The calculator uses functions using differential calculus methods. Complete function study example online

Instructions

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

Identify areas of continuity and points of discontinuity. Typically a function is continuous in the same region where it is defined. To detect discontinuities, one must calculate as the argument approaches isolated points within 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 isolated point it is not defined.

Find vertical asymptotes, if any. The calculations from the previous step will help you here, since the vertical asymptote is almost always located at the discontinuity point of the second kind. However, sometimes it is not individual points that are excluded from the definition domain, 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 periodicity.
The function will be even if for any x in the domain f(x) = f(-x). For example, cos(x) and x^2 - even functions.

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

Find the points. To do this, calculate the derivative of given function and find those values ​​of x where it becomes zero. For example, the function f(x) = x^3 + 9x^2 -15 has a derivative g(x) = 3x^2 + 18x, which 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 at the found zeros. g(x) changes sign from plus at the point x = -6, and at the point x = 0 back from minus to plus. Consequently, the function f(x) has a minimum at the first point and a minimum at the second.

Thus, you have also found regions of monotonicity: f(x) monotonically increases on the interval -∞;-6, monotonically decreases by -6;0 and increases again by 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 goes to zero at x = -3, changing sign from minus to plus. Consequently, the graph of 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 besides 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 horizontal asymptote.

Oblique asymptote- straight line of the form kx + b. To find k, calculate the limit of f(x)/x as x→∞. To find the b - limit (f(x) – kx) for the same x→∞.

Plot a graph of the function based on the calculated data. Label the asymptotes, if any. Mark the extremum points and the function values ​​at them. For greater graph accuracy, calculate the function values ​​at several more intermediate points. The study is completed.

One of most important tasks differential calculus is the development common examples studies of function behavior.

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

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 has a discontinuity are called critical.

Theorem 1 (1st sufficient condition 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 interval and 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 coincide, 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 satisfied, then x 0 is the maximum point; if the derivative changes sign from minus to plus (to the right of x 0 executed f"(x)<0, то х 0 - точка минимума.

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

Theorem 2 (a necessary sign of 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 the 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 at which the function is continuous and the derivative is zero or does not exist.
3) Consider the neighborhood of each point, 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, substitute the values ​​of the critical points into this function. Using sufficient conditions for the extremum, draw the appropriate conclusions.

Example 18. Examine the function y=x 3 -9x 2 +24x for an extremum

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; This means that apart from the two points found, 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 through the point x=4 - from minus to plus.
4) At point x=2 the function has a maximum y max =20, and at point x=4 - a minimum y min =16.

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

Let f"(x 0) and at the point x 0 there exists f""(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 a segment, the function y=f(x) can reach the smallest (y the least) or the greatest (y the highest) value either at the critical points of the function lying in the interval (a;b), or at the ends of the segment.

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=f(x) at the points obtained in step 2), as well as at the ends of the segment and select the largest and smallest from them: they are, respectively, the largest (y the largest) and the smallest (y the least) values ​​of the function on the interval.

Example 19. Find the largest value of the 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 at which y"=0; we get:
3x 2 -6x-45=0
x 2 -2x-15=0
x 1 =-3; x 2 =5
3) Calculate the value of the function at points x=0 y=225, x=5 y=50, x=6 y=63
The segment contains only the point x=5. The largest of the found values ​​of the function is 225, and the smallest is the number 50. So, y max = 225, y min = 50.

Study of a function on convexity

The figure shows graphs of two functions. The first of them is convex upward, the second is convex downward.

The function y=f(x) is continuous on the segment and differentiable in the interval (a;b), is called convex upward (downward) 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 holds on the interval (a;b), then the function is convex downward 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 the points at which f""(x) does not exist or vanishes.
2) Examine the sign f""(x) to the left and right of each point found in the first step.
3) Based on Theorem 4, draw a conclusion.

Example 20. Find the extremum points and inflection points of the graph of the function 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 when x 1 =0, x 2 =1. When passing through the point x=0, the derivative changes sign from minus to plus, but 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 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 upward) and x=1 is also the inflection point (transition from convexity upward to convexity downward). If x=, then y=; if, then x=1, y=13.

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 a 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 a 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 the asymptote for a function

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

Scheme for studying a function and constructing its graph

I. Find the domain of definition of the function.
II. Find the intersection points of the function graph with the coordinate axes.
III. Find asymptotes.
IV. Find possible extremum points.
V. Find critical points.
VI. Using the auxiliary figure, explore the sign of the first and second derivatives. Determine the areas of increase and decrease of the function, find the direction of convexity of the graph, points of extrema and inflection points.
VII. Construct a graph, taking into account the research carried out in paragraphs 1-6.

Example 22: Construct a graph of the function according to the above diagram

Solution.
I. The domain of a function is the set of all real numbers except x=1.
II. Since the equation x 2 +1=0 has no real roots, the graph of the function has no 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. Let us study the behavior of the function near the discontinuity point x=1. Since y → ∞ as x → -∞, y → +∞ as x → 1+, then the line x=1 is the 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 obtain two possible extremum points:
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. Let us examine 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 domain 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 find that the function increases at (-∞;1-√2), decreases at (1-√2;1+√2), and increases again at (1+√2;+∞). Extremum 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. At (-∞;1) the graph is convex upward, and at (1;+∞) it is convex downwards.
VII Let's make a table of the obtained values

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

The reference points when studying functions and constructing their graphs are characteristic points - points of discontinuity, extremum, inflection, intersection with coordinate axes. Using differential calculus, it is possible to establish the characteristic features of changes in functions: increase and decrease, maximums and minimums, the direction of convexity and concavity of the graph, the presence of asymptotes.

A sketch of the graph of the function can (and should) be drawn after finding the asymptotes and extremum points, and it is convenient to fill out the summary table of the study of the function as the study progresses.

The following function study scheme is usually used.

1.Find the domain of definition continuity intervals And function break points .

2.Examine the function for evenness or oddness (axial or central symmetry of the graph.

3.Find asymptotes(vertical, horizontal or inclined).

4.Find and explore intervals of increasing and decreasing functions, its points extremum.

5.Find intervals convexity and concavity of a curve, its inflection points.

6.Find the intersection points of the curve with the coordinate axes, if they exist.

7.Compile a summary table of the study.

8.A graph is constructed, taking into account the study of the function carried out according to the points described above.

Example. Explore function

and build its graph.

7. Let’s compile a summary table for studying the function, where we will enter all the characteristic points and the intervals between them. Taking into account the parity of the function, we obtain the following table:

To fully study the function and plot its graph, it is recommended to use the following scheme:

1) find the domain of definition of the function;

2) find the discontinuity points of the function and vertical asymptotes (if they exist);

3) investigate the behavior of the function at infinity, find horizontal and oblique asymptotes;

4) examine the function for parity (oddness) and periodicity (for trigonometric functions);

5) find extrema and intervals of monotonicity of the function;

6) determine the convexity intervals and inflection points;

7) find the points of intersection with the coordinate axes, and, if possible, some additional points that clarify the graph.

The study of the function is carried out simultaneously with the construction of its graph.

Example 9 Explore the function and build a graph.

1. Scope of definition: ;

2. The function suffers discontinuity at points
,
;

We examine the function for the presence of vertical asymptotes.

;
,
─ vertical asymptote.

;
,
─ vertical asymptote.

3. We examine 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 relative to the ordinate.

5. Find the monotonicity intervals 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 divide the entire real axis into four intervals. 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; +∞) ─ it decreases. When passing through a point
the derivative changes sign from plus to minus, therefore, at this point the function has a maximum
.

6. Find the intervals of convexity and inflection points.

Let's find the points at which is 0, or does not exist.

has no real roots.
,
,

Points
And
divide the real axis into three intervals. Let's define the sign at every interval.

Thus, the curve on the intervals
And
convex downwards, on the interval (-1;1) convex upwards; there are no inflection points, since the function is at points
And
not defined.

7. Find the points of intersection with the axes.

With axle
the graph of the function intersects at the point (0; -1), and with the axis
the graph does not intersect, 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 derivative in economics. Elasticity function

To study economic processes and solve other applied problems, the concept of elasticity of a function is often used.

Definition. Elasticity 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 approximately how many percent the function will change
when the independent variable changes by 1%.

The elasticity function is used in the analysis of demand and consumption. If the elasticity of demand (in absolute value)
, then demand is considered elastic if
─ neutral if
─ inelastic relative to price (or income).

Example 10 Calculate the elasticity of the function
and find the value of the elasticity index for = 3.

Solution: according to formula (VII), the elasticity of the function is:

Let x=3, then
.This means that if the independent variable increases by 1%, then the value of the dependent variable will increase by 1.42%.

Example 11 Let the demand function regarding price looks like
, Where ─ constant coefficient. Find the value of the elasticity indicator of the demand function at price x = 3 den. units

Solution: calculate the elasticity of the demand function using formula (VII)

Believing
monetary units, we get
. This means that at a price
monetary units a 1% increase in price will cause a 6% decrease in demand, i.e. demand is elastic.