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Let a function f be given, which at some point x 0 has a finite derivative f (x 0). Then the line passing through the point (x 0 ; f (x 0)) has slope f ’(x 0) is called the tangent.
What happens if the derivative does not exist at the point x 0? There are two options:
- There is no tangent to the graph either. Classic example- function y = |x | at point (0; 0).
- The tangent becomes vertical. This is true, for example, for the function y = arcsin x at the point (1; π /2).
Tangent equation
Any non-vertical straight line is given by an equation of the form y = kx + b, where k is the slope. The tangent is no exception, and in order to create its equation at some point x 0, it is enough to know the value of the function and the derivative at this point.
So, let a function y = f (x) be given, which has a derivative y = f ’(x) on the segment. Then at any point x 0 ∈ (a ; b) a tangent can be drawn to the graph of this function, which is given by the equation:
y = f ’(x 0) (x − x 0) + f (x 0)
Here f ’(x 0) is the value of the derivative at point x 0, and f (x 0) is the value of the function itself.
Task. Given the function y = x 3 . Write an equation for the tangent to the graph of this function at the point x 0 = 2.
Tangent equation: y = f ’(x 0) · (x − x 0) + f (x 0). The point x 0 = 2 is given to us, but the values f (x 0) and f ’(x 0) will have to be calculated.
First, let's find the value of the function. Everything is easy here: f (x 0) = f (2) = 2 3 = 8;
Now let’s find the derivative: f ’(x) = (x 3)’ = 3x 2;
We substitute x 0 = 2 into the derivative: f ’(x 0) = f ’(2) = 3 2 2 = 12;
In total we get: y = 12 · (x − 2) + 8 = 12x − 24 + 8 = 12x − 16.
This is the tangent equation.
Task. Write an equation for the tangent to the graph of the function f (x) = 2sin x + 5 at point x 0 = π /2.
This time we will not describe each action in detail - we will only indicate the key steps. We have:
f (x 0) = f (π /2) = 2sin (π /2) + 5 = 2 + 5 = 7;
f ’(x) = (2sin x + 5)’ = 2cos x;
f ’(x 0) = f ’(π /2) = 2cos (π /2) = 0;
Tangent equation:
y = 0 · (x − π /2) + 7 ⇒ y = 7
IN the latter case the straight line turned out to be horizontal, because its angular coefficient k = 0. There is nothing wrong with this - we just stumbled upon an extremum point.
Consider the following figure:
It depicts a certain function y = f(x), which is differentiable at point a. Point M with coordinates (a; f(a)) is marked. Through arbitrary point P(a + ∆x; f(a + ∆x)) graph is drawn by a secant MR.
If now point P is shifted along the graph to point M, then straight line MR will rotate around point M. In this case, ∆x will tend to zero. From here we can formulate the definition of a tangent to the graph of a function.
Tangent to the graph of a function
The tangent to the graph of a function is the limiting position of the secant as the increment of the argument tends to zero. It should be understood that the existence of the derivative of the function f at the point x0 means that at this point of the graph there is tangent to him.
In this case, the angular coefficient of the tangent will be equal to the derivative of this function at this point f’(x0). This is geometric meaning derivative. The tangent to the graph of a function f differentiable at point x0 is a certain straight line passing through the point (x0;f(x0)) and having an angular coefficient f’(x0).
Tangent equation
Let's try to obtain the equation of the tangent to the graph of some function f at point A(x0; f(x0)). The equation of a straight line with slope k has next view:
Since our slope coefficient is equal to the derivative f’(x0), then the equation will take the following form: y = f’(x0)*x + b.
Now let's calculate the value of b. To do this, we use the fact that the function passes through point A.
f(x0) = f’(x0)*x0 + b, from here we express b and get b = f(x0) - f’(x0)*x0.
We substitute the resulting value into the tangent equation:
y = f’(x0)*x + b = f’(x0)*x + f(x0) - f’(x0)*x0 = f(x0) + f’(x0)*(x - x0).
y = f(x0) + f’(x0)*(x - x0).
Consider the following example: find the equation of the tangent to the graph of the function f(x) = x 3 - 2*x 2 + 1 at point x = 2.
2. f(x0) = f(2) = 2 2 - 2*2 2 + 1 = 1.
3. f’(x) = 3*x 2 - 4*x.
4. f’(x0) = f’(2) = 3*2 2 - 4*2 = 4.
5. Substitute the obtained values into the tangent formula, we get: y = 1 + 4*(x - 2). Opening the brackets and bringing similar terms we get: y = 4*x - 7.
Answer: y = 4*x - 7.
General scheme for composing the tangent equation to the graph of the function y = f(x):
1. Determine x0.
2. Calculate f(x0).
3. Calculate f’(x)
In this article we will analyze all types of problems to find
Let's remember geometric meaning of derivative: if a tangent is drawn to the graph of a function at a point, then the slope coefficient of the tangent ( equal to tangent angle between the tangent and the positive direction of the axis) is equal to the derivative of the function at the point.
Let's take an arbitrary point on the tangent with coordinates:
And consider a right triangle:
In this triangle 
From here 
This is the equation of the tangent drawn to the graph of the function at the point.
To write the tangent equation, we only need to know the equation of the function and the point at which the tangent is drawn. Then we can find and .
There are three main types of tangent equation problems.
1. Given a point of contact
2. The tangent slope coefficient is given, that is, the value of the derivative of the function at the point.
3. Given are the coordinates of the point through which the tangent is drawn, but which is not the point of tangency.
Let's look at each type of task.
1 . Write the equation of the tangent to the graph of the function 
at the point .
.
b) Find the value of the derivative at point . First let's find the derivative of the function
Let's substitute the found values into the tangent equation:
Let's open the brackets on the right side of the equation. We get:
Answer: .
2. Find the abscissa of the points at which the functions are tangent to the graph 
parallel to the x-axis.
If the tangent is parallel to the x-axis, therefore the angle between the tangent and the positive direction of the axis equal to zero, therefore the tangent of the tangent angle is zero. This means that the value of the derivative of the function at the points of contact is zero.
a) Find the derivative of the function .
b) Let’s equate the derivative to zero and find the values in which the tangent is parallel to the axis:
Equating each factor to zero, we get:
Answer: 0;3;5
3. Write equations for tangents to the graph of a function , parallel straight .
A tangent is parallel to a line. The slope of this line is -1. Since the tangent is parallel to this line, therefore, the slope of the tangent is also -1. That is we know the slope of the tangent, and, thereby, derivative value at the point of tangency.
This is the second type of problem to find the tangent equation.
So, we are given the function and the value of the derivative at the point of tangency.
a) Find the points at which the derivative of the function is equal to -1.
First, let's find the derivative equation.
Let's equate the derivative to the number -1.
Let's find the value of the function at the point.
(by condition)
.
b) Find the equation of the tangent to the graph of the function at point .
Let's find the value of the function at the point.
(by condition).
Let's substitute these values into the tangent equation:
.
Answer:
4 . Write the equation of the tangent to the curve , passing through a point
First, let's check if the point is a tangent point. If a point is a tangent point, then it belongs to the graph of the function, and its coordinates must satisfy the equation of the function. Let's substitute the coordinates of the point into the equation of the function.
Title="1sqrt(8-3^2)">. Мы получили под корнем !} a negative number, the equality is not true, and the point does not belong to the graph of the function and is not a point of contact.
This is the last type of problem to find the tangent equation. First thing we need to find the abscissa of the tangent point.
Let's find the value.
Let be the point of contact. The point belongs to the tangent to the graph of the function. If we substitute the coordinates of this point into the tangent equation, we get the correct equality:
.
The value of the function at a point is 
.
Let's find the value of the derivative of the function at the point.
First, let's find the derivative of the function. This .
The derivative at a point is equal to 
.
Let's substitute the expressions for and into the tangent equation. We get the equation for:
Let's solve this equation.
Reduce the numerator and denominator of the fraction by 2:
Let us reduce the right side of the equation to common denominator. We get:
Let's simplify the numerator of the fraction and multiply both sides by - this expression is strictly greater than zero.
We get the equation
Let's solve it. To do this, let's square both parts and move on to the system.
Title="delim(lbrace)(matrix(2)(1)((64-48(x_0)+9(x_0)^2=8-(x_0)^2) (8-3x_0>=0 ) ))( )">!}
Let's solve the first equation.
Let's decide quadratic equation, we get
The second root does not satisfy the condition title="8-3x_0>=0">, следовательно, у нас только одна точка касания и её абсцисса равна .!}
Let's write the equation of the tangent to the curve at the point. To do this, substitute the value into the equation 
- We already recorded it.
Answer:
.
You will need
- - mathematical reference book;
- - notebook;
- - a simple pencil;
- - pen;
- - protractor;
- - compass.
Instructions
Please note that the graph of the differentiable function f(x) at the point x0 is no different from the tangent segment. Therefore, it is quite close to the segment l, to the one passing through the points (x0; f(x0)) and (x0+Δx; f(x0 + Δx)). To specify a straight line passing through point A with coefficients (x0; f(x0)), specify its slope. Moreover, it is equal to Δy/Δx secant tangent (Δх→0), and also tends to the number f‘(x0).
If there are no values for f‘(x0), then there is no tangent, or it runs vertically. Based on this, the derivative of the function at the point x0 is explained by the existence of a non-vertical tangent, which is in contact with the graph of the function at the point (x0, f(x0)). IN in this case the angular coefficient of the tangent is equal to f "(x0). The geometric derivative, that is, the angular coefficient of the tangent, becomes clear.
That is, in order to find the slope of the tangent, you need to find the value of the derivative of the function at the point of tangency. Example: find the angular coefficient of the tangent to the function y = x³ at the point with abscissa X0 = 1. Solution: Find the derivative of this function y΄(x) = 3x²; find the value of the derivative at the point X0 = 1. у΄(1) = 3 × 1² = 3. The angle coefficient of the tangent at the point X0 = 3.
Draw additional tangents in the figure so that they touch the graph of the function at the points: x1, x2 and x3. Mark the angles formed by these tangents with the abscissa axis (the angle is counted in the positive direction - from the axis to the tangent line). For example, angle α1 will be acute, angle (α2) will be obtuse, and the third (α3) will be equal to zero, since the tangent line drawn is parallel axis OH. In this case, tangent obtuse angle There is negative meaning, and tangent acute angle– positive, at tg0 and the result is zero.
A tangent to a given circle is a straight line that has only one common point with this circle. A tangent to a circle is always perpendicular to its radius drawn to the point of tangency. If two tangents are drawn from one point that does not belong to the circle, then the distances from this point to the points of tangency will always be the same. Tangents to circles are being built different ways, depending on their location relative to each other.
Instructions
Constructing a tangent to one circle.
1. Construct a circle of radius R and take A, which the tangent will pass through.
2. A circle is constructed with a center in the middle of the segment OA and radii equal to this segment.
3. The intersection of two tangent points drawn through point A to a given circle.
External tangent to two circles.
2. Draw a circle of radius R – r with center at point O.
3. A tangent from O1 is drawn to the resulting circle, the point of tangency is designated M.
4. Radius R passing through point M to point T – the tangent point of the circle.
5. Through the center O1 of the small circle, a radius r is drawn parallel to R of the large circle. The radius r points to point T1 – the point of tangency of the small circle.
circles.
Internal tangent to two circles.
1. Two circles of radius R and r are constructed.
2. Draw a circle of radius R + r with center at point O.
3. A tangent is drawn to the resulting circle from point O1, the point of tangency is designated by the letter M.
4. Ray OM intersects the first circle at point T - at the point of tangency of the great circle.
5. Through the center O1 of the small circle, a radius r is drawn parallel to the ray OM. The radius r points to point T1 – the point of tangency of the small circle.
6. Straight line TT1 – tangent to the given circles.
Sources:
- internal tangent
Angular closet – perfect option for empty corners in the apartment. In addition, the corner configuration closet ov gives the interior a classic atmosphere. As a finishing for corners closet ov any material that is suitable for this purpose can be used.
You will need
- Fiberboard, MDF, screws, nails, saw blade, frieze.
Instructions
Cut a template 125 mm wide and 1065 mm long from plywood or fiberboard. The edges must be filed at an angle of 45 degrees. By ready-made template determine the dimensions of the side walls, as well as the place where it will be located closet.
Connect the lid to the side walls and triangular shelves. The cover must be secured to the upper edges of the side walls using screws. For structural strength, additional glue is used. Attach the shelves to the slats.
Angle the saw blade at a 45-degree angle and bevel the leading edge of the side walls along the guide bar. Attach fixed shelves to MDF strips. Connect the side walls with screws. Make sure there are no gaps.
Make marks in the wall, between which place the frame of the corner closet A. Attach using screws closet to Wall. The length of the dowel should be 75 mm.
Cut out the front frame from a solid MDF board. Using a circular saw, cut the openings in it using a ruler. Finish the corners.
Find the abscissa value of the tangent point, which is denoted by the letter “a”. If it coincides with a given tangent point, then "a" will be its x-coordinate. Determine the value functions f(a) by substituting into the equation functions abscissa value.
Determine the first derivative of the equation functions f’(x) and substitute the value of point “a” into it.
Take general equation tangent, which is defined as y = f(a) = f (a)(x – a), and substitute the found values of a, f(a), f "(a) into it. As a result, the solution to the graph and the tangent will be found.
Solve the problem in a different way if the given tangent point does not coincide with the tangent point. In this case, it is necessary to substitute “a” instead of numbers in the tangent equation. After that, instead of the letters “x” and “y”, substitute the coordinate value given point. Solve the resulting equation in which “a” is the unknown. Plug the resulting value into the tangent equation.
Write an equation for a tangent with the letter “a” if the problem statement specifies the equation functions and equation parallel line relative to the desired tangent. After this we need the derivative functions, to the coordinate at point “a”. Substitute the appropriate value into the tangent equation and solve the function.
When composing the equation of a tangent to the graph of a function, the concept of “abscissa of the point of tangency” is used. This value may be specified initially in the conditions of the task or it must be determined independently.
Instructions
Draw the x and y coordinate axes on a piece of paper. Explore given equation for the graph of a function. If it is , then it is enough to have two values for the parameter y for any x, then plot the found points on the coordinate axis and connect them with a line. If the graph is nonlinear, then make a table of the dependence of y on x and select at least five points to construct the graph.
Determine the value of the abscissa of the tangent point for the case when the given tangent point does not coincide with the graph of the function. We set the third parameter with the letter “a”.
Write down the equation of the function f(a). To do this, substitute a instead of x in the original equation. Find the derivative of the function f(x) and f(a). Substitute the required data into the general tangent equation, which has the form: y = f(a) + f "(a)(x – a). As a result, obtain an equation that consists of three unknown parameters.
Substitute into it, instead of x and y, the coordinates of the given point through which the tangent passes. After this, find the solution to the resulting equation for all a. If it is square, then there will be two values for the abscissa of the tangent point. This is that the tangent passes twice near the graph of the function.
Draw a graph given function and , which are specified according to the conditions of the problem. In this case, it is also necessary to specify the unknown parameter a and substitute it into the equation f(a). Equate the derivative f(a) to the derivative of the equation of a parallel line. This comes from the condition of parallelism of the two. Find the roots of the resulting equation, which will be the abscissa of the point of tangency.
The straight line y=f(x) will be tangent to the graph shown in the figure at point x0 if it passes through the point with coordinates (x0; f(x0)) and has an angular coefficient f"(x0). Find such a coefficient, Knowing the features of a tangent, it’s not difficult.
You will need
- - mathematical reference book;
- - a simple pencil;
- - notebook;
- - protractor;
- - compass;
- - pen.
Instructions
If the value f‘(x0) does not exist, then either there is no tangent, or it runs vertically. In view of this, the presence of a derivative of the function at the point x0 is due to the existence of a non-vertical tangent tangent to the graph of the function at the point (x0, f(x0)). In this case, the angular coefficient of the tangent will be equal to f "(x0). Thus, the geometric meaning of the derivative becomes clear - the calculation of the angular coefficient of the tangent.
Determine the general one. This kind of information can be obtained by referring to census data. To determine the general fertility, mortality, marriage and divorce rates, you will need to find the product general population and billing period. Write the resulting number into the denominator.
Put on the numerator the indicator corresponding to the desired relative. For example, if you are faced with determining the total fertility rate, then in place of the numerator there should be a number that reflects the total number of births for the period of interest to you. If your goal is the mortality rate or marriage rate, then in place of the numerator put the number of deaths in the calculation period or the number of marriages, respectively.
Multiply the resulting number by 1000. This will be the overall coefficient you are looking for. If you are faced with the task of finding the overall growth rate, then subtract the mortality rate from the birth rate.
Video on the topic
Sources:
- General vital rates
The main indicator of extraction efficiency is coefficient distribution. It is calculated by the formula: Co/Sw, where Co is the concentration of the extracted substance in the organic solvent (extractor), and St is the concentration of the same substance in water, after equilibrium has reached. How can you experimentally find the distribution coefficient?