In this paper, we consider the scheme of the investigation of a function, and also give examples of research on extrema, monotonicity, and asymptotes of a given function.
- Area of existence (ODZ) function.
- Intersection of function (if any) with coordinate axes, function signs, parity, periodicity.
- Points of discontinuity (their kind). Continuity. Asymptotes are vertical.
- Monotonicity and extremum points.
- Inflection points. Convex.
- The study of the function at infinity, on the asymptotes: horizontal and oblique.
- Drawing a graph.
Study for monotony
Theorem. If the functiong continuous on[a, b]. differentiated by(a, b) andg? (x) ≥ 0 (g \u0026 rsquo; (x) ≤0). xє (a; b). theng ascending (decreasing) by[a, b] .
y = 1. 3x3 \u0026 ndash; 6. 2x 2 + 5x.
We find the intervals of constant signsy \u0026 rsquo;. Because they \u0026 rsquo; - an elementary function, then it can change signs only at points where it turns to zero or does not exist. Her ODZ:хєR .
Let us find the points whose derivative is equal to 0 (zero):
Research on extremes
T.x0 is called the maximum point (max) on the setA functiong when the value at the point is taken as the largestg (x0 ) ≥ g (x), xєA .
T.x0 are called the minimum point (min) of the functiong on the setA when the value at the lowest pointg (x0 ) ≤ g (x), xєA.
On the setA The points of maximum (max) and minimum (min) are called extremum pointsg. Such extrema are also called absolute extrema on a set.
Ifx0 - extremum point of the functiong in some of its constituencies, thenx0 is called a local or local extremum point (max or min) of the functiong.
Theorem (condition necessary). Ifx0 - extremum point (local) functiong. then the derivative does not exist or is equal to this zero (zero).
Definition. Critical are the points with a non-existent or equal to 0 (zero) derivative. It is these points that are suspicious of an extremum.
Theorem (sufficient condition No. 1). If the functiong is continuous in some neighborhood of m.x0 and the sign changes through this point on passing the derivative, then the given point is an m extremumg .
Theorem (sufficient condition No. 2). Let the function in a certain neighborhood of a point be differentiable twice andg \u0026 rsquo; = 0, and g \u0026 rsquo; \u0026 rsquo; \u0026 gt; 0 (g \u0026 rsquo; \u0026 rsquo; \u0026 lt; 0). then this point is the maximum (max) or minimum (min) of the function.
Study on convexity
A function is called convex down (or concave) on the interval(a, b) then, when the graph of the function is located not above the secant on the interval for any x with(a, b). which passes through these points.
The function will be convex strictly downwards on(a, b). if - the graph lies below the secant on the interval.
A function is called convex up (convex) on the interval(a, b). if for any mschech from(a, b) the graph of the function on the interval lies not less than the secant line passing through the abscissas at these points.
The function is strictly convex upwards on(a, b ), if - the graph on the interval lies above the secant.
If a function in a certain neighborhood of a point is continuous andt. x0 when the function changes the convexity, this point is called the inflection point of the function.
Investigation into asymptotes
Definition. A straight line is called an asymptoteg (x). if at an infinite distance from the origin of coordinates the point of the graph of the function approaches it:d (M, l).
Asymptotes can be vertical, horizontal and inclined.
Vertical straight line with equationx = x0 will be the asymptote of the vertical graph of g. if in t. x0 an infinite discontinuity, that is, at least one left or right boundary at this point is infinity.
Investigation of a function on a segment by the value of the smallest and largest
If the function is continuous on[a, b]. then by the Weierstrass theorem there exists a value of the greatest and the smallest value on this interval, that is, there exist mglasses that belong to[a, b] such thatg (x1 ) ≤ g (x) \u0026 lt; g (x2 ), x2 [[a, b]. From theorems on monotonicity and extremum, we obtain the following scheme for investigating the function on the segment by the smallest and largest value.
- Find the derivativeg \u0026 rsquo; (x) .
- Search for function valueg at these points and at the ends of the segment.
- Compare the values found and choose the smallest and largest.
Comment. If it is necessary to investigate a function on a finite interval(a, b). or on an infinite(- \u0026 gt;; b); (- ∞) + ∞) on max and min value, then in the plan instead of the values of the function at the ends of the interval, the corresponding one-sided boundaries are searched: instead off (a) are looking forf (a +) = limf (x). insteadf (b) are looking forf (-b). So you can find the ODZ functions on the gap, because absolute extremes do not necessarily exist in this case.
The application of the derivative to the solution of applied problems on the extremum of certain quantities
- Express this value through other quantities from the condition of the problem so that it is a function of only one variable (if possible).
- Determine the interval of change of this variable.
- Perform an investigation of the function on the gap by max and min values.
A task. It is necessary to build a square of rectangular shape, using a meter of grid, against the wall so that on one side it adjoins the wall, and from the other three it was fenced with a grid. At what ratio of sides will the area of such a site be the largest?
S = xy Is a function of 2 variables.
S = x (a-2x) - function of the 1st variable; x є [0; a: 2].
S = ax - 2x 2; S '= a - 4x = 0, xєR, x = a. 4.
S (a. 4) = a 2. 8 - the greatest value;
Let us find the other side of the rectangle:the= a. 2.
Aspect ratio:y. x = 2.
Answer. The greatest area will be equal toa 2/8. If the side that is parallel to the wall is 2 times larger than the other side.
Investigation of function. Examples
There isy = x 3. (1-x) 2. Do the research.
- LDU:xe (- ∞) 1 U (1; ∞).
- The general form of the function (neither even nor odd), with respect to the point 0 (zero), is not symmetric.
- Signs of function. The function is elementary, so it can change sign only at points where it is 0 (zero), or does not exist.
- The function is elementary, therefore continuous on the DGS:(- ∞) 1 U (1; ∞).
limx 3. (1 - x) 2 = \u0026 lt; - A break of the second kind (infinite), therefore there is a vertical asymptote at point 1;
x = 1 - the asymptote equation is vertical.
5. y \u0026 rsquo; = x 2 (3 - x). (1 - x) 3;
x = 1 - the point is critical.
0; 3 - points are critical.
6. y \u0026 rsquo; rsquo; = 6x. (1 - x) 4;
Critical t.1, 0;
7. limx 3. (1 - 2x + x 2) = \u0026 infin; - there is no horizontal asymptote, but it can be inclined.
k = 1 \u0026 ndash; number;
b = 2 \u0026 ndash; number.
Consequently, there is an asymptote obliquey = x + 2 at + \u0026 infin; and on -.
Giveny = (x2 + 1). (x-1). Produce and research. Construct a graph.
1. The domain of existence is the whole number line, except for m.x = 1.
2. y crosses OY (if possible) in t.(0; g (0)). We findy (0) = -1 - the intersection of OY.
The intersection points of the graph withOX we find, solving equationy = 0. The equation of roots has no real roots, so this function does not intersectOX .
3. The function is non-periodical. Consider the expression
g (-x) \u0026 lt; ne; g (x), and g (-x) ne; -g (x). This means that this is a general kind of function (neither even, nor odd).
4. T.x = 1 the gap is of the second kind. At all other points the function is continuous.
5. Investigation of the function at an extremum:
and solve equationy '= 0.
So,1 -? 2, 1 +? 2, 1 - critical points or points of possible extremum. These points divide the number line into four intervals.
At each interval, the derivative has a definite sign, which can be established by the method of intervals or the calculation of the values of the derivative at individual points. On the intervals(- ∞ ;;1 -? 2 ) U(1 + 2 ; ∞). positive derivative, so the function grows; ifxє(1 -? 2 ; 1) U(1; 1 + 2 ). then the function decreases, because at these intervals the derivative is negative. Through m.x1 at the transition (the movement follows from the left to the right) changes the derivative from "+" to "-", therefore, at this point there is a local maximum, we find
When passing throughx2 changes the derived sign from "-" to "+", therefore, at this point there is a local minimum, and
T.x = 1 not an extremum.
On(- ∞ ;;1 ) 0 > y " . consequently, on this interval the curve is convex; if xє(1 ; ∞) The curve is concave. In tpoint 1 A function is not defined, so this point is not an inflection point.
7. From the results of paragraph 4 it follows thatx = 1 - asymptote vertical curve.
There are no horizontal asymptotes.
x + 1 =y - the asymptote is oblique by a given curve. There are no other asymptotes.
8. Taking into account the conducted studies, we construct a graph (see the figure above).