Quite often in the know mathematical analysis You can find a task with the following wording: “examine a function and plot a graph”. This formulation speaks for itself and breaks the task into two stages:
- Stage 1: study of function;
- Stage 2: plotting the graph of the function under study.
The first stage is the most extensive and includes finding domains of definition and values, extrema of the function, inflection points of the graph, etc.
A complete research plan for the function $y=f(x)$, preceding the goal of plotting a graph, has the following points:
- Search for the domain of definition of the function $D_(y) $ and the domain of permissible values $E_(y) $ of the function.
- Determining the type of function: even, odd, general view.
- Determining the points of intersection of the graph of a function with the coordinate axes.
- Finding asymptotes of the graph of a function (vertical, inclined, horizontal).
- Finding intervals of monotonicity of a function and extremum points.
- Finding the intervals of convexity, concavity of the graph and inflection points.
Finding the domain of definition of the function $D_(y) $ implies finding the intervals on which this function exists (is defined). As a rule, this task comes down to finding the VA (region of permissible values), on the basis of which $D_(y) $ is formed.
Example 1
Find the domain of definition of the function $y=\frac(x)(x-1) $.
Let us find the ODZ of the function under consideration, i.e. values of a variable at which the denominator does not go to zero.
ODZ: $x-1\ne 0\Rightarrow x\ne 1$
Let's write down the domain of definition: $D_(y) =\( x\in R|x\ne 1\) $.
Definition 1
The function $y=f(x)$ is even if it holds the following equality$f(-x)=f(x)$ $\forall x\in D_(y) $.
Definition 2
The function $y=f(x)$ is odd if the following equality holds: $f(-x)=-f(x)$ $\forall x\in D_(y) $.
Definition 3
A function that is neither even nor odd is called a function of general form.
Example 2
Determine the type of functions: 1) $y=\frac(x)(x-1) $, 2) $y=\frac(x^(2) )(x^(2) -1) $; 3) $y=\frac(x)(x^(2) -1) $.
1) $y=\frac(x)(x-1) $
$f(-x)\ne f(x);f(-x)\ne -f(x)$, therefore, we have a function of general form.
2) $y=\frac(x^(2) )(x^(2) -1) $
$f(-x)=f(x)$, therefore, we have an even function.
3) $y=\frac(x)(x^(2) -1) $.
$f(-x)\ne -f(x)$, therefore, we have an odd function.
Determining the intersection points of the function graph with the coordinate axes includes finding the intersection points: with the OX axis ($y=0$), with the OY axis ($x=0$).
Example 3
Find the points of intersection with the coordinate axes of the function $y=\frac(x+2)(x-1) $.
- with OX axis ($y=0$)
$\frac(x+2)(x-1) =0\Rightarrow x+2=0\Rightarrow x=-2$; we get the point (-2;0)
- with OY axis ($x=0$)
$y(0)=\frac(0+2)(0-1) =-2$, we get the point (0;-2)
Based on the results obtained at the function research stage, a graph is constructed. Sometimes the points obtained at the first stage are not enough to construct a graph of a function; then it is necessary to find additional points.
Example 4
Explore the function and build its graph: $y=x^(3) -6x^(2) +2x+1$.
- Domain: $D_(y) =\( x|x\in R\) $.
- Range of values: $E_(y) =\( y|y\in R\) $.
- Even, odd function :\ \
A general function, i.e. is neither even nor odd.
4) Intersection with coordinate axes:
with the OY axis: $y(0)=0^(3) -6\cdot 0^(2) +2\cdot 0+1=1$, therefore, the graph passes through the point (0;1).
with the OX axis: $x^(3) -6x^(2) +2x+1=0$ (no rational roots)
5) Asymptotes of the graph:
There are no vertical asymptotes, since $D_(y) =\( x|x\in R\) $
We will look for oblique asymptotes in the form $y=kx+b$.
$k=\mathop(\lim )\limits_(x\to \infty ) \frac(y(x))(x) =\mathop(\lim )\limits_(x\to \infty ) \frac(x^ (3) -6x^(2) +2x+1)(x) =\infty $. Therefore, there are no oblique asymptotes.
6) Increasing, decreasing functions; extremes:
\ \[\begin(array)(l) (y"=0\Rightarrow 3x^(2) -12x+2=0) \\ (D=144-24=120) \\ (x_(1,2) =\frac(12\pm \sqrt(120) )(6) ) \end(array)\]
Let's mark the points on the number axis, arrange the signs of the first derivative and note the behavior of the function:
Picture 1.
The function increases by $\left(-\infty ;\frac(12-\sqrt(120) )(6) \right]$ and $\left[\frac(12+\sqrt(120) )(6) ;\ infty \right)$, decreases by $\left[\frac(12-\sqrt(120) )(6) ;\frac(12+\sqrt(120) )(6) \right]$.
$x=\frac(12-\sqrt(120) )(6) $ - maximum point; $y\left(\frac(12-\sqrt(120) )(6) \right)=1,172$
$x=\frac(12+\sqrt(120) )(6) $ - minimum point; $y\left(\frac(12+\sqrt(120) )(6) \right)=-23,172$
7) Convexity, concavity of the graph:
\ \[\begin(array)(l) (y""=(3x^(2) -12x+2)"=6x-12) \\ (y""=0\Rightarrow 6x-12=0\Rightarrow x=2) \end(array)\]
Let's mark the points on the number axis, arrange the signs of the second derivative and note the behavior of the function graph:
Figure 2.
The graph is convex upward by $(-\infty ;2]$, downwards by $
8) Function graph:
Figure 3.
The function research process consists of several stages. For the most complete understanding of the behavior of the function and the nature of its graph, it is necessary to find:
The domain of existence of a function.
This concept includes both the domain of values and the domain of definition of a function.
Breaking points. (If available).
Intervals of increasing and decreasing.
Maximum and minimum points.
The maximum and minimum value of a function on its domain of definition.
Areas of convexity and concavity.
Inflection points (if any).
Asymptotes (if any).
Building a graph.
Let's look at the application of this scheme using an example.
Example. Explore the function and construct its graph.
We find the domain of existence of the function. It's obvious that domain of definition function is the area (-; -1) (-1; 1) (1; ).
In turn, it is clear that the straight lines x = 1, x = -1 are vertical asymptotes crooked.
Range of values of this function is the interval (-; ).
Break points functions are points x = 1, x = -1.
We find critical points.
Let's find the derivative of the function
Critical points: x = 0; x = -;x = ;x = -1; x = 1.
Let's find the second derivative of the function
Let us determine the convexity and concavity of the curve at intervals.
- < x < -,y < 0, кривая выпуклая
- 1 < x
< 0, y
>0, concave curve 0 < x
< 1, y
< 0, кривая выпуклая 1 < x
<
,y
>0, concave curve < x
< ,
y
>0, concave curve Finding the gaps increasing And descending functions. To do this, we determine the signs of the derivative of the function on intervals. -
< x
< -,y
>0, function is increasing - 1 < x
< 0, y
< 0, функция убывает 0 < x
< 1, y
< 0, функция убывает 1 < x
<
,y
< 0, функция убывает < x
< ,
y
>0, function is increasing It can be seen that the point x = - is a point maximum, and point x = is a point minimum. The function values at these points are equal to 3/2 and -3/2, respectively. About vertical asymptotes has already been said above. Now let's find oblique asymptotes. In total, the equation of the oblique asymptote is y = x. Let's build schedule Features: Below we will consider several examples of studying various types of functions using differential calculus methods. Example: Methods of differential calculus 1. The domain of definition of this function is all real numbers (-; ). 3. Points of intersection with coordinate axes: with Oy axis: x = 0; y = 1; with the Ox axis: y = 0; x = 1; 4. Breakpoints and asymptotes: There are no vertical asymptotes. Sloping asymptotes: general equation y = kx + b; Total: y = -x – oblique asymptote. 5. Increasing and decreasing function, extremum points. It can be seen that y 0 for any x 0, therefore, the function decreases over the entire domain of definition and has no extrema. At the point x = 0, the first derivative of the function is equal to zero, but at this point the decrease does not change to an increase, therefore, at the point x = 0 the function most likely has an inflection. To find inflection points, we find the second derivative of the function. y = 0 for x =0 and y = for x = 1. Points (0,1) and (1,0) are inflection points, because y(1-h)< 0; y(1+h)
>0; y(-h) > 0; y(h)< 0 для любого h
> 0. 6. Let's build a graph of the function. Example: Explore the function and construct its graph. 1. The domain of definition of the function is all values of x except x = 0. 2. The function is a function of general form in the sense of even and odd. 3. Points of intersection with coordinate axes: with the Ox axis: y = 0; x = with Oy axis: x = 0; y – does not exist. 4. The point x = 0 is a discontinuity point, therefore, the straight line x = 0 is a vertical asymptote. We look for oblique asymptotes in the form: y = kx + b. Oblique asymptote y = x. 5. Find the extremum points of the function. ; y = 0 for x = 2, y = for x = 0. y > 0 for x (-, 0) – the function increases, y< 0 при х
(0, 2) – функция убывает, y > 0 for x (2, ) – the function increases. Thus, point (2, 3) is a minimum point. To determine the nature of the convexity/concavity of a function, we find the second derivative. > 0 for any x 0, therefore, the function is concave throughout the entire domain of definition. 6. Let's build a graph of the function. Example: Explore the function and construct its graph. The domain of definition of this function is the interval x (-, ). In the sense of even and odd, the function is a function of general form. Points of intersection with the coordinate axes: with the Oy axis: x = 0, y = 0; with the Ox axis: y = 0, x = 0, x = 1. Asymptotes of the curve. There are no vertical asymptotes. Let's try to find oblique asymptotes in the form y = kx + b. Finding extremum points. To find critical points you should solve the equation 4x 3 – 9x 2 + 6x –1 = 0. To do this, let us factorize this polynomial of the third degree. By selection we can determine that one of the roots of this equation is the number x = 1. Then: 4x 3 – 9x 2 + 6x – 1 x - 1 4x 3 – 4x 2 4x 2 – 5x + 1 Then we can write (x – 1)(4x 2 – 5x + 1) = 0. Finally, we obtain two critical points: x = 1 and x = ¼. Note. The operation of dividing polynomials could be avoided if, when finding the derivative, we used the formula for the derivative of a product: Let's find the second derivative of the function: 12x 2 – 18x + 6. Equating to zero, we find: Let's systematize the information received in the table: issue down increases issue down increases issue up increases issue down Let's build a graph of the function. We offer to your attention a service for constructing graphs of functions online, all rights to which belong to the company Desmos. Use the left column to enter functions. You can enter manually or using the virtual keyboard at the bottom of the window. To enlarge the window with the graph, you can hide both the left column and the virtual keyboard. With us it’s easy to build charts of varying complexity online. Construction is done instantly. The service is in demand for finding intersection points of functions, for depicting graphs for further moving them into a Word document as illustrations when solving problems, and for analyzing the behavioral features of function graphs. The optimal browser for working with charts on this website page is Google Chrome. Correct operation is not guaranteed when using other browsers.
- There are no oblique asymptotes.
Build function
Benefits of online charting