Plot the function using the general study design. How to Examine and Graph a Function
Let us examine the function \ (y = \ frac (x ^ 3) (1-x) \) and build its graph.
1. Domain of definition.
The scope of rational function(fraction) will be: the denominator is not is zero, i.e. \ (1 -x \ ne 0 => x \ ne 1 \). Scope $$ D_f = (- \ infty; 1) \ cup (1; + \ infty) $$
2. Breakpoints of a function and their classification.
The function has one breakpoint x = 1
investigate the point x = 1. Find the limit of the function to the right and left of the discontinuity point, to the right $$ \ lim_ (x \ to 1 + 0) (\ frac (x ^ 3) (1-x)) = - \ infty $$ and to the left of the point $$ \ lim_ (x \ to 1-0) (\ frac (x ^ 3) (1-x)) = + \ infty $$ This is a discontinuity point of the second kind because one-sided limits are \ (\ infty \).
The straight line \ (x = 1 \) is the vertical asymptote.
3. The parity of the function.
Check for evenness \ (f (-x) = \ frac ((- x) ^ 3) (1 + x) \) the function is neither even nor odd.
4. Zeros of the function (points of intersection with the Ox axis). Sign intervals of the function.
The zeros of the function ( point of intersection with the Ox-axis): we equate \ (y = 0 \), we get \ (\ frac (x ^ 3) (1-x) = 0 => x = 0 \). The curve has one point of intersection with the Ox axis with coordinates \ ((0; 0) \).
Intervals of constancy of the function.
On the considered intervals \ ((- \ infty; 1) \ cup (1; + \ infty) \) the curve has one point of intersection with the Ox axis, therefore we will consider the domain of definition on three intervals.
Let us determine the sign of the function on the intervals of the domain of definition:
interval \ ((- \ infty; 0) \) find the value of the function at any point \ (f (-4) = \ frac (x ^ 3) (1-x)< 0 \), на этом интервале функция отрицательная \(f(x) < 0 \), т.е. находится ниже оси Ox
interval \ ((0; 1) \) find the value of the function at any point \ (f (0.5) = \ frac (x ^ 3) (1-x)> 0 \), on this interval the function is positive \ (f (x )> 0 \), i.e. is located above the Ox axis.
interval \ ((1; + \ infty) \) find the value of the function at any point \ (f (4) = \ frac (x ^ 3) (1-x)< 0 \), на этом интервале функция отрицательная \(f(x) < 0 \), т.е. находится ниже оси Ox
5. Points of intersection with the Oy axis: we equate \ (x = 0 \), we get \ (f (0) = \ frac (x ^ 3) (1-x) = 0 \). Coordinates of the point of intersection with the Oy axis \ ((0; 0) \)
6. Intervals of monotony. Function extrema.
Find the critical (stationary) points, for this we find the first derivative and equate it to zero $$ y "= (\ frac (x ^ 3) (1-x))" = \ frac (3x ^ 2 (1-x) + x ^ 3) ((1-x) ^ 2) = \ frac (x ^ 2 (3-2x)) ((1-x) ^ 2) $$ equate to 0 $$ \ frac (x ^ 2 (3 -2x)) ((1-x) ^ 2) = 0 => x_1 = 0 \ quad x_2 = \ frac (3) (2) $$ Find the value of the function at this point \ (f (0) = 0 \) and \ (f (\ frac (3) (2)) = -6.75 \). We got two critical points with coordinates \ ((0; 0) \) and \ ((1.5; -6.75) \)
Monotonic intervals.
The function has two critical points (points of possible extremum), therefore, monotonicity will be considered on four intervals:
interval \ ((- \ infty; 0) \) find the value of the first derivative at any point in the interval \ (f (-4) = \ frac (x ^ 2 (3-2x)) ((1-x) ^ 2)>
interval \ ((0; 1) \) find the value of the first derivative at any point in the interval \ (f (0.5) = \ frac (x ^ 2 (3-2x)) ((1-x) ^ 2)> 0 \) , the function increases on this interval.
interval \ ((1; 1.5) \) find the value of the first derivative at any point in the interval \ (f (1.2) = \ frac (x ^ 2 (3-2x)) ((1-x) ^ 2)> 0 \) , the function increases on this interval.
interval \ ((1.5; + \ infty) \) find the value of the first derivative at any point in the interval \ (f (4) = \ frac (x ^ 2 (3-2x)) ((1-x) ^ 2)< 0\), на этом интервале функция убывает.
Function extrema.
In the study of the function, two critical (stationary) points were obtained on the interval of the domain of definition. Determine if they are extrema. Let us consider the change in the sign of the derivative when passing through the critical points:
point \ (x = 0 \) derivative changes sign with \ (\ quad + \ quad 0 \ quad + \ quad \) - the point is not an extremum.
point \ (x = 1.5 \) derivative changes sign with \ (\ quad + \ quad 0 \ quad - \ quad \) - the point is a maximum point.
7. Intervals of convexity and concavity. Inflection points.
To find the intervals of convexity and concavity, we find the second derivative of the function and equate it to zero $$ y "" = (\ frac (x ^ 2 (3-2x)) ((1-x) ^ 2)) "= \ frac (2x (x ^ 2-3x + 3)) ((1-x) ^ 3) $$ Equate to zero $$ \ frac (2x (x ^ 2-3x + 3)) ((1-x) ^ 3) = 0 => 2x (x ^ 2-3x + 3) = 0 => x = 0 $$ The function has one critical point of the second kind with coordinates \ ((0; 0) \).
Let us define the convexity on the intervals of the domain of definition, taking into account the critical point of the second kind (the point of possible inflection).
interval \ ((- \ infty; 0) \) find the value of the second derivative at any point \ (f "" (- 4) = \ frac (2x (x ^ 2-3x + 3)) ((1-x) ^ 3)< 0 \), на этом интервале вторая производная функции отрицательная \(f""(x) < 0 \) - функция выпуклая вверх (вогнутая).
interval \ ((0; 1) \) find the value of the second derivative at any point \ (f "" (0.5) = \ frac (2x (x ^ 2-3x + 3)) ((1-x) ^ 3)> 0 \), on this interval the second derivative of the function is positive \ (f "" (x)> 0 \) the function is convex downward (convex).
interval \ ((1; \ infty) \) find the value of the second derivative at any point \ (f "" (4) = \ frac (2x (x ^ 2-3x + 3)) ((1-x) ^ 3)< 0 \), на этом интервале вторая производная функции отрицательная \(f""(x) < 0 \) - функция выпуклая вверх (вогнутая).
Inflection points.
Let us consider the change in the sign of the second derivative upon passing through the critical point of the second kind:
At the point \ (x = 0 \) the second derivative changes sign from \ (\ quad - \ quad 0 \ quad + \ quad \), the graph of the function changes convexity, i.e. it is an inflection point with coordinates \ ((0; 0) \).
8. Asymptotes.
Vertical asymptote... The graph of the function has one vertical asymptote \ (x = 1 \) (see item 2).
Oblique asymptote.
In order for the graph of the function \ (y = \ frac (x ^ 3) (1-x) \) with \ (x \ to \ infty \) to have an oblique asymptote \ (y = kx + b \), it is necessary and sufficient so that there are two limits $$ \ lim_ (x \ to + \ infty) = \ frac (f (x)) (x) = k $$ find it $$ \ lim_ (x \ to \ infty) (\ frac ( x ^ 3) (x (1-x))) = \ infty => k = \ infty $$ and second limit $$ \ lim_ (x \ to + \ infty) (f (x) - kx) = b $ $, because \ (k = \ infty \) - there is no oblique asymptote.
Horizontal asymptote: for the horizontal asymptote to exist, the limit must exist $$ \ lim_ (x \ to \ infty) f (x) = b $$ find it $$ \ lim_ (x \ to + \ infty) (\ frac ( x ^ 3) (1-x)) = - \ infty $$$$ \ lim_ (x \ to - \ infty) (\ frac (x ^ 3) (1-x)) = - \ infty $$
There is no horizontal asymptote.
9. Function graph.
Do a full research and plot the function
y (x) = x2 + 81 − x.y (x) = x2 + 81 − x.
1) Function definition area. Since the function is a fraction, you need to find the zeros of the denominator.
1 − x = 0, ⇒x = 1.1 − x = 0, ⇒x = 1.
We exclude the only point x = 1x = 1 from the domain of the function and get:
D (y) = (- ∞; 1) ∪ (1; + ∞). D (y) = (- ∞; 1) ∪ (1; + ∞).
2) Let us investigate the behavior of the function in the vicinity of the discontinuity point. Let's find one-sided limits:
Since the limits are equal to infinity, the point x = 1x = 1 is a discontinuity of the second kind, the straight line x = 1x = 1 is the vertical asymptote.
3) Let's define the points of intersection of the graph of the function with the coordinate axes.
Find the points of intersection with the ordinate axis OyOy, for which we equate x = 0x = 0:
Thus, the point of intersection with the OyOy axis has coordinates (0; 8) (0; 8).
Find the points of intersection with the abscissa axis OxOx, for which we put y = 0y = 0:
The equation has no roots, so there are no points of intersection with the OxOx axis.
Note that x2 + 8> 0x2 + 8> 0 for any xx. Therefore, for x∈ (−∞; 1) x∈ (−∞; 1) the function y> 0y> 0 (takes positive values, the graph is located above the abscissa axis), for x∈ (1; + ∞) x∈ (1; + ∞) the function y<0y<0 (принимает отрицательные значения, график находится ниже оси абсцисс).
4) The function is neither even nor odd because:
5) Let us examine the function for periodicity. The function is not periodic, since it is a fractional rational function.
6) Let us examine the function for extrema and monotonicity. To do this, we find the first derivative of the function:
Let us equate the first derivative to zero and find stationary points (at which y ′ = 0y ′ = 0):
We got three critical points: x = −2, x = 1, x = 4x = −2, x = 1, x = 4. We split the entire domain of the function into intervals with given points and determine the signs of the derivative in each interval:
For x∈ (−∞; −2), (4; + ∞) x∈ (−∞; −2), (4; + ∞) the derivative y ′<0y′<0, поэтому функция убывает на данных промежутках.
For x∈ (−2; 1), (1; 4) x∈ (−2; 1), (1; 4) the derivative y> 0y> 0, the function increases on these intervals.
In this case, x = −2x = −2 is a local minimum point (the function decreases and then increases), x = 4x = 4 is a local maximum point (the function increases and then decreases).
Let's find the values of the function at these points: 
Thus, the minimum point is (−2; 4) (- 2; 4), the maximum point is (4; −8) (4; −8).
7) Let us examine the function for inflections and convexity. Let's find the second derivative of the function:
Let's equate the second derivative to zero:
The resulting equation has no roots, so there are no inflection points. Moreover, when x∈ (−∞; 1) x∈ (−∞; 1) y ′ ′> 0y ″> 0 holds, that is, the function is concave when x∈ (1; + ∞) x∈ (1; + ∞) y ′ ′<0y″<0, то есть функция выпуклая.
8) Let us investigate the behavior of the function at infinity, that is, at.
Since the limits are infinite, there are no horizontal asymptotes.
Let's try to determine the oblique asymptotes of the form y = kx + by = kx + b. We calculate the values of k, bk, b according to the well-known formulas:
We got that the function has one oblique asymptote y = −x − 1y = −x − 1.
9) Additional points. Let's calculate the value of the function at some other points in order to build a graph more accurately.
y (−5) = 5.5; y (2) = - 12; y (7) = - 9.5.y (−5) = 5.5; y (2) = - 12; y (7) = - 9.5.
10) Based on the data obtained, we construct a graph, supplement it with the asymptotes x = 1x = 1 (blue), y = −x − 1y = −x − 1 (green) and mark the characteristic points (purple intersection with the ordinate axis, orange extrema, black additional points) :
Task 4: Geometric, Economic problems (I have no idea which ones, here is an approximate selection of problems with a solution and formulas) 
Example 3.23. a
Solution. x and y y
y = a - 2 × a / 4 = a / 2. Since x = a / 4 is the only critical point, let us check whether the sign of the derivative changes when passing through this point. For xa / 4 S "> 0, and for x> a / 4 S"< 0, значит, в точке x=a/4 функция S имеет максимум. Значение функции S(a/4) = a/4(a - a/2) = a 2 /8 (кв. ед).Поскольку S непрерывна на и ее значения на концах S(0) и S(a/2) равны нулю, то найденное значение будет наибольшим значением функции. Таким образом, наиболее выгодным соотношением сторон площадки при данных условиях задачи является y = 2x.
Example 3.24.
Solution.
R = 2, H = 16/4 = 4.
Example 3.22. Find the extrema of the function f (x) = 2x 3 - 15x 2 + 36x - 14.
Solution. Since f "(x) = 6x 2 - 30x +36 = 6 (x -2) (x - 3), then the critical points of the function x 1 = 2 and x 2 = 3. Extrema can be only at these points. So as when passing through the point x 1 = 2 the derivative changes its sign plus to minus, then the function has a maximum at this point. When passing through the point x 2 = 3, the derivative changes its sign minus to plus, so at the point x 2 = 3 the function has a minimum. Calculating the values of the function in points
x 1 = 2 and x 2 = 3, we find the extrema of the function: maximum f (2) = 14 and minimum f (3) = 13.
Example 3.23. It is necessary to build a rectangular area near the stone wall so that on three sides it is fenced off with a wire mesh, and on the fourth side it is adjacent to the wall. For this there is a running meters of mesh. At what aspect ratio will the site have the largest area?
Solution. We denote the sides of the site by x and y... The area of the site is S = xy. Let be y is the length of the side adjacent to the wall. Then, by condition, the equality 2x + y = a must be fulfilled. Therefore, y = a - 2x and S = x (a - 2x), where
0 ≤ x ≤ a / 2 (the length and width of the site cannot be negative). S "= a - 4x, a - 4x = 0 for x = a / 4, whence
y = a - 2 × a / 4 = a / 2. Since x = a / 4 is the only critical point, let us check whether the sign of the derivative changes when passing through this point. For xa / 4 S "> 0, and for x> a / 4 S"< 0, значит, в точке x=a/4 функция S имеет максимум. Значение функции S(a/4) = a/4(a - a/2) = a 2 /8 (кв. ед).Поскольку S непрерывна на и ее значения на концах S(0) и S(a/2) равны нулю, то найденное значение будет наибольшим значением функции. Таким образом, наиболее выгодным соотношением сторон площадки при данных условиях задачи является y = 2x.
Example 3.24. It is required to manufacture a closed cylindrical tank with a capacity of V = 16p ≈ 50 m 3. What are the dimensions of the tank (radius R and height H) so that the least amount of material is used to make it?
Solution. The total surface area of the cylinder is S = 2pR (R + H). We know the volume of the cylinder V = pR 2 H Þ H = V / pR 2 = 16p / pR 2 = 16 / R 2. Hence, S (R) = 2p (R 2 + 16 / R). Find the derivative of this function:
S "(R) = 2p (2R- 16 / R 2) = 4p (R- 8 / R 2). S" (R) = 0 when R 3 = 8, therefore,
R = 2, H = 16/4 = 4.
Similar information.
For a complete study of the function and plotting its graph, it is recommended to use the following scheme:
1) find the domain of the function;
2) find the points of discontinuity of the function and vertical asymptotes (if they exist);
3) investigate the behavior of the function at infinity, find horizontal and oblique asymptotes;
4) investigate the function for evenness (oddness) and periodicity (for trigonometric functions);
5) find the extrema and intervals of monotonicity of the function;
6) determine the intervals of convexity and points of inflection;
7) find the points of intersection with the coordinate axes, if possible, and some additional points that refine the graph.
The study of the function is carried out simultaneously with the construction of its graph.
Example 9 Explore the function and plot the graph.
1. Scope of definition:;
2. The function is broken at points 
,
;
Let us examine the function for the presence of vertical asymptotes.
;
,
─ vertical asymptote.
;
,
─ vertical asymptote.
3. Let us investigate the function for the presence of oblique and horizontal asymptotes.
Straight 
─ oblique asymptote if 
,
.
,
.
Straight 
─ horizontal asymptote.
4. The function is even because 
... The parity of the function indicates the symmetry of the graph about the ordinate axis.
5. Find the intervals of monotonicity and extrema of the function.
Let's find the critical points, i.e. points at which the derivative is 0 or does not exist: 
;
... We have three points 
;
... These points split the entire valid axis into four spaces. Let's define the signs 
on each of them.
On the intervals (-∞; -1) and (-1; 0) the function increases, on the intervals (0; 1) and (1; + ∞) ─ decreases. When crossing a point 
the derivative changes sign from plus to minus, therefore, at this point the function has a maximum 
.
6. Find the convexity intervals, inflection points.
Find the points at which 
is 0, or does not exist.
has no valid roots. 
,
,
Points 
and 
split the real axis into three intervals. Let's define the sign 
at each interval.
Thus, the curve at intervals 
and 
convex downward, on the interval (-1; 1) convex upward; there are no inflection points, since the function at the points 
and 
unspecified.
7. Find the points of intersection with the axes.
With axis 
the graph of the function intersects at the point (0; -1), and with the axis 
the graph does not overlap, because the numerator of this function has no real roots.
The graph of the given function is shown in Figure 1.
Figure 1 ─ Function graph 
Application of the concept of a derivative in economics. Elasticity of function
To study economic processes and solve other applied problems, the concept of elasticity of a function is often used.
Definition. Elasticity of function 
is called the limit of the ratio of the relative increment of the function 
to the relative increment of the variable 
at 
,. (Vii)
The elasticity of a function shows an approximate percentage of the change in the function 
when changing the independent variable 
by 1%.
The elasticity of the function is applied in the analysis of demand and consumption. If the elasticity of demand (in absolute value) 
, then the demand is considered elastic if 
─ neutral if 
─ inelastic with respect to price (or income).
Example 10 Calculate the elasticity of a function 
and find the value of the elasticity index for 
= 3.
Solution: according to formula (VII) elasticity of function:
Let x = 3, then 
This means that if the explanatory variable increases by 1%, then the value of the dependent variable increases by 1.42%.
Example 11 Let the demand function 
regarding the price 
has the form 
, where 
─ constant coefficient. Find the value of the elasticity index of the demand function at a price x = 3 den. units
Solution: calculate the elasticity of the demand function by formula (VII)
Assuming 
monetary units, we get 
... This means that at a price 
monetary units a 1% price increase will cause a 6% decrease in demand, i.e. demand is elastic.
Today we invite you to explore and graph the function with us. After carefully studying this article, you will not have to sweat for a long time to complete this kind of task. Exploring and plotting a function is not easy, the work is voluminous, requiring maximum attention and accuracy of calculations. To facilitate the perception of the material, we will study the same function step by step, explain all our actions and calculations. Welcome to the amazing and exciting world of mathematics! Go!
Domain
In order to explore and plot a function, you need to know several definitions. Function is one of the basic (basic) concepts in mathematics. It reflects the relationship between several variables (two, three or more) with changes. The function also shows the dependence of the sets.
Imagine that we have two variables that have a certain range of variation. So, y is a function of x, provided that each value of the second variable corresponds to one value of the second. In this case, the variable y is dependent, and it is called a function. It is customary to say that the variables x and y are in. For greater clarity of this dependence, a function graph is plotted. What is a function graph? This is a set of points on the coordinate plane, where each value of x corresponds to one value of y. Graphs can be different - straight line, hyperbola, parabola, sinusoid and so on.
It is impossible to plot a function graph without research. Today we will learn how to conduct research and plot a function graph. It is very important to make notes during the research. This will make the task much easier. The most convenient research plan:
- Domain.
- Continuity.
- Even or odd parity.
- Periodicity.
- Asymptotes.
- Zeros.
- Constancy of signs.
- Increasing and decreasing.
- Extremes.
- Convexity and concavity.
Let's start with the first point. Let's find the domain of definition, that is, on what intervals our function exists: y = 1/3 (x ^ 3-14x ^ 2 + 49x-36). In our case, the function exists for any values of x, that is, the domain is equal to R. It can be written as follows xÎR.
Continuity
Now we are going to investigate the break function. In mathematics, the term "continuity" appeared as a result of the study of the laws of motion. What is infinite? Space, time, some dependencies (an example is the dependence of the variables S and t in motion problems), the temperature of the heated object (water, frying pan, thermometer, etc.), a continuous line (that is, one that can be drawn without tearing it off the sheet pencil).
A chart is considered to be continuous if it does not break at some point. One of the clearest examples of such a graph is a sine wave, which you can see in the picture in this section. The function is continuous at some point x0 if a number of conditions are met:
- a function is defined at this point;
- the right and left limits at the point are equal;
- the limit is equal to the value of the function at the point x0.
If at least one condition is not met, the function is said to be broken. And the points at which the function is discontinuous are usually called discontinuity points. An example of a function that will "break" when displayed graphically is: y = (x + 4) / (x-3). Moreover, y does not exist at the point x = 3 (since it is impossible to divide by zero).
In the function that we are examining (y = 1/3 (x ^ 3-14x ^ 2 + 49x-36)), everything turned out to be simple, since the graph will be continuous.
Even, odd
Now examine the function for parity. First, a little theory. An even function is one that satisfies the condition f (-x) = f (x) for any value of the variable x (from the range of values). Examples include:
- module x (the graph looks like a daw, the bisector of the first and second quarters of the graph);
- x squared (parabola);
- cosine x (cosine).
Note that all of these plots are symmetrical when viewed in relation to the ordinate (i.e. y).
What, then, is called an odd function? These are those functions that satisfy the condition: f (-x) = - f (x) for any value of the variable x. Examples:
- hyperbola;
- cubic parabola;
- sinusoid;
- tangentoid and so on.
Please note that these functions are symmetrical about the point (0: 0), that is, the origin. Based on what has been said in this section of the article, an even and an odd function must have the property: x belongs to the set of definitions and -x too.
Let us examine the function for parity. We can see that it doesn't fit any of the descriptions. Therefore, our function is neither even nor odd.
Asymptotes
Let's start with the definition. An asymptote is a curve that is as close to the graph as possible, that is, the distance from a point tends to zero. In total, there are three types of asymptotes:
- vertical, that is, parallel to the y-axis;
- horizontal, that is, parallel to the x axis;
- inclined.
As for the first type, data straight lines should be looked for at some points:
- break;
- ends of the domain of definition.
In our case, the function is continuous, and the domain is equal to R. Therefore, there are no vertical asymptotes.
The graph of a function has a horizontal asymptote, which meets the following requirement: if x tends to infinity or minus infinity, and the limit is equal to a certain number (for example, a). In this case, y = a - this is the horizontal asymptote. In the function we are investigating, there are no horizontal asymptotes.
The oblique asymptote only exists if two conditions are met:
- lim (f (x)) / x = k;
- lim f (x) -kx = b.
Then it can be found by the formula: y = kx + b. Again, in our case there are no oblique asymptotes.
Function zeros
The next step is to examine the graph of the function at zeros. It is also very important to note that the task associated with finding the zeros of a function occurs not only in the study and plotting of a function graph, but also as an independent task and as a way to solve inequalities. You may be required to find the zeros of a function on a graph, or use mathematical notation.
Finding these values will help you plot the function more accurately. In simple terms, the zero of a function is the value of the variable x, at which y = 0. If you are looking for the zeros of a function on a graph, then you should pay attention to the points at which the graph crosses the abscissa axis.
To find the zeros of a function, you need to solve the following equation: y = 1/3 (x ^ 3-14x ^ 2 + 49x-36) = 0. After performing the necessary calculations, we get the following answer:
Constancy
The next stage of research and construction of a function (graph) is to find intervals of constancy. This means that we must determine at which intervals the function takes a positive value, and at which - negative. The function zeros found in the previous section will help us to do this. So, we need to build a straight line (separately from the graph) and distribute the zeros of the function from smallest to largest on it in the correct order. Now you need to determine which of the resulting intervals has a "+" sign, and which "-".
In our case, the function takes a positive value in the intervals:
- from 1 to 4;
- from 9 to infinity.
Negative meaning:
- from minus infinity to 1;
- 4 to 9.
This is easy to define. Plug in any number from the interval into the function and see what sign the answer is (minus or plus).
Increasing and decreasing functions
In order to explore and build a function, we need to find out where the graph will increase (go up along Oy), and where it will fall (crawl down along the ordinate).
The function increases only if the larger value of the variable x corresponds to the larger value of y. That is, x2 is greater than x1, and f (x2) is greater than f (x1). And we observe a completely opposite phenomenon in a decreasing function (the more x, the less y). To determine the intervals of increase and decrease, you need to find the following:
- scope (we already have it);
- derivative (in our case: 1/3 (3x ^ 2-28x + 49);
- Solve the equation 1/3 (3x ^ 2-28x + 49) = 0.
After the calculations, we get the result:
We get: the function increases in the intervals from minus infinity to 7/3 and from 7 to infinity, and decreases in the interval from 7/3 to 7.
Extremes
The investigated function y = 1/3 (x ^ 3-14x ^ 2 + 49x-36) is continuous and exists for any values of the variable x. The extremum point shows the maximum and minimum of this function. In our case, there are none, which greatly simplifies the construction task. Otherwise, they are also found using the derivative of the function. After finding, do not forget to mark them on the chart.
Convexity and concavity
We continue to investigate the function y (x) further. Now we need to check it for convexity and concavity. The definitions of these concepts are quite difficult to perceive, it is better to analyze everything with examples. For the test: the function is convex if it is a non-decreasing function. Agree, this is incomprehensible!
We need to find the derivative of a second-order function. We get: y = 1/3 (6x-28). Now let's set the right side to zero and solve the equation. Answer: x = 14/3. We found the inflection point, that is, the place where the graph changes from convexity to concavity, or vice versa. In the interval from minus infinity to 14/3, the function is convex, and from 14/3 to plus infinity, it is concave. It is also very important to note that the inflection point on the graph should be smooth and soft, no sharp corners should be present.
Definition of additional points
Our task is to investigate and plot the function. We have finished the research, it will not be difficult to plot the function now. For more accurate and detailed reproduction of a curve or straight line on the coordinate plane, you can find several auxiliary points. It is quite easy to calculate them. For example, we take x = 3, solve the resulting equation, and find y = 4. Or x = 5 and y = -5 and so on. You can take as many additional points as you need to build. At least 3-5 of them are found.
Plotting a graph
We needed to investigate the function (x ^ 3-14x ^ 2 + 49x-36) * 1/3 = y. All the necessary notes during the calculations were made on the coordinate plane. All that remains to be done is to build a graph, that is, to connect all the points to each other. Connecting the dots should be smooth and neat, it's a matter of skill - a little practice and your schedule will be perfect.
y = _______ ax n + b cx m + d
And build its graph according to the general scheme.
The general study of functions and the construction of graphs is performed according to the following scheme:
- Find the domain of the function.
- Find out if the function is even, odd, periodic.
- Examine the function for continuity, find the points of discontinuity and find out the nature of the discontinuities.
- Find the asymptotes of the graph of the function.
- Find the extremum points of the function, calculate the values of the function at these points. Set the intervals of monotonicity of the function at these points.
- Find the inflection points of the graph of the function, calculate the values of the function and the values of the derivative at these points. Set the intervals of the convexity of the graph of the function.
- Using the research results, build a graph of the function. If you need to refine individual parts of the curve, calculate the coordinates of several additional points. In particular, it is recommended to calculate the coordinates of the points of intersection of the graph with the coordinate axes, the so-called "zeros" of the function.
Set the numerical parameters of your variant and press the "Enter."
y = ______ a x n + b c x m + d
Exponents n and m must be whole positive single digits. Odds a, b, c, d can take any integer values from the range [-99.99]. If there is a "-" in front of the fraction, refer it to the numerator. Do not get carried away with too large and small values of the coefficients. Remember that "infinity" will not fit on the screen.
a = b = c = d =
n = m =
Let's apply this scheme to the function
y = _____ 2x 3 x 2 − 4
(a = 2; b = 0; c = 1; d = −4; n = 3; m = 2).
1.
The function is defined on the entire number axis, except for points x
= ± 2, where the denominator of the fraction vanishes. Thus, its domain of definition
D
(f
) = (−∞;−2)∪(−2;+2)∪(+2;+∞)
.
2.
The function is odd because
,
therefore, its graph will be symmetrical about the origin, so it is enough to examine the function in the interval)
