Draw a graph using the derivative. How to Conduct a Full Function Study

For a complete study of the function and plotting its graph, it is recommended to use the following scheme:

1) find the scope 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 the horizontal and oblique asymptotes;

4) investigate the function for evenness (oddity) and for periodicity (for trigonometric functions);

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

6) determine the intervals of convexity and inflection points;

7) find points of intersection with the coordinate axes, if possible, and some additional points that refine 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. Domain of definition: ;

2. The function breaks at points
,
;

We investigate the function for the presence of vertical asymptotes.

;
,
─ vertical asymptote.

3. We investigate 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 with respect to the y-axis.

5. Find the intervals of monotonicity and extrema of the function.

Let's find the critical points, i.e. points where 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. Let's find convexity intervals, inflection points.

Let's find the points where 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 at the points
and
not determined.

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 ─ Graph of the function

Application of the concept of derivative in economics. Function elasticity

To study economic processes and solve other applied tasks The concept of elasticity of a function is often used.

Definition. Function elasticity
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 changing the independent variable by 1%.

The elasticity of a 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 with respect to price (or income).

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

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

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 the price has the form
, where ─ constant coefficient. Find the value of the elasticity index of the demand function at the price x = 3 den. units

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

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

Let's examine the function $y= \frac(x^3)(1-x) $ and build its graph.

1. Domain of definition.
The domain of definition of a rational function (fraction) will be: the denominator is not equal to zero, i.e. $1 -x \ne 0 => x \ne 1$. Domain $$D_f= (-\infty; 1) \cup (1;+\infty)$$

2. Breakpoints of a function and their classification.
The function has one break point x = 1
examine the point x= 1. Find the limit of the function to the right and left of the discontinuity point, to the right $$ \lim_(x \to 1+0) (\frac(x^3)(1-x)) = -\infty $$ and to the left of the point $$ \lim_(x \to 1-0)(\frac(x^3)(1-x)) = +\infty $$ one-sided limits are $\infty$.

The straight line $x = 1$ is a vertical asymptote.

3. Evenness of the function.
Checking for parity $f(-x) = \frac((-x)^3)(1+x) $ the function is neither even nor odd.

4. Zeros of the function (points of intersection with the Ox axis). Function constancy intervals.
Function zeros ( point of intersection with the Ox axis): equate $y=0$, we get $\frac(x^3)(1-x) = 0 => x=0 $. The curve has one point of intersection with the Ox axis with coordinates $(0;0)$.

Function constancy intervals.
On the considered intervals $(-\infty; 1) \cup (1;+\infty)$ the curve has one point of intersection with the axis Ox , so we will consider the domain of definition on three intervals.

Let us determine the sign of the function on the intervals of the domain of definition:
interval $(-\infty; 0) $ find the value of the function at any point $f(-4) = \frac(x^3)(1-x)< 0 $, на этом интервале функция отрицательная $f(x) < 0 $, т.е. находится ниже оси Ox
interval $(0; 1) $ find the value of the function at any point $f(0.5) = \frac(x^3)(1-x) > 0 $, on this interval the function is positive $f(x ) > 0 $, i.e. is above the x-axis.
interval $(1;+\infty) $ find the value of the function at any point $f(4) = \frac(x^3)(1-x)< 0 $, на этом интервале функция отрицательная $f(x) < 0 $, т.е. находится ниже оси Ox

5. Points of intersection with the axis Oy: equate $x=0 $, we get $f(0) = \frac(x^3)(1-x) = 0$. Coordinates of the point of intersection with the Oy axis $(0; 0)$

6. Intervals of monotonicity. Function extremes.
Let's find the critical (stationary) points, for this we find the first derivative and equate it to zero $$ y" = (\frac(x^3)(1-x))" = \frac(3x^2(1-x) + x^3)( (1-x)^2) = \frac(x^2(3-2x))( (1-x)^2) $$ equate to 0 $$ \frac(x^2(3 -2x))( (1-x)^2) = 0 => x_1 = 0 \quad x_2= \frac(3)(2)$$ Find the value of the function at this point $f(0) = 0$ and $f(\frac(3)(2)) = -6.75$. Got two critical points with coordinates $(0;0)$ and $(1.5;-6.75)$

Intervals of monotonicity.
The function has two critical points (possible extremum points), so we will consider monotonicity on four intervals:
interval $(-\infty; 0) $ find the value of the first derivative at any point of the interval $f(-4) = \frac(x^2(3-2x))( (1-x)^2) >
interval \((0;1)$ find the value of the first derivative at any point of the interval $f(0.5) = \frac(x^2(3-2x))( (1-x)^2) > 0$ , the function increases on this interval.
interval $(1;1.5)$ find the value of the first derivative at any point of the interval $f(1.2) = \frac(x^2(3-2x))( (1-x)^2) > 0$ , the function increases on this interval.
interval $(1.5; +\infty)$ find the value of the first derivative at any point of the interval $f(4) = \frac(x^2(3-2x))( (1-x)^2)< 0$, на этом интервале функция убывает.

Function extremes.

In the study of the function, two critical (stationary) points were obtained on the interval of the domain of definition. Let us determine whether they are extremums. Consider the change in the sign of the derivative when passing through the critical points:

point $x = 0$ derivative changes sign from $\quad +\quad 0 \quad + \quad$ - the point is not an extremum.
the point $x = 1.5$ the derivative changes sign from $\quad +\quad 0 \quad - \quad$ - the point is the maximum point.

7. Intervals of convexity and concavity. Inflection points.

To find the intervals of convexity and concavity, we find the second derivative of the function and equate it to zero $$y"" = (\frac(x^2(3-2x))( (1-x)^2))"= \frac(2x (x^2-3x+3))((1-x)^3) $$Set $$ equal to zero \frac(2x(x^2-3x+3))((1-x)^3)= 0 => 2x(x^2-3x+3) =0 => x=0$$ The function has one critical point of the second kind with coordinates $(0;0)$.
Let us define the convexity on the intervals of the domain of definition, taking into account the critical point of the second kind (the point of possible inflection).

interval $(-\infty; 0)$ find the value of the second derivative at any point $f""(-4) = \frac(2x(x^2-3x+3))((1-x)^ 3)< 0 $, на этом интервале вторая производная функции отрицательная $f""(x) < 0 $ - функция выпуклая вверх (вогнутая).
interval $(0; 1)$ find the value of the second derivative at any point $f""(0.5) = \frac(2x(x^2-3x+3))((1-x)^3) > 0 $, on this interval the second derivative of the function is positive $f""(x) > 0 $ the function is downward convex (convex).
interval $(1; \infty)$ find the value of the second derivative at any point $f""(4) = \frac(2x(x^2-3x+3))((1-x)^3)< 0 $, на этом интервале вторая производная функции отрицательная $f""(x) < 0 $ - функция выпуклая вверх (вогнутая).

Inflection points.

Consider the change in sign of the second derivative when passing through a critical point of the second kind:
At the point $x =0$, the second derivative changes sign from $\quad - \quad 0 \quad + \quad$, the graph of the function changes convexity, i.e. this is the inflection point with coordinates $(0;0)$.

8. Asymptotes.

Vertical asymptote. The graph of the function has one vertical asymptote $x =1$ (see item 2).
Oblique asymptote.
In order for the graph of the function $y= \frac(x^3)(1-x) $ for $x \to \infty$ to have an oblique asymptote $y = kx+b$, it is necessary and sufficient , so that there are two limits $$\lim_(x \to +\infty)=\frac(f(x))(x) =k $$ find it $$ \lim_(x \to \infty) (\frac( x^3)(x(1-x))) = \infty => k= \infty $$ and second limit $$ \lim_(x \to +\infty)(f(x) - kx) = b$ $, because $k = \infty$ - there is no oblique asymptote.

Horizontal asymptote: in order for the horizontal asymptote to exist, it is necessary that the limit $$\lim_(x \to \infty)f(x) = b$$ exist, find it $$ \lim_(x \to +\infty)(\frac( x^3)(1-x))= -\infty$$$$ \lim_(x \to -\infty)(\frac(x^3)(1-x))= -\infty$$
There is no horizontal asymptote.

9. Graph of the function.

Conduct a complete study and plot a function graph

y(x)=x2+81−x.y(x)=x2+81−x.

1) Function scope. Since the function is a fraction, you need to find the zeros of the denominator.

1−x=0,⇒x=1.1−x=0,⇒x=1.

We exclude the only point x=1x=1 from the function definition area and get:

D(y)=(−∞;1)∪(1;+∞).D(y)=(−∞;1)∪(1;+∞).

2) Let us study the behavior of the function in the vicinity of the discontinuity point. Find one-sided limits:

Since the limits are equal to infinity, the point x=1x=1 is a discontinuity of the second kind, the line x=1x=1 is a vertical asymptote.

3) Let's determine the intersection points of the graph of the function with the coordinate axes.

Let's find the points of intersection with the ordinate axis OyOy, for which we equate x=0x=0:

Thus, the point of intersection with the axis OyOy has coordinates (0;8)(0;8).

Let's find the points of intersection with the abscissa axis OxOx, for which we set y=0y=0:

The equation has no roots, so there are no points of intersection with the OxOx axis.

Note that x2+8>0x2+8>0 for any xx. Therefore, for x∈(−∞;1)x∈(−∞;1) the function y>0y>0(takes positive values, the graph is above the x-axis), for x∈(1;+∞)x∈(1;+∞) the function y<0y<0 (принимает отрицательные значения, график находится ниже оси абсцисс).

4) The function is neither even nor odd because:

5) We investigate the function for periodicity. The function is not periodic, as it is a fractional rational function.

6) We investigate the function for extremums and monotonicity. To do this, we find the first derivative of the function:

Let us equate the first derivative to zero and find the stationary points (at which y′=0y′=0):

We got three critical points: x=−2,x=1,x=4x=−2,x=1,x=4. We divide the entire domain of the function into intervals by these points and determine the signs of the derivative in each interval:

For x∈(−∞;−2),(4;+∞)x∈(−∞;−2),(4;+∞) the derivative y′<0y′<0, поэтому функция убывает на данных промежутках.

For x∈(−2;1),(1;4)x∈(−2;1),(1;4) the derivative y′>0y′>0, the function increases on these intervals.

In this case, x=−2x=−2 is a local minimum point (the function decreases and then increases), x=4x=4 is a local maximum point (the function increases and then decreases).

Let's find the values of the function at these points:

Thus, the minimum point is (−2;4)(−2;4), the maximum point is (4;−8)(4;−8).

7) We examine the function for kinks and convexity. Let's find the second derivative of the function:

Equate the second derivative to zero:

The resulting equation has no roots, so there are no inflection points. Moreover, when x∈(−∞;1)x∈(−∞;1) y′′>0y″>0 is satisfied, that is, the function is concave when x∈(1;+∞)x∈(1;+ ∞) y′′<0y″<0, то есть функция выпуклая.

8) We investigate the behavior of the function at infinity, that is, at .

Since the limits are infinite, there are no horizontal asymptotes.

Let's try to determine oblique asymptotes of the form y=kx+by=kx+b. We calculate the values of k,bk,b according to the known formulas:

We found that the function has one oblique asymptote y=−x−1y=−x−1.

9) Additional points. Let's calculate the value of the function at some other points in order to build a graph more accurately.

y(−5)=5.5;y(2)=−12;y(7)=−9.5.y(−5)=5.5;y(2)=−12;y(7)=−9.5.

10) Based on the data obtained, we will build a graph, supplement it with asymptotes x=1x=1 (blue), y=−x−1y=−x−1 (green) and mark the characteristic points (the intersection with the y-axis is purple, extrema are orange, additional points are black) :

Task 4: Geometric, Economic problems (I have no idea what, here is an approximate selection of problems with a solution and formulas)

Example 3.23. a

Solution. x and y y
y \u003d a - 2 × a / 4 \u003d a / 2. Since x = a/4 is the only critical point, let's check whether the sign of the derivative changes when passing through this point. For xa/4 S "> 0, and for x >a/4 S "< 0, значит, в точке x=a/4 функция S имеет максимум. Значение функции S(a/4) = a/4(a - a/2) = a 2 /8 (кв. ед).Поскольку S непрерывна на и ее значения на концах S(0) и S(a/2) равны нулю, то найденное значение будет наибольшим значением функции. Таким образом, наиболее выгодным соотношением сторон площадки при данных условиях задачи является y = 2x.

Example 3.24.

Solution.
R = 2, H = 16/4 = 4.

Example 3.22. Find the extrema of the function f(x) = 2x 3 - 15x 2 + 36x - 14.

Solution. Since f "(x) \u003d 6x 2 - 30x +36 \u003d 6 (x - 2) (x - 3), then the critical points of the function x 1 \u003d 2 and x 2 \u003d 3. Extreme points can only be at these points. So as when passing through the point x 1 \u003d 2, the derivative changes sign from plus to minus, then at this point the function has a maximum.When passing through the point x 2 \u003d 3, the derivative changes sign from minus to plus, therefore, at the point x 2 \u003d 3, the function has a minimum. Calculating the values of the function in points
x 1 = 2 and x 2 = 3, we find the extrema of the function: maximum f(2) = 14 and minimum f(3) = 13.

Example 3.23. It is necessary to build a rectangular area near the stone wall so that it is fenced off with wire mesh on three sides, and adjoins the wall on the fourth side. For this there is a linear meters of the grid. At what aspect ratio will the site have the largest area?

Solution. Denote the sides of the site through x and y. The area of the site is S = xy. Let y is the length of the side adjacent to the wall. Then, by condition, the equality 2x + y = a must hold. Therefore y = a - 2x and S = x(a - 2x), where
0 ≤ x ≤ a/2 (the length and width of the area cannot be negative). S "= a - 4x, a - 4x = 0 for x = a/4, whence
y \u003d a - 2 × a / 4 \u003d a / 2. Since x = a/4 is the only critical point, let's check whether the sign of the derivative changes when passing through this point. For xa/4 S "> 0, and for x >a/4 S "< 0, значит, в точке x=a/4 функция S имеет максимум. Значение функции S(a/4) = a/4(a - a/2) = a 2 /8 (кв. ед).Поскольку S непрерывна на и ее значения на концах S(0) и S(a/2) равны нулю, то найденное значение будет наибольшим значением функции. Таким образом, наиболее выгодным соотношением сторон площадки при данных условиях задачи является y = 2x.

Example 3.24. It is required to make a closed cylindrical tank with a capacity of V=16p ≈ 50 m 3 . What should be the dimensions of the tank (radius R and height H) in order to use the least amount of material for its manufacture?

Solution. The total surface area of the cylinder is S = 2pR(R+H). We know the volume of the cylinder V = pR 2 H Þ H = V/pR 2 =16p/ pR 2 = 16/ R 2 . Hence, S(R) = 2p(R 2 +16/R). We find the derivative of this function:
S "(R) \u003d 2p (2R- 16 / R 2) \u003d 4p (R- 8 / R 2). S " (R) \u003d 0 for R 3 \u003d 8, therefore,
R = 2, H = 16/4 = 4.

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If in the problem it is necessary to carry out a complete study of the function f (x) \u003d x 2 4 x 2 - 1 with the construction of its graph, then we will consider this principle in detail.

To solve a problem of this type, one should use the properties and graphs of the main elementary functions. The research algorithm includes the following steps:

Yandex.RTB R-A-339285-1

Finding the domain of definition

Since research is carried out on the domain of the function, it is necessary to start with this step.

Example 1

Per given example involves finding the zeros of the denominator in order to exclude them from the DPV.

4 x 2 - 1 = 0 x = ± 1 2 ⇒ x ∈ - ∞ ; - 1 2 ∪ - 1 2 ; 1 2 ∪ 1 2 ; +∞

As a result, you can get roots, logarithms, and so on. Then the ODZ can be searched for the root of an even degree of type g (x) 4 by the inequality g (x) ≥ 0 , for the logarithm log a g (x) by the inequality g (x) > 0 .

Investigation of ODZ boundaries and finding vertical asymptotes

There are vertical asymptotes on the boundaries of the function, when the one-sided limits at such points are infinite.

Example 2

For example, consider the border points equal to x = ± 1 2 .

Then it is necessary to study the function to find the one-sided limit. Then we get that: lim x → - 1 2 - 0 f (x) = lim x → - 1 2 - 0 x 2 4 x 2 - 1 = = lim x → - 1 2 - 0 x 2 (2 x - 1 ) (2 x + 1) = 1 4 (- 2) - 0 = + ∞ lim x → - 1 2 + 0 f (x) = lim x → - 1 2 + 0 x 2 4 x - 1 = = lim x → - 1 2 + 0 x 2 (2 x - 1) (2 x + 1) = 1 4 (- 2) (+ 0) = - ∞ lim x → 1 2 - 0 f (x) = lim x → 1 2 - 0 x 2 4 x 2 - 1 = = lim x → 1 2 - 0 x 2 (2 x - 1) (2 x + 1) = 1 4 (- 0) 2 = - ∞ lim x → 1 2 - 0 f (x) = lim x → 1 2 - 0 x 2 4 x 2 - 1 = = lim x → 1 2 - 0 x 2 (2 x - 1) (2 x + 1) = 1 4 ( + 0) 2 = + ∞

This shows that the one-sided limits are infinite, which means that the lines x = ± 1 2 are the vertical asymptotes of the graph.

Investigation of the function and for even or odd

When the condition y (- x) = y (x) is met, the function is considered to be even. This suggests that the graph is located symmetrically with respect to O y. When the condition y (- x) = - y (x) is met, the function is considered odd. This means that the symmetry goes with respect to the origin of coordinates. If at least one inequality fails, we obtain a function of general form.

The fulfillment of the equality y (- x) = y (x) indicates that the function is even. When constructing, it is necessary to take into account that there will be symmetry with respect to O y.

To solve the inequality, intervals of increase and decrease are used with the conditions f "(x) ≥ 0 and f" (x) ≤ 0, respectively.

Definition 1

Stationary points are points that turn the derivative to zero.

Critical points are interior points from the domain where the derivative of the function is equal to zero or does not exist.

When making a decision, the following points should be taken into account:

for the existing intervals of increase and decrease of the inequality of the form f "(x) > 0, the critical points are not included in the solution;
points at which the function is defined without a finite derivative must be included in the intervals of increase and decrease (for example, y \u003d x 3, where the point x \u003d 0 makes the function defined, the derivative has the value of infinity at this point, y " \u003d 1 3 x 2 3 , y " (0) = 1 0 = ∞ , x = 0 is included in the increase interval);
in order to avoid disagreements, it is recommended to use mathematical literature, which is recommended by the Ministry of Education.

The inclusion of critical points in the intervals of increasing and decreasing in the event that they satisfy the domain of the function.

Definition 2

For determining the intervals of increase and decrease of the function, it is necessary to find:

derivative;
critical points;
break the domain of definition with the help of critical points into intervals;
determine the sign of the derivative at each of the intervals, where + is an increase and - is a decrease.

Example 3

Find the derivative on the domain f "(x) = x 2" (4 x 2 - 1) - x 2 4 x 2 - 1 "(4 x 2 - 1) 2 = - 2 x (4 x 2 - 1) 2 .

Solution

To solve you need:

find stationary points, this example has x = 0 ;
find the zeros of the denominator, the example takes the value zero at x = ± 1 2 .

We expose points on the numerical axis to determine the derivative on each interval. To do this, it is enough to take any point from the interval and make a calculation. At a positive result on the graph, we depict +, which means an increase in the function, and - means its decrease.

For example, f "(- 1) \u003d - 2 (- 1) 4 - 1 2 - 1 2 \u003d 2 9\u003e 0, which means that the first interval on the left has a + sign. Consider on the number line.

Answer:

there is an increase in the function on the interval - ∞ ; - 1 2 and (- 1 2 ; 0 ] ;
there is a decrease on the interval [ 0 ; 1 2) and 1 2 ; +∞ .

In the diagram, using + and -, the positivity and negativity of the function are depicted, and the arrows indicate decreasing and increasing.

The extremum points of a function are the points where the function is defined and through which the derivative changes sign.

Example 4

If we consider an example where x \u003d 0, then the value of the function in it is f (0) \u003d 0 2 4 0 2 - 1 \u003d 0. When the sign of the derivative changes from + to - and passes through the point x \u003d 0, then the point with coordinates (0; 0) is considered the maximum point. When the sign is changed from - to +, we get the minimum point.

Convexity and concavity are determined by solving inequalities of the form f "" (x) ≥ 0 and f "" (x) ≤ 0 . Less often they use the name bulge down instead of concavity, and bulge up instead of bulge.

Definition 3

For determining the gaps of concavity and convexity necessary:

find the second derivative;
find the zeros of the function of the second derivative;
break the domain of definition by the points that appear into intervals;
determine the sign of the gap.

Example 5

Find the second derivative from the domain of definition.

Solution

f "" (x) = - 2 x (4 x 2 - 1) 2 " = = (- 2 x) " (4 x 2 - 1) 2 - - 2 x 4 x 2 - 1 2 " (4 x 2 - 1) 4 = 24 x 2 + 2 (4 x 2 - 1) 3

We find the zeros of the numerator and denominator, where, using our example, we have that the zeros of the denominator x = ± 1 2

Now you need to put points on the number line and determine the sign of the second derivative from each interval. We get that

Answer:

the function is convex from the interval - 1 2 ; 12 ;
the function is concave from the gaps - ∞ ; - 1 2 and 1 2 ; +∞ .

Definition 4

inflection point is a point of the form x 0 ; f(x0) . When it has a tangent to the graph of the function, then when it passes through x 0, the function changes sign to the opposite.

In other words, this is such a point through which the second derivative passes and changes sign, and at the points themselves is equal to zero or does not exist. All points are considered to be the domain of the function.

In the example, it was seen that there are no inflection points, since the second derivative changes sign while passing through the points x = ± 1 2 . They, in turn, are not included in the domain of definition.

Finding horizontal and oblique asymptotes

When defining a function at infinity, one must look for horizontal and oblique asymptotes.

Definition 5

Oblique asymptotes are drawn using lines given by the equation y = k x + b, where k = lim x → ∞ f (x) x and b = lim x → ∞ f (x) - k x .

For k = 0 and b not equal to infinity, we find that the oblique asymptote becomes horizontal.

In other words, the asymptotes are the lines that the graph of the function approaches at infinity. This contributes to the rapid construction of the graph of the function.

If there are no asymptotes, but the function is defined at both infinities, it is necessary to calculate the limit of the function at these infinities in order to understand how the graph of the function will behave.

Example 6

As an example, consider that

k = lim x → ∞ f (x) x = lim x → ∞ x 2 4 x 2 - 1 x = 0 b = lim x → ∞ (f (x) - k x) = lim x → ∞ x 2 4 x 2 - 1 = 1 4 ⇒ y = 1 4

is a horizontal asymptote. After researching the function, you can start building it.

Calculating the value of a function at intermediate points

To make the plotting the most accurate, it is recommended to find several values of the function at intermediate points.

Example 7

From the example we have considered, it is necessary to find the values of the function at the points x \u003d - 2, x \u003d - 1, x \u003d - 3 4, x \u003d - 1 4. Since the function is even, we get that the values coincide with the values at these points, that is, we get x \u003d 2, x \u003d 1, x \u003d 3 4, x \u003d 1 4.

Let's write and solve:

F (- 2) = f (2) = 2 2 4 2 2 - 1 = 4 15 ≈ 0, 27 f (- 1) - f (1) = 1 2 4 1 2 - 1 = 1 3 ≈ 0 , 33 f - 3 4 = f 3 4 = 3 4 2 4 3 4 2 - 1 = 9 20 = 0 , 45 f - 1 4 = f 1 4 = 1 4 2 4 1 4 2 - 1 = - 1 12 ≈ - 0.08

To determine the maxima and minima of the function, inflection points, intermediate points, it is necessary to build asymptotes. For convenient designation, intervals of increase, decrease, convexity, concavity are fixed. Consider the figure below.

It is necessary to draw graph lines through the marked points, which will allow you to get closer to the asymptotes, following the arrows.

This concludes the complete study of the function. There are cases of constructing some elementary functions for which geometric transformations are used.

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