Conduct a complete study of the functions of the calculator. Function research online

Instruction

Find the scope of the function. For example, the function sin(x) is defined on the entire interval from -∞ to +∞, and the function 1/x is defined from -∞ to +∞, except for the point x = 0.

Define areas of continuity and break points. Usually a function is continuous in the same domain where it is defined. To detect discontinuities, you need to calculate when the argument approaches isolated points inside the domain of definition. For example, the function 1/x tends to infinity when x→0+ and to minus infinity when x→0-. This means that at the point x = 0 it has a discontinuity of the second kind.
If the limits at the discontinuity point are finite but not equal, then this is a discontinuity of the first kind. If they are equal, then the function is considered continuous, although it is not defined at an isolated point.

Find the vertical asymptotes, if any. The calculations from the previous step will help you here, since the vertical asymptote is almost always at the discontinuity point of the second kind. However, sometimes not individual points are excluded from the domain of definition, but entire intervals of points, and then the vertical asymptotes can be located at the edges of these intervals.

Check if the function has special properties: even, odd, and periodic.
The function will be even if for any x in the domain f(x) = f(-x). For example, cos(x) and x^2 are even functions.

Periodicity is a property that says that there is a certain number T called a period, which for any x f(x) = f(x + T). For example, all basic trigonometric functions (sine, cosine, tangent) are periodic.

Find points. To do this, calculate the derivative of the given function and find those x values ​​where it vanishes. For example, the function f(x) = x^3 + 9x^2 -15 has a derivative g(x) = 3x^2 + 18x that vanishes at x = 0 and x = -6.

To determine which extremum points are maxima and which are minima, track the change in the signs of the derivative in the found zeros. g(x) changes sign from plus at x = -6 and back from minus to plus at x = 0. Therefore, the function f(x) has a minimum at the first point and a minimum at the second.

Thus, you have also found areas of monotonicity: f(x) increases monotonically on the interval -∞;-6, decreases monotonically on -6;0 and increases again on 0;+∞.

Find the second derivative. Its roots will show where the graph of a given function will be convex, and where it will be concave. For example, the second derivative of the function f(x) will be h(x) = 6x + 18. It vanishes at x = -3, changing its sign from minus to plus. Therefore, the graph f (x) before this point will be convex, after it - concave, and this point itself will be an inflection point.

A function may have other asymptotes, except for vertical ones, but only if its domain of definition includes . To find them, calculate the limit of f(x) when x→∞ or x→-∞. If it is finite, then you have found the horizontal asymptote.

The oblique asymptote is a straight line of the form kx + b. To find k, calculate the limit of f(x)/x as x→∞. To find b - limit (f(x) – kx) with the same x→∞.

Plot the function on the computed data. Label the asymptotes, if any. Mark the extremum points and the function values ​​in them. For greater accuracy of the graph, calculate the function values ​​at several more intermediate points. Research completed.

One of the most important tasks of differential calculus is the development of general examples of the study of the behavior of functions.

If the function y \u003d f (x) is continuous on the segment , and its derivative is positive or equal to 0 on the interval (a, b), then y \u003d f (x) increases by (f "(x) 0). If the function y \u003d f (x) is continuous on the segment , and its derivative is negative or equal to 0 on the interval (a,b), then y=f(x) decreases by (f"(x)0)

The intervals in which the function does not decrease or increase are called intervals of monotonicity of the function. The nature of the monotonicity of a function can change only at those points of its domain of definition, at which the sign of the first derivative changes. The points at which the first derivative of a function vanishes or breaks are called critical points.

Theorem 1 (1st sufficient condition for the existence of an extremum).

Let the function y=f(x) be defined at the point x 0 and let there be a neighborhood δ>0 such that the function is continuous on the segment , differentiable on the interval (x 0 -δ, x 0)u(x 0 , x 0 + δ) , and its derivative retains a constant sign on each of these intervals. Then if on x 0 -δ, x 0) and (x 0, x 0 + δ) the signs of the derivative are different, then x 0 is an extremum point, and if they match, then x 0 is not an extremum point. Moreover, if, when passing through the point x0, the derivative changes sign from plus to minus (to the left of x 0, f "(x)> 0 is performed, then x 0 is the maximum point; if the derivative changes sign from minus to plus (to the right of x 0 is executed by f"(x)<0, то х 0 - точка минимума.

The maximum and minimum points are called the extremum points of the function, and the maxima and minima of the function are called its extreme values.

Theorem 2 (necessary criterion for a local extremum).

If the function y=f(x) has an extremum at the current x=x 0, then either f'(x 0)=0 or f'(x 0) does not exist.
At the extremum points of a differentiable function, the tangent to its graph is parallel to the Ox axis.

Algorithm for studying a function for an extremum:

1) Find the derivative of the function.
2) Find critical points, i.e. points where the function is continuous and the derivative is zero or does not exist.
3) Consider the neighborhood of each of the points, and examine the sign of the derivative to the left and right of this point.
4) Determine the coordinates of the extreme points, for this value of the critical points, substitute into this function. Using sufficient extremum conditions, draw appropriate conclusions.

Example 18. Investigate the function y=x 3 -9x 2 +24x

Solution.
1) y"=3x 2 -18x+24=3(x-2)(x-4).
2) Equating the derivative to zero, we find x 1 =2, x 2 =4. In this case, the derivative is defined everywhere; hence, apart from the two found points, there are no other critical points.
3) The sign of the derivative y "=3(x-2)(x-4) changes depending on the interval as shown in Figure 1. When passing through the point x=2, the derivative changes sign from plus to minus, and when passing through the point x=4 - from minus to plus.
4) At the point x=2, the function has a maximum y max =20, and at the point x=4 - a minimum y min =16.

Theorem 3. (2nd sufficient condition for the existence of an extremum).

Let f "(x 0) and f "" (x 0) exist at the point x 0. Then if f "" (x 0)> 0, then x 0 is the minimum point, and if f "" (x 0)<0, то х 0 – точка максимума функции y=f(x).

On the segment, the function y \u003d f (x) can reach the smallest (at least) or largest (at most) value either at the critical points of the function lying in the interval (a; b), or at the ends of the segment.

The algorithm for finding the largest and smallest values ​​of a continuous function y=f(x) on the segment :

1) Find f "(x).
2) Find the points at which f "(x) = 0 or f" (x) - does not exist, and select from them those that lie inside the segment.
3) Calculate the value of the function y \u003d f (x) at the points obtained in paragraph 2), as well as at the ends of the segment and choose the largest and smallest of them: they are, respectively, the largest (for the largest) and the smallest (for the smallest) function values ​​on the interval .

Example 19. Find the largest value of a continuous function y=x 3 -3x 2 -45+225 on the segment .

1) We have y "=3x 2 -6x-45 on the segment
2) The derivative y" exists for all x. Let's find the points where y"=0; we get:
3x2 -6x-45=0
x 2 -2x-15=0
x 1 \u003d -3; x2=5
3) Calculate the value of the function at the points x=0 y=225, x=5 y=50, x=6 y=63
Only the point x=5 belongs to the segment. The largest of the found values ​​of the function is 225, and the smallest is the number 50. So, at max = 225, at max = 50.

Investigation of a function on convexity

The figure shows the graphs of two functions. The first of them is turned with a bulge up, the second - with a bulge down.

The function y=f(x) is continuous on the segment and differentiable in the interval (a;b), is called convex up (down) on this segment, if for axb its graph lies no higher (not lower) than the tangent drawn at any point M 0 (x 0 ;f(x 0)), where axb.

Theorem 4. Let the function y=f(x) have a second derivative at any interior point x of the segment and be continuous at the ends of this segment. Then if the inequality f""(x)0 is satisfied on the interval (а;b), then the function is downward convex on the interval ; if the inequality f""(x)0 holds on the interval (a;b), then the function is convex upward on .

Theorem 5. If the function y=f(x) has a second derivative on the interval (a;b) and if it changes sign when passing through the point x 0 , then M(x 0 ;f(x 0)) is an inflection point.

Rule for finding inflection points:

1) Find points where f""(x) does not exist or vanishes.
2) Examine the sign f""(x) to the left and right of each point found at the first step.
3) Based on Theorem 4, draw a conclusion.

Example 20. Find extremum points and inflection points of the function graph y=3x 4 -8x 3 +6x 2 +12.

We have f"(x)=12x 3 -24x 2 +12x=12x(x-1) 2. Obviously, f"(x)=0 for x 1 =0, x 2 =1. The derivative, when passing through the point x=0, changes sign from minus to plus, and when passing through the point x=1, it does not change sign. This means that x=0 is the minimum point (y min =12), and there is no extremum at the point x=1. Next, we find . The second derivative vanishes at the points x 1 =1, x 2 =1/3. The signs of the second derivative change as follows: On the ray (-∞;) we have f""(x)>0, on the interval (;1) we have f""(x)<0, на луче (1;+∞) имеем f""(x)>0. Therefore, x= is the inflection point of the function graph (transition from convexity down to convexity up) and x=1 is also an inflection point (transition from convexity up to convexity down). If x=, then y= ; if, then x=1, y=13.

An algorithm for finding the asymptote of a graph

I. If y=f(x) as x → a , then x=a is a vertical asymptote.
II. If y=f(x) as x → ∞ or x → -∞ then y=A is the horizontal asymptote.
III. To find the oblique asymptote, we use the following algorithm:
1) Calculate . If the limit exists and is equal to b, then y=b is the horizontal asymptote; if , then go to the second step.
2) Calculate . If this limit does not exist, then there is no asymptote; if it exists and is equal to k, then go to the third step.
3) Calculate . If this limit does not exist, then there is no asymptote; if it exists and is equal to b, then go to the fourth step.
4) Write down the equation of the oblique asymptote y=kx+b.

Example 21: Find an asymptote for a function

1)
2)
3)
4) The oblique asymptote equation has the form

The scheme of the study of the function and the construction of its graph

I. Find the domain of the function.
II. Find the points of intersection of the graph of the function with the coordinate axes.
III. Find asymptotes.
IV. Find points of possible extremum.
V. Find critical points.
VI. Using the auxiliary drawing, investigate the sign of the first and second derivatives. Determine the areas of increase and decrease of the function, find the direction of the convexity of the graph, extremum points and inflection points.
VII. Build a graph, taking into account the study conducted in paragraphs 1-6.

Example 22: Plot a function graph according to the above scheme

Solution.
I. The domain of the function is the set of all real numbers, except for x=1.
II. Since the equation x 2 +1=0 does not have real roots, then the graph of the function does not have points of intersection with the Ox axis, but intersects the Oy axis at the point (0; -1).
III. Let us clarify the question of the existence of asymptotes. We investigate the behavior of the function near the discontinuity point x=1. Since y → ∞ for x → -∞, y → +∞ for x → 1+, then the line x=1 is a vertical asymptote of the graph of the function.
If x → +∞(x → -∞), then y → +∞(y → -∞); therefore, the graph does not have a horizontal asymptote. Further, from the existence of limits

Solving the equation x 2 -2x-1=0, we get two points of a possible extremum:
x 1 =1-√2 and x 2 =1+√2

V. To find the critical points, we calculate the second derivative:

Since f""(x) does not vanish, there are no critical points.
VI. We investigate the sign of the first and second derivatives. Possible extremum points to be considered: x 1 =1-√2 and x 2 =1+√2, divide the area of ​​existence of the function into intervals (-∞;1-√2),(1-√2;1+√2) and (1+√2;+∞).

In each of these intervals, the derivative retains its sign: in the first - plus, in the second - minus, in the third - plus. The sequence of signs of the first derivative will be written as follows: +, -, +.
We get that the function on (-∞;1-√2) increases, on (1-√2;1+√2) it decreases, and on (1+√2;+∞) it increases again. Extreme points: maximum at x=1-√2, and f(1-√2)=2-2√2, minimum at x=1+√2, and f(1+√2)=2+2√2. On (-∞;1) the graph is convex upwards, and on (1;+∞) - downwards.
VII Let's make a table of the obtained values

VIII Based on the data obtained, we build a sketch of the graph of the function

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