = x+2, namely f(x) = x+2.
Draw two intersecting perpendicular lines. The horizontal line is the X-axis. The vertical line is the Y-axis.
Label the coordinate axes. Break each axis into equal segments and number them. The intersection point of the axes is 0. For the X axis: positive numbers are plotted on the right (from 0), and negative numbers on the left. For the Y-axis: positive numbers are plotted on top (from 0), and negative numbers on the bottom.
Find the "y" values from the "x" values. In our example f(x) = x+2. Substitute certain "x" values into this formula to calculate the corresponding "y" values. If given a complex function, simplify it by isolating the "y" on one side of the equation.
- -1: -1 + 2 = 1
- 0: 0 +2 = 2
- 1: 1 + 2 = 3
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Draw points on the coordinate plane. For each pair of coordinates, do the following: find the corresponding value on the x-axis and draw a vertical line (dotted line); find the corresponding value on the y-axis and draw a horizontal line (dotted line). Mark the point of intersection of the two dotted lines; thus, you have plotted a graph point.
Erase the dotted lines. Do this after plotting all the graph points on the coordinate plane. Note: the graph of the function f(x) = x is a straight line passing through the center of coordinates [point with coordinates (0,0)]; the graph f(x) = x + 2 is a line parallel to the line f(x) = x, but shifted up by two units and therefore passing through the point with coordinates (0,2) (because the constant is 2).
Plotting a complex function
Find the zeros of the function. The zeros of a function are the values of the variable "x" at which y = 0, that is, these are the points of intersection of the graph with the x-axis. Keep in mind that not all functions have zeros, but this is the first step in the process of plotting any function. To find the zeros of a function, set it equal to zero. For example:
Find and label the horizontal asymptotes. An asymptote is a line that the graph of a function approaches but never crosses (that is, the function is not defined in this area, for example, when dividing by 0). Mark the asymptote with a dotted line. If the variable "x" is in the denominator of a fraction (for example, y = 1 4 − x 2 (\displaystyle y=(\frac (1)(4-x^(2))))), set the denominator to zero and find "x". In the obtained values of the variable "x", the function is not defined (in our example, draw dashed lines through x = 2 and x = -2), because you cannot divide by 0. But asymptotes exist not only in cases where the function contains a fractional expression. Therefore, it is recommended to use common sense:
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Determine if the function is linear. A linear function is given by a formula of the form F (x) = k x + b (\displaystyle F(x)=kx+b) or y = k x + b (\displaystyle y=kx+b)(for example, ), and its graph is a straight line. Thus, the formula includes one variable and one constant (constant) without any exponents, root signs, and the like. Given a function of a similar form, plotting such a function is quite simple. Here are other examples of linear functions:
Use a constant to mark a point on the y-axis. The constant (b) is the “y” coordinate of the intersection point of the graph with the Y-axis. That is, it is a point whose “x” coordinate is 0. Thus, if x = 0 is substituted into the formula, then y = b (constant). In our example y = 2x + 5 (\displaystyle y=2x+5) the constant is 5, that is, the point of intersection with the Y-axis has coordinates (0,5). Plot this point on the coordinate plane.
Find the slope of the line. It is equal to the multiplier of the variable. In our example y = 2x + 5 (\displaystyle y=2x+5) with the variable "x" is a factor of 2; thus, the slope is 2. The slope determines the angle of inclination of the straight line to the X-axis, that is, the larger the slope, the faster the function increases or decreases.
Write the slope as a fraction. The slope is equal to the tangent of the angle of inclination, that is, the ratio of the vertical distance (between two points on a straight line) to the horizontal distance (between the same points). In our example, the slope is 2, so we can say that the vertical distance is 2 and the horizontal distance is 1. Write this as a fraction: 2 1 (\displaystyle (\frac (2)(1))).
Construct a curve given by parametric equations \
Let us first study the graphs of the functions \(x\left(t \right)\) and \(x\left(t \right)\). Both functions are cubic polynomials that 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 \( x"\left(t \right) = 0,\) define the stationary points of the function \(x\left(t \right):\) \[ (x"\left(t \right) = 0,)\;\ ; (\Rightarrow 3(t^2) + 2t - 1 = 0,)\;\; (\Rightarrow (t_(1,2)) = \frac(( - 2 \pm \sqrt (16) ))(6) = - 1;\;\frac(1)(3).) \] (t = 1\) the function \(x\left(t \right)\) reaches a maximum equal to \ and at the point \(t = \large\frac(1)(3)\normalsize\) 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.) \ ] Find the stationary points of the function \(y\left(t \right):\) \[ (y"\left(t \right) = 0,)\;\; (\Rightarrow 3(t^2) + 4t - 4 = 0,)\;\;(\Rightarrow (t_(1,2)) = \frac(( - 4 \pm \sqrt (64) ))(6) = - 2;\;\frac(2) (3).) \] Here, similarly, the function \(y\left(t \right)\) reaches its maximum at the point \(t = -2:\) \ and its minimum at the point \(t = \large\frac (2)(3)\normalsize:\) \[ (y\left((\frac(2)(3)) \right) ) = ((\left((\frac(2)(3)) \righ t)^3) + 2(\left((\frac(2)(3)) \right)^2) - 4 \cdot \frac(2)(3) ) = (\frac(8)((27 )) + \frac(8)(9) - \frac(8)(3) ) = ( - \frac((40))((27)).) \] Graphs of functions \(x\left(t \ right)\), \(y\left(t \right)\) are schematically shown in the figure \(15a.\)
Fig.15a |
Fig.15b |
Fig.15c |
Note that since \[ (\lim\limits_(t \to \pm \infty ) x\left(t \right) = \pm \infty ,)\;\;\; (\lim\limits_(t \to \pm \infty ) y\left(t \right) = \pm \infty ,) \] then the curve \(y\left(x \right)\) has neither vertical, no horizontal asymptotes. Moreover, since \[ (k = \lim\limits_(t \to \pm \infty ) \frac((y\left(t \right)))((x\left(t \right))) ) = (\lim\limits_(t \to \pm \infty ) \frac(((t^3) + 2(t^2) - 4t))(((t^3) + (t^2) - t) ) ) = (\lim\limits_(t \to \pm \infty ) \frac((1 + \frac(2)(t) - \frac(4)(((t^2)))))(( 1 + \frac(1)(t) - \frac(1)(((t^2))))) = 1,) \] \[ (b = \lim\limits_(t \to \pm \infty ) \left[ (y\left(t \right) - kx\left(t \right)) \right] ) = (\lim\limits_(t \to \pm \infty ) \left((\cancel(\ color(blue)(t^3)) + \color(red)(2(t^2)) - \color(green)(4t) - \cancel(\color(blue)(t^3)) - \ color(red)(t^2) + \color(green)(t)) \right) ) = (\lim\limits_(t \to \pm \infty ) \left((\color(red)(t^ 2) - \color(green)(3t)) \right) = + \infty ,) \] then the curve \(y\left(x \right)\) also has no oblique asymptotes.
Let's determine the intersection points of the graph \(y\left(x \right)\) with the coordinate axes. The intersection with the x-axis occurs at the following points: \[ (y\left(t \right) = (t^3) + 2(t^2) - 4t = 0,)\;\; (\Rightarrow t\left(((t^2) + 2t - 4) \right) = 0;) \]
\(((t^2) + 2t - 4 = 0,)\;\; (\Rightarrow D = 4 - 4 \cdot \left(( - 4) \right) = 20,)\;\; (\ Rightarrow (t_(2,3)) = \large\frac(( - 2 \pm \sqrt (20) ))(2)\normalsize = - 1 \pm \sqrt 5 .) \)
\(((t^2) + t - 1 = 0,)\;\; (\Rightarrow D = 1 - 4 \cdot \left(( - 1) \right) = 5,)\;\; (\ Rightarrow (t_(2,3)) = \large\frac(( - 1 \pm \sqrt (5) ))(2)\normalsize.) \)
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 section of the decreasing curve \(y\left(x \right).\) It intersects the y-axis at the point \(\left((0,3 + 2\sqrt 5 ) \right).\)
On the third interval \(\left(( - 1,\large\frac(1)(3)\normalsize) \right)\) both variables decrease. \(x\) changes from \(x\left(( - 1) \right) = 1\) to \(x\left((\large\frac(1)(3)\normalsize) \right) = - \large\frac(5)((27))\normalsize.\) Accordingly, \(y\) decreases from \(y\left(( - 1) \right) = 5\) to \(y\ left((\large\frac(1)(3)\normalsize) \right) = - \large\frac(29)((27))\normalsize.\) Curve \(y\left(x \right)\ ) intersects the origin of coordinates.
On the fourth interval \(\left((\large\frac(1)(3)\normalsize,\large\frac(2)(3)\normalsize) \right)\) the variable \(x\) increases from \( x\left((\large\frac(1)(3)\normalsize) \right) = - \large\frac(5)((27))\normalsize\) to \(x\left((\large\ frac(2)(3)\normalsize) \right) = \large\frac(2)((27))\normalsize,\) and the variable \(y\) decreases from \(y\left((\large\ frac(1)(3)\normalsize) \right) = - \large\frac(29)((27))\normalsize\) to \(y\left((\large\frac(2)(3)\ normalsize) \right) = - \large\frac(40)((27))\normalsize.\) In this section, 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 \(\left((\large\frac(2)(3)\normalsize, + \infty ) \right)\) 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 refine the shape of the curve \(y\left(x \right)\), we calculate the maximum and minimum points. The derivative \(y"\left(x \right)\) is expressed as \[ (y"\left(x \right) = (y"_x) ) = (\frac(((y"_t)))( ((x"_t))) ) = (\frac((((\left(((t^3) + 2(t^2) - 4t) \right))^\prime )))(((( \left(((t^3) + (t^2) - t) \right))^\prime ))) ) = (\frac((3(t^2) + 4t - 4))((3 (t^2) + 2t - 1)) ) = (\frac((\cancel(3)\left((t + 2) \right)\left((t - \frac(2)(3)) \ right)))((\cancel(3)\left((t + 1) \right)\left((t - \frac(1)(3)) \right))) ) = (\frac((\ left((t + 2) \right)\left((t - \frac(2)(3)) \right)))((\left((t + 1) \right)\left((t - \ frac(1)(3)) \right))).) \] The change in the sign of the derivative \(y"\left(x \right)\) is shown in the figure \(15c.\) It can be seen that at the point \(t = - 2,\) i.e. on the boundary of the \(I\)th and \(II\)th intervals, the curve has a maximum, and for \(t = \large\frac(2)(3)\normalsize\) (on the boundary \(IV\) th and \(V\)th intervals) there is a minimum. When passing through the point \(t = \large\frac(1)(3)\normalsize\) the derivative also changes sign from plus to minus, but in this region the curve \(y\left(x \right)\) is not unambiguous function. Therefore, the indicated point is not an extremum.
We also investigate the convexity of this curve. Second derivative\(y""\left(x \right)\) has the form: \[ y""\left(x \right) = (y""_(xx)) = \frac((((\left(( (y"_x)) \right))"_t)))(((x"_t))) = \frac((((\left((\frac((3(t^2) + 4t - 4) )((3(t^2) + 2t - 1))) \right))^\prime )))((((\left(((t^3) + (t^2) - t) \right ))^\prime ))) = \frac((\left((6t + 4) \right)\left((3(t^2) + 2t - 1) \right) - \left((3(t ^2) + 4t - 4) \right)\left((6t + 2) \right)))((((\left((3(t^2) + 2t - 1) \right))^3) )) = \frac((18(t^3) + 12(t^2) + 12(t^2) + 8t - 6t - 4 - \left((18(t^3) + 24(t^2 ) - 24t + 6(t^2) + 8t - 8) \right)))((((\left((3(t^2) + 2t - 1) \right))^3))) = \ frac((\cancel(\color(blue)(18(t^3))) + \color(red)(24(t^2)) + \color(green)(2t) - \color(maroon)( 4) - \cancel(\color(blue)(18(t^3))) - \color(red)(30(t^2)) + \color(green)(16t) + \color(maroon)( 8)))((((\left((3(t^2) + 2t - 1) \right))^3))) = \frac(( - \color(red)(6(t^2) ) + \color(green)(18t) + \color(maroon)(4)))((((\left((3(t^2) + 2t - 1) \right))^3))) = \frac(( - 6\left((t - \frac((9 - \sqrt (105) ))(6)) \right)\left((t - \ frac((9 + \sqrt (105) ))(6)) \right)))((((\left((t + 1) \right))^3)((\left((3t - 1) \right))^3))). \] Consequently, the second derivative changes its sign to the opposite when passing through the following points (Fig.\(15с\)): \[ ((t_1) = - 1:\;\;x\left(( - 1) \right ) = 1,)\;\; (y\left(( - 1) \right) = 5;) \] \[ ((t_2) = \frac((9 - \sqrt (105) ))(6):)\;\; (x\left((\frac((9 - \sqrt (105) ))(6)) \right) \approx 0.24;)\;\; (y\left((\frac((9 - \sqrt (105) ))(6)) \right) \approx 0.91;) \] \[ ((t_3) = \frac(1)(3) :)\;\; (x\left((\frac(1)(3)) \right) = - \frac(5)((27)),)\;\; (y\left((\frac(1)(3)) \right) = - \frac((29))((27));) \] \[ ((t_4) = \frac((9 + \ sqrt(105) ))(6):)\;\; (x\left((\frac((9 + \sqrt (105) ))(6)) \right) \approx 40,1;)\;\; (y\left((\frac((9 + \sqrt (105) ))(6)) \right) \approx 40,8.) \] Therefore, these points are inflection points of the curve \(y\left(x \right).\)
A schematic plot of the curve \(y\left(x \right)\) is shown above in the figure \(15b.\)
