In this article, we will look at the scheme for studying a function, and also give examples of studies on extremes, monotony, and asymptotes of this function.
- Area of existence (DHS) function.
- The intersection of the function (if any) with the axes of coordinates, signs of the function, parity, periodicity.
- Break points (their genus). Continuity. Vertical asymptotes.
- Monotony and extremum points.
- Points of inflection Convex.
- The study of the function at infinity, on asymptotes: horizontal and oblique.
Theorem. If the functiong continuous on[a, b]. differentiated by(a; b) andg \u0026 rsquo; (x) ≥ 0 (g \u0026 rsquo; (x) ≤0). xє (a; b). thatg increasing (decreasing) by[a, b] .
y = 1. 3x 3 \u0026 ndash; 6. 2x 2 + 5x.
Find the intervals of permanent characters.y \u0026 rsquo;. Insofar asy \u0026 rsquo; - an elementary function, it can change signs only at points where it turns into zero or does not exist. Her DHS:xєR .
Find the points in which the derivative is 0 (zero):
T.x0 called the maximum point (max) on the setBUT functionsg when the value is accepted at this point as the largestg (x0 ) ≥ g (x), xєA .
T.x0 called the minimum point (min) functiong on setBUT when the smallest value is taken at this point as a functiong (x0 ) ≤ g (x), xєA.
On setBUT maximum (max) and minimum (min) points are called extremum pointsg. Such extremes are also called absolute extremes on the set.
If ax0 - extremum point functiong in some district thenx0 referred to as a local or local extremum point (max or min) functiong.
Theorem (necessary condition). If ax0 - point of extremum (local) functiong. then the derivative does not exist or is equal in this t. 0 (zero).
Definition Critical points are points with a nonexistent or equal to 0 (zero) derivative. These data points are suspicious at the extremum.
Theorem (condition no. 1). If the functiong continuous in some district t.x0 and the sign changes through this point during the transition, then this point is the extremumg .
Theorem (condition sufficient number 2). Let the function in a district be differentiated 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 point of maximum (max) or minimum (min) function.
The function is called convex down (or concave) on the interval(a, b) when the graph of the function is not higher than the secant on the interval for any x with(a, b). which passes through these points.
The function will be convex strictly down on(a, b). if - the chart lies below the secant on the interval.
The function is called convex up (convex) on the gap(a, b). if for any tglasses with(a, b) the graph of the function on the interval is not lower than the secant passing through the abscissas at these points.
The function will be strictly convex up on(a, b ), if - the graph on the interval lies above the secant.
If the function in some district of the point is continuous and throught. x0 during the transition, the function changes the convexity, then this point is called the inflection point of the function.
Definition Direct is called asymptote.g (x). if at an infinite distance from the origin of coordinates the point of the function graph approaches it:d (M, l).
Asymptotes can be vertical, horizontal and oblique.
Vertical line with equationx = x0 will be the asymptote of the vertical graph of the function g. if in t. x0 infinite discontinuity, that is, at least one left or right boundary at this point is infinity.
The study of the function on the segment on the value of the smallest and largest
If the function is continuous on[a, b]. then by the Weierstrass theorem there is the largest value and the smallest value on this segment, that is, there are tpoints that belong[a, b] such thatg (x1 ) ≤ g (x) \u0026 lt; g (x2 ), x2 є [a, b]. From the theorems on monotonicity and extrema, we obtain the following scheme for studying a function on a segment for the smallest and largest value.
- Find derivativeg \u0026 rsquo; (x) .
- Search function valueg at these points and at the ends of the segment.
- The values found compare and select the smallest and largest.
Comment. If you want to make a study of the function on a finite interval(a, b). or at infinite(- \u0026 infin ;; b); (- \u0026 infin ;; + \u0026 infin;) on max and min value, then in the plan, instead of the values of the function at the ends of the gap, they search for the corresponding one-sided boundaries: instead off (a) are looking forf (a +) = limf (x). insteadf (b) are looking forf (-b). So you can find the LDU function on the interval, because absolute extremes do not necessarily exist in this case.
Application of the derivative to the solution of applied problems on the extremum of certain quantities
- Express this value in terms of other values from the condition of the problem so that it is a function of only one variable (if possible).
- Determine the period of change of this variable.
- Conduct a study of the function on the interval at max and min values.
Task. It is necessary to build a rectangular platform, using grid meters, against the wall so that on one side it fits to the wall, and on the other three it is fenced with a grid. At what aspect ratio will the area of such a site be greatest?
S = xy - 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. four.
S (a. 4) = a 2. 8 - the greatest value;
Find the other side of the rectangle:at= a. 2
Aspect ratio:y. x = 2.
Answer. The largest area will be equal toa 2/8. if the side that is parallel to the wall is 2 times larger than the other side.
Research function. Examples
There isy = x 3. (1-x) 2. Perform research.
- DHS:xє (- inf; 1) U (1; inf);
- The general form of the function (neither even nor odd), is not symmetric about the point 0 (zero).
- Signs of function. The function is elementary, so it can only change sign at the points where it is 0 (zero), or does not exist.
- The function is elementary, therefore continuous on the DHS:(- \u0026 infin ;; 1) U (1; \u0026 infin;).
limx 3. (1- x) 2 = \u0026 infin; - Gap of the 2nd kind (infinite), therefore there is a vertical asymptote at point 1;
x = 1 - the equation of the vertical asymptote.
5. y \u0026 rsquo; = x 2 (3 - x). (1 - x) 3;
x = 1 - The point is critical.
0; 3 - critical points.
6. y \u0026 rsquo; \u0026 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 may be oblique.
k = 1 \u0026 ndash; number;
b = 2 \u0026 ndash; number.
Therefore, there is an asymptote inclinedy = x + 2 by + \u0026 infin; and on - \u0026 infin ;.
Giveny = (x 2 + 1). (x - 1). Produce and investigation Build a graph.
1. The domain of existence is the whole numerical line, except for m.x = 1.
2. y Crosses OY (if possible) in t.(0; g (0)). Findy (0) = -1 - t. Intersection OY.
Points of intersection of the graph withOX find by solving the equationy = 0. The root equation has no real, therefore this function does not intersectOX .
3. The function is non-periodic. Consider the expression
g (-x) ne g (x) and g (-x)? ne -g (x). This means that this is a generic function (neither even nor odd).
4. T.x = 1 the gap has a second kind. At all other points, the function is continuous.
5. The study of the function on the extremum:
and solve the equationy ’= 0.
So,1 - \u0026 radic; 2, 1 + \u0026 radic; 2, 1 - Critical points or points of possible extremum. These points break the number line into four intervals..
At each interval, the derivative has a certain sign, which can be set by the method of intervals or by calculating the values of the derivative at individual points. At intervals(- \u0026 infin ;;1 - \u0026 radic; 2 ) U(1 + rad 2 ; \u0026 infin;). positive derivative, so the function grows; if axє(1 - \u0026 radic; 2 ; 1) U(1; 1 + rad 2 ). then the function decreases, because at these intervals the derivative is negative. Through t.x1 during the transition (the movement follows from left to right) changes the derivative sign from “+” to “-“, therefore, at this point there is a local maximum, we find
When going throughx2 changes the derivative sign from “-” to “+”, therefore, at this point there is a local minimum, and
T.x = 1 not t extremum.
On(- \u0026 infin ;;1 ) 0 > y ” . consequently, on this interval the curve is convex; if xє(1 ; \u0026 infin;) - The curve is concave. In t1 point no function is defined, so this point is not an inflection point.
7. From the results of paragraph 4 it follows thatx = 1 - asymptote vertical curve.
Horizontal asymptotes are absent.
x + 1 =y - the asymptote is inclined by this curve. There are no other asymptotes.
8. Considering the conducted studies, we build a graph (see the figure above).