Write the equation of the tangent to the graph of the function f. Tangent to the graph of a function at a point
The article provides a detailed explanation of the definitions, the geometric meaning of the derivative with graphic notations. The equation of a tangent line will be considered with examples, the equations of a tangent to 2nd order curves will be found.
Yandex.RTB R-A-339285-1 Definition 1
The angle of inclination of the straight line y = k x + b is called angle α, which is measured from the positive direction of the x axis to the straight line y = k x + b in the positive direction.
In the figure, the x direction is indicated by a green arrow and a green arc, and the angle of inclination by a red arc. The blue line refers to the straight line.
Definition 2
The slope of the straight line y = k x + b is called the numerical coefficient k.
The angular coefficient is equal to the tangent of the straight line, in other words k = t g α.
- The angle of inclination of a straight line is equal to 0 only if it is parallel about x and the slope is equal to zero, because the tangent of zero is equal to 0. This means that the form of the equation will be y = b.
- If the angle of inclination of the straight line y = k x + b is acute, then the conditions 0 are satisfied< α < π 2 или 0 ° < α < 90 ° . Отсюда имеем, что значение углового коэффициента k считается positive number, because the tangent value satisfies the condition t g α > 0, and there is an increase in the graph.
- If α = π 2, then the location of the line is perpendicular to x. Equality is specified by x = c with the value c being a real number.
- If the angle of inclination of the straight line y = k x + b is obtuse, then it corresponds to the conditions π 2< α < π или 90 ° < α < 180 ° , значение углового коэффициента k принимает negative meaning, and the graph is decreasing.
A secant is a line that passes through 2 points of the function f (x). In other words, a secant is a straight line that passes through any two points on the graph of a given function.
The figure shows that A B is a secant, and f (x) is a black curve, α is a red arc, indicating the angle of inclination of the secant.
When the angular coefficient of a straight line is equal to the tangent of the angle of inclination, it is clear that the tangent of a right triangle A B C can be found by the ratio of the opposite side to the adjacent one.
Definition 4
We get a formula for finding a secant of the form:
k = t g α = B C A C = f (x B) - f x A x B - x A, where the abscissas of points A and B are the values x A, x B, and f (x A), f (x B) are the values functions at these points.
Obviously, the angular coefficient of the secant is determined using the equality k = f (x B) - f (x A) x B - x A or k = f (x A) - f (x B) x A - x B, and the equation must be written as y = f (x B) - f (x A) x B - x A x - x A + f (x A) or
y = f (x A) - f (x B) x A - x B x - x B + f (x B) .
The secant divides the graph visually into 3 parts: to the left of point A, from A to B, to the right of B. The figure below shows that there are three secants that are considered coincident, that is, they are set using a similar equation.
By definition, it is clear that a straight line and its secant in in this case match up.
A secant can intersect the graph of a given function multiple times. If there is an equation of the form y = 0 for a secant, then the number of points of intersection with the sinusoid is infinite.
Definition 5
Tangent to the graph of the function f (x) at point x 0 ; f (x 0) is a straight line passing through a given point x 0; f (x 0), with the presence of a segment that has many x values close to x 0.
Example 1
Let's take a closer look at the example below. Then it is clear that the line defined by the function y = x + 1 is considered tangent to y = 2 x at the point with coordinates (1; 2). For clarity, it is necessary to consider graphs with values close to (1; 2). The function y = 2 x is shown in black, the blue line is the tangent line, and the red dot is the intersection point.
Obviously, y = 2 x merges with the line y = x + 1.
To determine the tangent, we should consider the behavior of the tangent A B as point B approaches point A infinitely. For clarity, we present a drawing.
The secant A B, indicated by the blue line, tends to the position of the tangent itself, and the angle of inclination of the secant α will begin to tend to the angle of inclination of the tangent itself α x.
Definition 6
The tangent to the graph of the function y = f (x) at point A is considered to be the limiting position of the secant A B as B tends to A, that is, B → A.
Now let's move on to consider the geometric meaning of the derivative of a function at a point.
Let's move on to considering the secant A B for the function f (x), where A and B with coordinates x 0, f (x 0) and x 0 + ∆ x, f (x 0 + ∆ x), and ∆ x is denoted as the increment of the argument . Now the function will take the form ∆ y = ∆ f (x) = f (x 0 + ∆ x) - f (∆ x) . For clarity, let's give an example of a drawing.
Let's consider the resulting right triangle A B C. We use the definition of tangent to solve, that is, we obtain the relation ∆ y ∆ x = t g α . From the definition of a tangent it follows that lim ∆ x → 0 ∆ y ∆ x = t g α x . According to the rule of the derivative at a point, we have that the derivative f (x) at the point x 0 is called the limit of the ratio of the increment of the function to the increment of the argument, where ∆ x → 0, then we denote it as f (x 0) = lim ∆ x → 0 ∆ y ∆ x .
It follows that f " (x 0) = lim ∆ x → 0 ∆ y ∆ x = t g α x = k x, where k x is denoted as the slope of the tangent.
That is, we find that f ' (x) can exist at point x 0, and like the tangent to a given graph of the function at the point of tangency equal to x 0, f 0 (x 0), where the value of the slope of the tangent at the point is equal to the derivative at point x 0 . Then we get that k x = f " (x 0) .
Geometric meaning derivative of a function at a point is that the concept of the existence of a tangent to the graph at the same point is given.
To write down the equation of any straight line on a plane, it is necessary to have an angular coefficient with the point through which it passes. Its notation is taken to be x 0 at intersection.
The tangent equation to the graph of the function y = f (x) at the point x 0, f 0 (x 0) takes the form y = f "(x 0) x - x 0 + f (x 0).
What is meant is that final value derivative f "(x 0) you can determine the position of the tangent, that is, vertically under the condition lim x → x 0 + 0 f " (x) = ∞ and lim x → x 0 - 0 f " (x) = ∞ or absence at all with condition lim x → x 0 + 0 f " (x) ≠ lim x → x 0 - 0 f " (x) .
The location of the tangent depends on the value of its angular coefficient k x = f "(x 0). When parallel to the o x axis, we obtain that k k = 0, when parallel to o y - k x = ∞, and the form of the tangent equation x = x 0 increases with k x > 0, decreases as k x< 0 .
Example 2
Compile an equation for the tangent to the graph of the function y = e x + 1 + x 3 3 - 6 - 3 3 x - 17 - 3 3 at the point with coordinates (1; 3) and determine the angle of inclination.
Solution
By condition we have that the function is defined for all real numbers. We find that the point with the coordinates specified by the condition, (1; 3) is a point of tangency, then x 0 = - 1, f (x 0) = - 3.
It is necessary to find the derivative at the point with value - 1. We get that
y " = e x + 1 + x 3 3 - 6 - 3 3 x - 17 - 3 3 " = = e x + 1 " + x 3 3 " - 6 - 3 3 x " - 17 - 3 3 " = e x + 1 + x 2 - 6 - 3 3 y " (x 0) = y " (- 1) = e - 1 + 1 + - 1 2 - 6 - 3 3 = 3 3
The value of f' (x) at the point of tangency is the slope of the tangent, which is equal to the tangent of the slope.
Then k x = t g α x = y " (x 0) = 3 3
It follows that α x = a r c t g 3 3 = π 6
Answer: the tangent equation takes the form
y = f " (x 0) x - x 0 + f (x 0) y = 3 3 (x + 1) - 3 y = 3 3 x - 9 - 3 3
For clarity, we give an example in a graphic illustration.
Black color is used for the graph of the original function, Blue colour– image of a tangent, red dot – point of tangency. The figure on the right shows an enlarged view.
Example 3
Determine the existence of a tangent to the graph of a given function
y = 3 · x - 1 5 + 1 at the point with coordinates (1 ; 1) . Write an equation and determine the angle of inclination.
Solution
By condition, we have that the domain of definition of a given function is considered to be the set of all real numbers.
Let's move on to finding the derivative
y " = 3 x - 1 5 + 1 " = 3 1 5 (x - 1) 1 5 - 1 = 3 5 1 (x - 1) 4 5
If x 0 = 1, then f' (x) is undefined, but the limits are written as lim x → 1 + 0 3 5 1 (x - 1) 4 5 = 3 5 1 (+ 0) 4 5 = 3 5 · 1 + 0 = + ∞ and lim x → 1 - 0 3 5 · 1 (x - 1) 4 5 = 3 5 · 1 (- 0) 4 5 = 3 5 · 1 + 0 = + ∞ , which means the existence vertical tangent at point (1; 1).
Answer: the equation will take the form x = 1, where the angle of inclination will be equal to π 2.
For clarity, let's depict it graphically.
Example 4
Find the points on the graph of the function y = 1 15 x + 2 3 - 4 5 x 2 - 16 5 x - 26 5 + 3 x + 2, where
- There is no tangent;
- The tangent is parallel to x;
- The tangent is parallel to the line y = 8 5 x + 4.
Solution
It is necessary to pay attention to the scope of definition. By condition, we have that the function is defined on the set of all real numbers. We expand the module and solve the system with intervals x ∈ - ∞ ; 2 and [ - 2 ; + ∞) . We get that
y = - 1 15 x 3 + 18 x 2 + 105 x + 176 , x ∈ - ∞ ; - 2 1 15 x 3 - 6 x 2 + 9 x + 12 , x ∈ [ - 2 ; + ∞)
It is necessary to differentiate the function. We have that
y " = - 1 15 x 3 + 18 x 2 + 105 x + 176 " , x ∈ - ∞ ; - 2 1 15 x 3 - 6 x 2 + 9 x + 12 ", x ∈ [ - 2 ; + ∞) ⇔ y " = - 1 5 (x 2 + 12 x + 35) , x ∈ - ∞ ; - 2 1 5 x 2 - 4 x + 3 , x ∈ [ - 2 ; + ∞)
When x = − 2, then the derivative does not exist because the one-sided limits are not equal at that point:
lim x → - 2 - 0 y " (x) = lim x → - 2 - 0 - 1 5 (x 2 + 12 x + 35 = - 1 5 (- 2) 2 + 12 (- 2) + 35 = - 3 lim x → - 2 + 0 y " (x) = lim x → - 2 + 0 1 5 (x 2 - 4 x + 3) = 1 5 - 2 2 - 4 - 2 + 3 = 3
We calculate the value of the function at the point x = - 2, where we get that
- y (- 2) = 1 15 - 2 + 2 3 - 4 5 (- 2) 2 - 16 5 (- 2) - 26 5 + 3 - 2 + 2 = - 2, that is, the tangent at the point (- 2; - 2) will not exist.
- The tangent is parallel to x when the slope is zero. Then k x = t g α x = f "(x 0). That is, it is necessary to find the values of such x when the derivative of the function turns it to zero. That is, the values of f ' (x) will be the points of tangency, where the tangent is parallel to x .
When x ∈ - ∞ ; - 2, then - 1 5 (x 2 + 12 x + 35) = 0, and for x ∈ (- 2; + ∞) we get 1 5 (x 2 - 4 x + 3) = 0.
1 5 (x 2 + 12 x + 35) = 0 D = 12 2 - 4 35 = 144 - 140 = 4 x 1 = - 12 + 4 2 = - 5 ∈ - ∞ ; - 2 x 2 = - 12 - 4 2 = - 7 ∈ - ∞ ; - 2 1 5 (x 2 - 4 x + 3) = 0 D = 4 2 - 4 · 3 = 4 x 3 = 4 - 4 2 = 1 ∈ - 2 ; + ∞ x 4 = 4 + 4 2 = 3 ∈ - 2 ; +∞
Calculate the corresponding function values
y 1 = y - 5 = 1 15 - 5 + 2 3 - 4 5 - 5 2 - 16 5 - 5 - 26 5 + 3 - 5 + 2 = 8 5 y 2 = y (- 7) = 1 15 - 7 + 2 3 - 4 5 (- 7) 2 - 16 5 - 7 - 26 5 + 3 - 7 + 2 = 4 3 y 3 = y (1) = 1 15 1 + 2 3 - 4 5 1 2 - 16 5 1 - 26 5 + 3 1 + 2 = 8 5 y 4 = y (3) = 1 15 3 + 2 3 - 4 5 3 2 - 16 5 3 - 26 5 + 3 3 + 2 = 4 3
Hence - 5; 8 5, - 4; 4 3, 1; 8 5, 3; 4 3 are considered to be the required points of the function graph.
Let's look at a graphical representation of the solution.
The black line is the graph of the function, the red dots are the tangency points.
- When the lines are parallel, the angular coefficients are equal. Then it is necessary to search for points on the function graph where the slope will be equal to the value 8 5. To do this, you need to solve an equation of the form y "(x) = 8 5. Then, if x ∈ - ∞; - 2, we obtain that - 1 5 (x 2 + 12 x + 35) = 8 5, and if x ∈ ( - 2 ; + ∞), then 1 5 (x 2 - 4 x + 3) = 8 5.
The first equation has no roots since the discriminant is less than zero. Let's write down that
1 5 x 2 + 12 x + 35 = 8 5 x 2 + 12 x + 43 = 0 D = 12 2 - 4 43 = - 28< 0
Another equation has two real roots, then
1 5 (x 2 - 4 x + 3) = 8 5 x 2 - 4 x - 5 = 0 D = 4 2 - 4 · (- 5) = 36 x 1 = 4 - 36 2 = - 1 ∈ - 2 ; + ∞ x 2 = 4 + 36 2 = 5 ∈ - 2 ; +∞
Let's move on to finding the values of the function. We get that
y 1 = y (- 1) = 1 15 - 1 + 2 3 - 4 5 (- 1) 2 - 16 5 (- 1) - 26 5 + 3 - 1 + 2 = 4 15 y 2 = y (5) = 1 15 5 + 2 3 - 4 5 5 2 - 16 5 5 - 26 5 + 3 5 + 2 = 8 3
Points with values - 1; 4 15, 5; 8 3 are the points at which the tangents are parallel to the line y = 8 5 x + 4.
Answer: black line – graph of the function, red line – graph of y = 8 5 x + 4, blue line – tangents at points - 1; 4 15, 5; 8 3.
There may be an infinite number of tangents for given functions.
Example 5
Write the equations of all available tangents of the function y = 3 cos 3 2 x - π 4 - 1 3, which are located perpendicular to the straight line y = - 2 x + 1 2.
Solution
To compile the tangent equation, it is necessary to find the coefficient and coordinates of the tangent point, based on the condition of perpendicularity of the lines. The definition is as follows: the product of angular coefficients that are perpendicular to straight lines is equal to - 1, that is, written as k x · k ⊥ = - 1. From the condition we have that the angular coefficient is located perpendicular to the line and is equal to k ⊥ = - 2, then k x = - 1 k ⊥ = - 1 - 2 = 1 2.
Now you need to find the coordinates of the touch points. You need to find x and then its value for a given function. Note that from the geometric meaning of the derivative at the point
x 0 we obtain that k x = y "(x 0). From this equality we find the values of x for the points of contact.
We get that
y " (x 0) = 3 cos 3 2 x 0 - π 4 - 1 3 " = 3 - sin 3 2 x 0 - π 4 3 2 x 0 - π 4 " = = - 3 sin 3 2 x 0 - π 4 3 2 = - 9 2 sin 3 2 x 0 - π 4 ⇒ k x = y " (x 0) ⇔ - 9 2 sin 3 2 x 0 - π 4 = 1 2 ⇒ sin 3 2 x 0 - π 4 = - 1 9
This trigonometric equation will be used to calculate the ordinates of the tangent points.
3 2 x 0 - π 4 = a r c sin - 1 9 + 2 πk or 3 2 x 0 - π 4 = π - a r c sin - 1 9 + 2 πk
3 2 x 0 - π 4 = - a r c sin 1 9 + 2 πk or 3 2 x 0 - π 4 = π + a r c sin 1 9 + 2 πk
x 0 = 2 3 π 4 - a r c sin 1 9 + 2 πk or x 0 = 2 3 5 π 4 + a r c sin 1 9 + 2 πk , k ∈ Z
Z is a set of integers.
x points of contact have been found. Now you need to move on to searching for the values of y:
y 0 = 3 cos 3 2 x 0 - π 4 - 1 3
y 0 = 3 1 - sin 2 3 2 x 0 - π 4 - 1 3 or y 0 = 3 - 1 - sin 2 3 2 x 0 - π 4 - 1 3
y 0 = 3 1 - - 1 9 2 - 1 3 or y 0 = 3 - 1 - - 1 9 2 - 1 3
y 0 = 4 5 - 1 3 or y 0 = - 4 5 + 1 3
From this we obtain that 2 3 π 4 - a r c sin 1 9 + 2 πk ; 4 5 - 1 3 , 2 3 5 π 4 + a r c sin 1 9 + 2 πk ; - 4 5 + 1 3 are the points of tangency.
Answer: the necessary equations will be written as
y = 1 2 x - 2 3 π 4 - a r c sin 1 9 + 2 πk + 4 5 - 1 3 , y = 1 2 x - 2 3 5 π 4 + a r c sin 1 9 + 2 πk - 4 5 + 1 3 , k ∈ Z
For a visual representation, consider a function and a tangent on a coordinate line.
The figure shows that the function is located on the interval [ - 10 ; 10 ], where the black line is the graph of the function, the blue lines are tangents, which are located perpendicular to the given line of the form y = - 2 x + 1 2. Red dots are touch points.
The canonical equations of 2nd order curves are not single-valued functions. Tangent equations for them are compiled according to known schemes.
Tangent to a circle
To define a circle with center at point x c e n t e r ; y c e n t e r and radius R, apply the formula x - x c e n t e r 2 + y - y c e n t e r 2 = R 2 .
This equality can be written as a union of two functions:
y = R 2 - x - x c e n t e r 2 + y c e n t e r y = - R 2 - x - x c e n t e r 2 + y c e n t e r
The first function is located at the top, and the second at the bottom, as shown in the figure.
To compile the equation of a circle at the point x 0; y 0 , which is located in the upper or lower semicircle, you should find the equation of the graph of a function of the form y = R 2 - x - x c e n t e r 2 + y c e n t e r or y = - R 2 - x - x c e n t e r 2 + y c e n t e r at the indicated point.
When at points x c e n t e r ; y c e n t e r + R and x c e n t e r ; y c e n t e r - R tangents can be given by the equations y = y c e n t e r + R and y = y c e n t e r - R , and at points x c e n t e r + R ; y c e n t e r and
x c e n t e r - R ; y c e n t e r will be parallel to o y, then we obtain equations of the form x = x c e n t e r + R and x = x c e n t e r - R .
Tangent to an ellipse
When the ellipse has a center at x c e n t e r ; y c e n t e r with semi-axes a and b, then it can be specified using the equation x - x c e n t e r 2 a 2 + y - y c e n t e r 2 b 2 = 1.
An ellipse and a circle can be denoted by combining two functions, namely the upper and lower half-ellipse. Then we get that
y = b a · a 2 - (x - x c e n t e r) 2 + y c e n t e r y = - b a · a 2 - (x - x c e n t e r) 2 + y c e n t e r
If the tangents are located at the vertices of the ellipse, then they are parallel about x or about y. Below, for clarity, consider the figure.
Example 6
Write the equation of the tangent to the ellipse x - 3 2 4 + y - 5 2 25 = 1 at points with values of x equal to x = 2.
Solution
It is necessary to find the tangent points that correspond to the value x = 2. We substitute into the existing equation of the ellipse and find that
x - 3 2 4 x = 2 + y - 5 2 25 = 1 1 4 + y - 5 2 25 = 1 ⇒ y - 5 2 = 3 4 25 ⇒ y = ± 5 3 2 + 5
Then 2 ; 5 3 2 + 5 and 2; - 5 3 2 + 5 are the tangent points that belong to the upper and lower half-ellipse.
Let's move on to finding and solving the equation of the ellipse with respect to y. We get that
x - 3 2 4 + y - 5 2 25 = 1 y - 5 2 25 = 1 - x - 3 2 4 (y - 5) 2 = 25 1 - x - 3 2 4 y - 5 = ± 5 1 - x - 3 2 4 y = 5 ± 5 2 4 - x - 3 2
Obviously, the upper half-ellipse is specified using a function of the form y = 5 + 5 2 4 - x - 3 2, and the lower half ellipse y = 5 - 5 2 4 - x - 3 2.
Let's apply a standard algorithm to create an equation for a tangent to the graph of a function at a point. Let us write that the equation for the first tangent at point 2; 5 3 2 + 5 will look like
y " = 5 + 5 2 4 - x - 3 2 " = 5 2 1 2 4 - (x - 3) 2 4 - (x - 3) 2 " = = - 5 2 x - 3 4 - ( x - 3) 2 ⇒ y " (x 0) = y " (2) = - 5 2 2 - 3 4 - (2 - 3) 2 = 5 2 3 ⇒ y = y " (x 0) x - x 0 + y 0 ⇔ y = 5 2 3 (x - 2) + 5 3 2 + 5
We find that the equation of the second tangent with a value at the point
2 ; - 5 3 2 + 5 takes the form
y " = 5 - 5 2 4 - (x - 3) 2 " = - 5 2 1 2 4 - (x - 3) 2 4 - (x - 3) 2 " = = 5 2 x - 3 4 - (x - 3) 2 ⇒ y " (x 0) = y " (2) = 5 2 2 - 3 4 - (2 - 3) 2 = - 5 2 3 ⇒ y = y " (x 0) x - x 0 + y 0 ⇔ y = - 5 2 3 (x - 2) - 5 3 2 + 5
Graphically, tangents are designated as follows:
Tangent to hyperbole
When a hyperbola has a center at x c e n t e r ; y c e n t e r and vertices x c e n t e r + α ; y c e n t e r and x c e n t e r - α ; y c e n t e r , the inequality x - x c e n t e r 2 α 2 - y - y c e n t e r 2 b 2 = 1 takes place, if with vertices x c e n t e r ; y c e n t e r + b and x c e n t e r ; y c e n t e r - b , then is specified using the inequality x - x c e n t e r 2 α 2 - y - y c e n t e r 2 b 2 = - 1 .
A hyperbola can be represented as two combined functions of the form
y = b a · (x - x c e n t e r) 2 - a 2 + y c e n t e r y = - b a · (x - x c e n t e r) 2 - a 2 + y c e n t e r or y = b a · (x - x c e n t e r) 2 + a 2 + y c e n t e r y = - b a · (x - x c e n t e r) 2 + a 2 + y c e n t e r
In the first case we have that the tangents are parallel to y, and in the second they are parallel to x.
It follows that in order to find the equation of the tangent to a hyperbola, it is necessary to find out which function the point of tangency belongs to. To determine this, it is necessary to substitute into the equations and check for identity.
Example 7
Write an equation for the tangent to the hyperbola x - 3 2 4 - y + 3 2 9 = 1 at point 7; - 3 3 - 3 .
Solution
It is necessary to transform the solution record for finding a hyperbola using 2 functions. We get that
x - 3 2 4 - y + 3 2 9 = 1 ⇒ y + 3 2 9 = x - 3 2 4 - 1 ⇒ y + 3 2 = 9 x - 3 2 4 - 1 ⇒ y + 3 = 3 2 x - 3 2 - 4 and y + 3 = - 3 2 x - 3 2 - 4 ⇒ y = 3 2 x - 3 2 - 4 - 3 y = - 3 2 x - 3 2 - 4 - 3
It is necessary to identify which function a given point with coordinates 7 belongs to; - 3 3 - 3 .
Obviously, to check the first function it is necessary y (7) = 3 2 · (7 - 3) 2 - 4 - 3 = 3 3 - 3 ≠ - 3 3 - 3, then the point does not belong to the graph, since the equality does not hold.
For the second function we have that y (7) = - 3 2 · (7 - 3) 2 - 4 - 3 = - 3 3 - 3 ≠ - 3 3 - 3, which means the point belongs to the given graph. From here you should find the slope.
We get that
y " = - 3 2 (x - 3) 2 - 4 - 3 " = - 3 2 x - 3 (x - 3) 2 - 4 ⇒ k x = y " (x 0) = - 3 2 x 0 - 3 x 0 - 3 2 - 4 x 0 = 7 = - 3 2 7 - 3 7 - 3 2 - 4 = - 3
Answer: the tangent equation can be represented as
y = - 3 x - 7 - 3 3 - 3 = - 3 x + 4 3 - 3
It is clearly depicted like this:
Tangent to a parabola
To create an equation for the tangent to the parabola y = a x 2 + b x + c at the point x 0, y (x 0), you must use a standard algorithm, then the equation will take the form y = y "(x 0) x - x 0 + y ( x 0).Such a tangent at the vertex is parallel to x.
You should define the parabola x = a y 2 + b y + c as the union of two functions. Therefore, we need to solve the equation for y. We get that
x = a y 2 + b y + c ⇔ a y 2 + b y + c - x = 0 D = b 2 - 4 a (c - x) y = - b + b 2 - 4 a (c - x) 2 a y = - b - b 2 - 4 a (c - x) 2 a
Graphically depicted as:
To find out whether a point x 0, y (x 0) belongs to a function, proceed gently according to the standard algorithm. Such a tangent will be parallel to o y relative to the parabola.
Example 8
Write the equation of the tangent to the graph x - 2 y 2 - 5 y + 3 when we have a tangent angle of 150 °.
Solution
We begin the solution by representing the parabola as two functions. We get that
2 y 2 - 5 y + 3 - x = 0 D = (- 5) 2 - 4 · (- 2) · (3 - x) = 49 - 8 x y = 5 + 49 - 8 x - 4 y = 5 - 49 - 8 x - 4
The value of the slope is equal to the value of the derivative at point x 0 of this function and is equal to the tangent of the angle of inclination.
We get:
k x = y "(x 0) = t g α x = t g 150 ° = - 1 3
From here we determine the x value for the points of contact.
The first function will be written as
y " = 5 + 49 - 8 x - 4 " = 1 49 - 8 x ⇒ y " (x 0) = 1 49 - 8 x 0 = - 1 3 ⇔ 49 - 8 x 0 = - 3
It's obvious that real roots no, because we got a negative value. We conclude that there is no tangent with an angle of 150° for such a function.
The second function will be written as
y " = 5 - 49 - 8 x - 4 " = - 1 49 - 8 x ⇒ y " (x 0) = - 1 49 - 8 x 0 = - 1 3 ⇔ 49 - 8 x 0 = - 3 x 0 = 23 4 ⇒ y (x 0) = 5 - 49 - 8 23 4 - 4 = - 5 + 3 4
We have that the points of contact are 23 4 ; - 5 + 3 4 .
Answer: the tangent equation takes the form
y = - 1 3 x - 23 4 + - 5 + 3 4
Let's depict it graphically this way:
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Instructions
We determine the angular coefficient of the tangent to the curve at point M.
The curve representing the graph of the function y = f(x) is continuous in a certain neighborhood of the point M (including the point M itself).
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.
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 this, instead of the letters “x” and “y”, substitute the value of the coordinates of the 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 the equation of a 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.
First level
Equation of a tangent to the graph of a function. Comprehensive guide (2019)
Do you already know what a derivative is? If not, read the topic first. So you say you know the derivative. Let's check it now. Find the increment of the function when the increment of the argument is equal to. Did you manage? It should work. Now find the derivative of the function at a point. Answer: . Happened? If you have any difficulties with any of these examples, I strongly recommend that you return to the topic and study it again. I know the topic is very big, but otherwise there is no point in going further. Consider the graph of some function:
Let's select a certain point on the graph line. Let its abscissa, then the ordinate is equal. Then we select the point with the abscissa close to the point; its ordinate is:
Let's draw a straight line through these points. It is called a secant (just like in geometry). Let us denote the angle of inclination of the straight line to the axis as. As in trigonometry, this angle is measured from the positive direction of the x-axis counterclockwise. What values can the angle take? No matter how you tilt this straight line, one half will still stick up. Therefore, the maximum possible angle is , and the minimum possible angle is . Means, . The angle is not included, since the position of the straight line in this case exactly coincides with, and it is more logical to choose a smaller angle. Let’s take a point in the figure such that the straight line is parallel to the abscissa axis and a is the ordinate axis:
From the figure it can be seen that, a. Then the increment ratio is:
(since it is rectangular).
Let's reduce it now. Then the point will approach the point. When it becomes infinitesimal, the ratio becomes equal to the derivative of the function at the point. What will happen to the secant? The point will be infinitely close to the point, so that they can be considered the same point. But a straight line that has only one common point with a curve is nothing more than tangent(in this case, this condition is met only in a small area - near the point, but this is enough). They say that in this case the secant takes limit position.
Let's call the angle of inclination of the secant to the axis. Then it turns out that the derivative
that is the derivative is equal to the tangent of the angle of inclination of the tangent to the graph of the function at a given point.
Since a tangent is a line, let's now remember the equation of a line:
What is the coefficient responsible for? For the slope of the straight line. This is what it's called: slope. What does it mean? And the fact that it is equal to the tangent of the angle between the straight line and the axis! So this is what happens:
But we got this rule by considering an increasing function. What will change if the function is decreasing? Let's see: 
Now the angles are obtuse. And the increment of the function is negative. Let's consider again: . On the other side, . We get: , that is, everything is the same as last time. Let us again direct the point to the point, and the secant will take a limiting position, that is, it will turn into a tangent to the graph of the function at the point. So, let’s formulate the final rule:
The derivative of a function at a given point is equal to the tangent of the angle of inclination of the tangent to the graph of the function at this point, or (which is the same) slope this tangent:
That's what it is geometric meaning of derivative. Okay, all this is interesting, but why do we need it? Here example:
The figure shows a graph of a function and a tangent to it at the abscissa point. Find the value of the derivative of the function at the point. 
Solution.
As we recently found out, the value of the derivative at the point of tangency is equal to the slope of the tangent, which in turn is equal to the tangent of the angle of inclination of this tangent to the abscissa axis: . This means that to find the value of the derivative we need to find the tangent of the tangent angle. In the figure we have marked two points lying on the tangent, the coordinates of which are known to us. So let's complete the construction of a right triangle passing through these points and find the tangent of the tangent angle! 
The angle of inclination of the tangent to the axis is. Let's find the tangent of this angle: . Thus, the derivative of the function at a point is equal to.
Answer:. Now try it yourself:
Answers:
Knowing geometric meaning of derivative, we can very simply explain the rule that the derivative at the point of a local maximum or minimum is equal to zero. Indeed, the tangent to the graph at these points is “horizontal”, that is, parallel to the x-axis: 
Why equal to the angle between parallel lines? Of course, zero! And the tangent of zero is also zero. So the derivative is equal to zero:
Read more about this in the topic “Monotonicity of functions. Extremum points."
Now let's focus on arbitrary tangents. Let's say we have some function, for example, . We have drawn its graph and want to draw a tangent to it at some point. For example, at a point. We take a ruler, attach it to the graph and draw:
What do we know about this line? What is the most important thing to know about a line on a coordinate plane? Because a straight line is an image linear function, it would be very convenient to know its equation. That is, the coefficients in the equation
But we already know! This is the slope of the tangent, which is equal to the derivative of the function at that point:
In our example it will be like this:
Now all that remains is to find it. It's as simple as shelling pears: after all - the value of. Graphically, this is the coordinate of the intersection of the line with the ordinate axis (after all, at all points of the axis):
Let's draw it (so it's rectangular). Then (to the same angle between the tangent and the x-axis). What are and equal to? The figure clearly shows that, a. Then we get:
We combine all the obtained formulas into the equation of a straight line:
Now decide for yourself:
- Find tangent equation to a function at a point.
- The tangent to a parabola intersects the axis at an angle. Find the equation of this tangent.
- The line is parallel to the tangent to the graph of the function. Find the abscissa of the tangent point.
- The line is parallel to the tangent to the graph of the function. Find the abscissa of the tangent point.
Solutions and answers:
EQUATION OF A TANGENT TO THE GRAPH OF A FUNCTION. BRIEF DESCRIPTION AND BASIC FORMULAS
The derivative of a function at a particular point is equal to the tangent of the tangent to the graph of the function at this point, or the slope of this tangent:
Equation of the tangent to the graph of a function at a point:
Algorithm for finding the tangent equation:
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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 the tangent of the 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 is 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's give right side equations to a 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:
.
The video lesson “Equation of a tangent to the graph of a function” demonstrates educational material to master the topic. During the video lesson, the theoretical material necessary to formulate the concept of the equation of a tangent to the graph of a function at a given point, an algorithm for finding such a tangent, and examples of solving problems using the studied theoretical material are described.
The video tutorial uses methods that improve the clarity of the material. The presentation contains drawings, diagrams, important voice comments, animation, highlighting and other tools.
The video lesson begins with a presentation of the topic of the lesson and an image of a tangent to the graph of some function y=f(x) at the point M(a;f(a)). It is known that the angular coefficient of the tangent plotted to the graph at a given point is equal to the derivative of the function f΄(a) at this point. Also from the algebra course we know the equation of the straight line y=kx+m. The solution to the problem of finding the tangent equation at a point is schematically presented, which reduces to finding the coefficients k, m. Knowing the coordinates of a point belonging to the graph of the function, we can find m by substituting the coordinate value into the tangent equation f(a)=ka+m. From it we find m=f(a)-ka. Thus, knowing the value of the derivative at a given point and the coordinates of the point, we can represent the tangent equation in this way y=f(a)+f΄(a)(x-a).
The following is an example of composing a tangent equation following the diagram. Given the function y=x 2 , x=-2. Taking a=-2, we find the value of the function at a given point f(a)= f(-2)=(-2) 2 =4. We determine the derivative of the function f΄(x)=2x. At this point the derivative is equal to f΄(a)= f΄(-2)=2·(-2)=-4. To compose the equation, all coefficients a=-2, f(a)=4, f΄(a)=-4 were found, so the tangent equation is y=4+(-4)(x+2). Simplifying the equation, we get y = -4-4x.
IN following example It is proposed to create an equation for the tangent at the origin to the graph of the function y=tgx. At a given point a=0, f(0)=0, f΄(x)=1/cos 2 x, f΄(0)=1. So the tangent equation looks like y=x.
As a generalization, the process of composing an equation tangent to the graph of a function at a certain point is formalized in the form of an algorithm consisting of 4 steps:
- Enter the designation a for the abscissa of the tangent point;
- f(a) is calculated;
- f΄(x) is determined and f΄(a) is calculated. The found values of a, f(a), f΄(a) are substituted into the tangent equation formula y=f(a)+f΄(a)(x-a).
Example 1 considers composing the tangent equation to the graph of the function y=1/x at point x=1. To solve the problem we use an algorithm. For a given function at point a=1, the value of the function f(a)=-1. Derivative of the function f΄(x)=1/x 2. At point a=1 the derivative f΄(a)= f΄(1)=1. Using the data obtained, the tangent equation y=-1+(x-1), or y=x-2, is drawn up.
In example 2, it is necessary to find the equation of the tangent to the graph of the function y=x 3 +3x 2 -2x-2. The main condition is the parallelism of the tangent and straight line y=-2x+1. First, we find the angular coefficient of the tangent, equal to the angular coefficient of the straight line y=-2x+1. Since f΄(a)=-2 for a given line, then k=-2 for the desired tangent. We find the derivative of the function (x 3 +3x 2 -2x-2)΄=3x 2 +6x-2. Knowing that f΄(a)=-2, we find the coordinates of point 3a 2 +6a-2=-2. Having solved the equation, we get a 1 =0, and 2 =-2. Using the found coordinates, you can find the tangent equation using a well-known algorithm. We find the value of the function at the points f(a 1)=-2, f(a 2)=-18. The value of the derivative at the point f΄(а 1)= f΄(а 2)=-2. Substituting the found values into the tangent equation, we obtain for the first point a 1 =0 y=-2x-2, and for the second point a 2 =-2 the tangent equation y=-2x-22.
Example 3 describes the composition of the tangent equation for drawing it at the point (0;3) to the graph of the function y=√x. The solution is made using a well-known algorithm. The tangent point has coordinates x=a, where a>0. The value of the function at the point f(a)=√x. The derivative of the function f΄(х)=1/2√х, therefore at a given point f΄(а)=1/2√а. Substituting all the obtained values into the tangent equation, we obtain y = √a + (x-a)/2√a. Transforming the equation, we get y=x/2√а+√а/2. Knowing that the tangent passes through the point (0;3), we find the value of a. We find a from 3=√a/2. Hence √a=6, a=36. We find the tangent equation y=x/12+3. The figure shows the graph of the function under consideration and the constructed desired tangent.
Students are reminded of the approximate equalities Δy=≈f΄(x)Δxand f(x+Δx)-f(x)≈f΄(x)Δx. Taking x=a, x+Δx=x, Δx=x-a, we get f(x)- f(a)≈f΄(a)(x-a), hence f(x)≈f(a)+ f΄(a)(x-a).
In example 4, it is necessary to find the approximate value of the expression 2.003 6. Since it is necessary to find the value of the function f(x)=x 6 at the point x=2.003, we can use the well-known formula, taking f(x)=x 6, a=2, f(a)= f(2)=64, f ΄(x)=6x 5. Derivative at the point f΄(2)=192. Therefore, 2.003 6 ≈65-192·0.003. Having calculated the expression, we get 2.003 6 ≈64.576.
The video lesson “Equation of a tangent to the graph of a function” is recommended for use in a traditional mathematics lesson at school. For a teacher teaching remotely, video material will help explain the topic more clearly. The video can be recommended for students to review independently if necessary to deepen their understanding of the subject.
TEXT DECODING:
We know that if a point M (a; f(a)) (em with coordinates a and ef from a) belongs to the graph of the function y = f (x) and if at this point it is possible to draw a tangent to the graph of the function that is not perpendicular to the axis abscissa, then the angular coefficient of the tangent is equal to f"(a) (eff prime from a).
Let a function y = f(x) and a point M (a; f(a)) be given, and it is also known that f´(a) exists. Let's create an equation for the tangent to the graph of the given function in given point. This equation, like the equation of any straight line that is not parallel to the ordinate axis, has the form y = kx+m (the y is equal to ka x plus em), so the task is to find the values of the coefficients k and m. (ka and em)
Angle coefficient k= f"(a). To calculate the value of m, we use the fact that the desired straight line passes through the point M(a; f (a)). This means that if we substitute the coordinates of the point M into the equation of the straight line, we obtain the correct equality : f(a) = ka+m, from where we find that m = f(a) - ka.
It remains to substitute the found values of the coefficients ki and m into the equation of the straight line:
y = kx+(f(a) -ka);
y = f(a)+k(x-a);
y= f(a)+ f"(a) (x- a). ( y is equal to ef from a plus ef prime from a, multiplied by x minus a).
We have obtained the equation for the tangent to the graph of the function y = f(x) at the point x=a.
If, say, y = x 2 and x = -2 (i.e. a = -2), then f (a) = f (-2) = (-2) 2 = 4; f´(x) = 2x, which means f"(a) = f´(-2) = 2·(-2) = -4. (then the ef of a is equal to four, the ef of the prime of x is equal to two x, which means ef prime from a equals minus four)
Substituting the found values a = -2, f(a) = 4, f"(a) = -4 into the equation, we obtain: y = 4+(-4)(x+2), i.e. y = -4x -4.
(E is equal to minus four x minus four)
Let's create an equation for the tangent to the graph of the function y = tanx (the y is equal to the tangent x) at the origin. We have: a = 0, f(0) = tan0=0;
f"(x)= , which means f"(0) = l. Substituting the found values a=0, f(a)=0, f´(a) = 1 into the equation, we get: y=x.
Let us summarize our steps in finding the equation of the tangent to the graph of a function at point x using an algorithm.
ALGORITHM FOR DEVELOPING AN EQUATION FOR A TANGENT TO THE GRAPH OF THE FUNCTION y = f(x):
1) Designate the abscissa of the tangent point with the letter a.
2) Calculate f(a).
3) Find f´(x) and calculate f´(a).
4) Substitute the found numbers a, f(a), f´(a) into the formula y= f(a)+ f"(a) (x- a).
Example 1. Create an equation for the tangent to the graph of the function y = - in
point x = 1.
Solution. Let us use the algorithm, taking into account that in in this example
2) f(a)=f(1)=- =-1
3) f´(x)=; f´(a)= f´(1)= =1.
4) Substitute the found three numbers: a = 1, f(a) = -1, f"(a) = 1 into the formula. We get: y = -1+(x-1), y = x-2.
Answer: y = x-2.
Example 2. Given the function y = x 3 +3x 2 -2x-2. Write down the equation of the tangent to the graph of the function y = f(x), parallel to the straight line y = -2x +1.
Using the algorithm for composing the tangent equation, we take into account that in this example f(x) = x 3 +3x 2 -2x-2, but the abscissa of the tangent point is not indicated here.
Let's start thinking like this. The desired tangent must be parallel to the straight line y = -2x+1. And parallel lines have equal angular coefficients. This means that the angular coefficient of the tangent is equal to the angular coefficient of the given straight line: k tangent. = -2. Hok cas. = f"(a). Thus, we can find the value of a from the equation f ´(a) = -2.
Let's find the derivative of the function y=f(x):
f"(x)= (x 3 +3x 2 -2x-2)´ =3x 2 +6x-2;f"(a)= 3a 2 +6a-2.
From the equation f"(a) = -2, i.e. 3a 2 +6a-2=-2 we find a 1 =0, a 2 =-2. This means that there are two tangents that satisfy the conditions of the problem: one at the point with abscissa 0, the other at the point with abscissa -2.
Now you can follow the algorithm.
1) a 1 =0, and 2 =-2.
2) f(a 1)= 0 3 +3·0 2 -2∙0-2=-2; f(a 2)= (-2) 3 +3·(-2) 2 -2·(-2)-2=6;
3) f"(a 1) = f"(a 2) = -2.
4) Substituting the values a 1 = 0, f(a 1) = -2, f"(a 1) = -2 into the formula, we get:
y=-2-2(x-0), y=-2x-2.
Substituting the values a 2 = -2, f(a 2) =6, f"(a 2) = -2 into the formula, we get:
y=6-2(x+2), y=-2x+2.
Answer: y=-2x-2, y=-2x+2.
Example 3. From the point (0; 3) draw a tangent to the graph of the function y = . Solution. Let's use the algorithm for composing the tangent equation, taking into account that in this example f(x) = . Note that here, as in example 2, the abscissa of the tangent point is not explicitly indicated. Nevertheless, we follow the algorithm.
1) Let x = a be the abscissa of the point of tangency; it is clear that a >0.
3) f´(x)=()´=; f´(a) =.
4) Substituting the values of a, f(a) = , f"(a) = into the formula
y=f (a) +f "(a) (x-a), we get:
By condition, the tangent passes through the point (0; 3). Substituting the values x = 0, y = 3 into the equation, we get: 3 = , and then =6, a =36.
As you can see, in this example, only at the fourth step of the algorithm we managed to find the abscissa of the tangent point. Substituting the value a =36 into the equation, we get: y=+3
In Fig. Figure 1 shows a geometric illustration of the considered example: a graph of the function y = is constructed, a straight line is drawn y = +3.
Answer: y = +3.
We know that for a function y = f(x), which has a derivative at point x, the approximate equality is valid: Δyf´(x)Δx (delta y is approximately equal to the eff prime of x multiplied by delta x)
or, in more detail, f(x+Δx)-f(x) f´(x) Δx (eff from x plus delta x minus ef from x is approximately equal to eff prime from x by delta x).
For the convenience of further discussion, let us change the notation:
instead of x we will write A,
instead of x+Δx we will write x
Instead of Δx we will write x-a.
Then the approximate equality written above will take the form:
f(x)-f(a)f´(a)(x-a)
f(x)f(a)+f´(a)(x-a). (eff from x is approximately equal to ef from a plus ef prime from a, multiplied by the difference between x and a).
Example 4. Find the approximate value of the numerical expression 2.003 6.
Solution. It's about about finding the value of the function y = x 6 at the point x = 2.003. Let's use the formula f(x)f(a)+f´(a)(x-a), taking into account that in this example f(x)=x 6, a = 2,f(a) = f(2) = 2 6 =64; x = 2.003, f"(x) = 6x 5 and, therefore, f"(a) = f"(2) = 6 2 5 =192.
As a result we get:
2.003 6 64+192· 0.003, i.e. 2.003 6 =64.576.
If we use a calculator, we get:
2,003 6 = 64,5781643...
As you can see, the approximation accuracy is quite acceptable.
