Job type: 7
Condition
The straight line y=3x+2 is tangent to the graph of the function y=-12x^2+bx-10. Find b, given that the abscissa of the tangent point is less than zero.
Show solutionSolution
Let x_0 be the abscissa of the point on the graph of the function y=-12x^2+bx-10 through which the tangent to this graph passes.
The value of the derivative at point x_0 is equal to the slope of the tangent, that is, y"(x_0)=-24x_0+b=3. On the other hand, the point of tangency belongs simultaneously to both the graph of the function and the tangent, that is, -12x_0^2+bx_0-10= 3x_0+2. We get a system of equations \begin(cases) -24x_0+b=3,\\-12x_0^2+bx_0-10=3x_0+2. \end(cases)
Solving this system, we get x_0^2=1, which means either x_0=-1 or x_0=1. According to the abscissa condition, the tangent points are less than zero, so x_0=-1, then b=3+24x_0=-21.
Answer
Job type: 7
Topic: Geometric meaning of derivatives. Tangent to the graph of a function
Condition
The straight line y=-3x+4 is parallel to the tangent to the graph of the function y=-x^2+5x-7. Find the abscissa of the tangent point.
Show solutionSolution
The angular coefficient of the straight line to the graph of the function y=-x^2+5x-7 at an arbitrary point x_0 is equal to y"(x_0). But y"=-2x+5, which means y"(x_0)=-2x_0+5. Angular the coefficient of the line y=-3x+4 specified in the condition is equal to -3. Parallel lines have the same angular coefficients. Therefore, we find a value of x_0 such that =-2x_0 +5=-3.
We get: x_0 = 4.
Answer
Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.
Job type: 7
Topic: Geometric meaning of derivatives. Tangent to the graph of a function
Condition
Show solutionSolution
From the figure we determine that the tangent passes through points A(-6; 2) and B(-1; 1). Let us denote by C(-6; 1) the point of intersection of the lines x=-6 and y=1, and by \alpha the angle ABC (you can see in the figure that it is acute). Then straight line AB forms an angle \pi -\alpha with the positive direction of the Ox axis, which is obtuse.
As is known, tg(\pi -\alpha) will be the value of the derivative of the function f(x) at point x_0. Note that tg \alpha =\frac(AC)(CB)=\frac(2-1)(-1-(-6))=\frac15. From here, using the reduction formulas, we get: tg(\pi -\alpha) =-tg \alpha =-\frac15=-0.2.
Answer
Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.
Job type: 7
Topic: Geometric meaning of derivatives. Tangent to the graph of a function
Condition
The straight line y=-2x-4 is tangent to the graph of the function y=16x^2+bx+12. Find b, given that the abscissa of the tangent point is greater than zero.
Show solutionSolution
Let x_0 be the abscissa of the point on the graph of the function y=16x^2+bx+12 through which
is tangent to this graph.
The value of the derivative at point x_0 is equal to the slope of the tangent, that is, y"(x_0)=32x_0+b=-2. On the other hand, the point of tangency belongs simultaneously to both the graph of the function and the tangent, that is, 16x_0^2+bx_0+12=- 2x_0-4. We obtain a system of equations \begin(cases) 32x_0+b=-2,\\16x_0^2+bx_0+12=-2x_0-4. \end(cases)
Solving the system, we get x_0^2=1, which means either x_0=-1 or x_0=1. According to the abscissa condition, the tangent points are greater than zero, so x_0=1, then b=-2-32x_0=-34.
Answer
Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.
Job type: 7
Topic: Geometric meaning of derivatives. Tangent to the graph of a function
Condition
The figure shows a graph of the function y=f(x), defined on the interval (-2; 8). Determine the number of points at which the tangent to the graph of the function is parallel to the straight line y=6.
Solution
The straight line y=6 is parallel to the Ox axis. Therefore, we find points at which the tangent to the graph of the function is parallel to the Ox axis. On this chart, such points are extremum points (maximum or minimum points). As you can see, there are 4 extremum points.
Answer
Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.
Job type: 7
Topic: Geometric meaning of derivatives. Tangent to the graph of a function
Condition
The straight line y=4x-6 is parallel to the tangent to the graph of the function y=x^2-4x+9. Find the abscissa of the tangent point.
Show solutionSolution
The slope of the tangent to the graph of the function y=x^2-4x+9 at an arbitrary point x_0 is equal to y"(x_0). But y"=2x-4, which means y"(x_0)=2x_0-4. The slope of the tangent y =4x-7, specified in the condition, is equal to 4. Parallel lines have the same angular coefficients. Therefore, we find a value of x_0 such that 2x_0-4=4.
Answer
Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.
Job type: 7
Topic: Geometric meaning of derivatives. Tangent to the graph of a function
Condition
The figure shows the graph of the function y=f(x) and the tangent to it at the point with the abscissa x_0. Find the value of the derivative of the function f(x) at point x_0.
Solution
From the figure we determine that the tangent passes through points A(1; 1) and B(5; 4). Let us denote by C(5; 1) the point of intersection of the lines x=5 and y=1, and by \alpha the angle BAC (you can see in the figure that it is acute). Then straight line AB forms an angle \alpha with the positive direction of the Ox axis.
Tangent is a straight line passing through a point on the curve and coinciding with it at this point up to first order (Fig. 1).
Another definition: this is the limiting position of the secant at Δ x→0.
Explanation: Take a straight line intersecting the curve at two points: A And b(see picture). This is a secant. We will rotate it clockwise until it finds only one common point with the curve. This will give us a tangent.
Strict definition of tangent:
Tangent to the graph of a function f, differentiable at the point xO, is a straight line passing through the point ( xO; f(xO)) and having slope f′( xO).
The slope has a straight line of the form y =kx +b. Coefficient k and is slope this straight line.
Slope factor equal to tangent acute angle, formed by this straight line with the abscissa axis:
|
Here angle α is the angle between the straight line y =kx +b and positive (that is, counterclockwise) direction of the x-axis. It's called angle of inclination of a straight line(Fig. 1 and 2). 
If the angle of inclination is straight y =kx +b acute, then the slope is a positive number. The graph is increasing (Fig. 1).
If the angle of inclination is straight y =kx +b is obtuse, then the slope is a negative number. The graph is decreasing (Fig. 2).
If the straight line is parallel to the x-axis, then the angle of inclination of the straight line is zero. In this case, the slope of the line is also zero (since the tangent of zero is zero). The equation of the straight line will look like y = b (Fig. 3).
If the angle of inclination of a straight line is 90º (π/2), that is, it is perpendicular to the abscissa axis, then the straight line is given by the equality x =c, Where c– some real number (Fig. 4).
Equation of the tangent to the graph of a functiony = f(x) at point xO:
Example: Find the equation of the tangent to the graph of the function f(x) = x 3 – 2x 2 + 1 at the point with abscissa 2.
Solution .
We follow the algorithm.
1) Touch point xO is equal to 2. Calculate f(xO):
f(xO) = f(2) = 2 3 – 2 ∙ 2 2 + 1 = 8 – 8 + 1 = 1
2) Find f′( x). To do this, we apply the differentiation formulas outlined in the previous section. According to these formulas, X 2 = 2X, A X 3 = 3X 2. Means:
f′( x) = 3X 2 – 2 ∙ 2X = 3X 2 – 4X.
Now, using the resulting value f′( x), calculate f′( xO):
f′( xO) = f′(2) = 3 ∙ 2 2 – 4 ∙ 2 = 12 – 8 = 4.
3) So, we have all the necessary data: xO = 2, f(xO) = 1, f ′( xO) = 4. Substitute these numbers into the tangent equation and find the final solution:
y = f(xO) + f′( xO) (x – x o) = 1 + 4 ∙ (x – 2) = 1 + 4x – 8 = –7 + 4x = 4x – 7.
Answer: y = 4x – 7.
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, it becomes clear geometric meaning derivative – calculation of the slope 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 that, 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.
Showing the connection between the sign of the derivative and the nature of the monotonicity of the function.
Please be extremely careful about the following. Look, the schedule of WHAT is given to you! Function or its derivative
If given a graph of the derivative, then we will be interested only in the function signs and zeros. We are not interested in any “hills” or “hollows” in principle!
Task 1.
The figure shows a graph of a function defined on the interval. Determine the number of integer points at which the derivative of the function is negative.
Solution:
In the figure, the areas of decreasing function are highlighted in color:
These decreasing regions of the function contain 4 integer values.
Task 2.
The figure shows a graph of a function defined on the interval. Find the number of points at which the tangent to the graph of the function is parallel to or coincides with the line.
Solution:
Once the tangent to the graph of a function is parallel (or coincides) with a straight line (or, which is the same thing), having slope, equal to zero, then the tangent has an angular coefficient .
This in turn means that the tangent is parallel to the axis, since the slope is the tangent of the angle of inclination of the tangent to the axis.
Therefore, we find extremum points (maximum and minimum points) on the graph - it is at these points that the functions tangent to the graph will be parallel to the axis.
There are 4 such points.
Task 3.
The figure shows a graph of the derivative of a function defined on the interval. Find the number of points at which the tangent to the graph of the function is parallel to or coincides with the line.
Solution:
Since the tangent to the graph of a function is parallel (or coincides) with a line that has an slope, then the tangent also has a slope.
This in turn means that at the touch points.
Therefore, we look at how many points on the graph have an ordinate equal to .
As you can see, there are four such points.
Task 4.
The figure shows a graph of a function defined on the interval. Find the number of points at which the derivative of the function is 0.
Solution:
The derivative is equal to zero at extremum points. We have 4 of them:
Task 5.
The figure shows a graph of a function and eleven points on the x-axis:. At how many of these points is the derivative of the function negative?
Solution:
On intervals of decreasing function, its derivative takes negative values. And the function decreases at points. There are 4 such points.
Task 6.
The figure shows a graph of a function defined on the interval. Find the sum of the extremum points of the function.
Solution:
Extremum points– these are the maximum points (-3, -1, 1) and minimum points (-2, 0, 3).
Sum of extremum points: -3-1+1-2+0+3=-2.
Task 7.
The figure shows a graph of the derivative of a function defined on the interval. Find the intervals of increase of the function. In your answer, indicate the sum of integer points included in these intervals.
Solution:
The figure highlights the intervals where the derivative of the function is non-negative.
There are no integer points on the small increasing interval; on the increasing interval there are four integer values: , , and .
Their sum:
Task 8.
The figure shows a graph of the derivative of a function defined on the interval. Find the intervals of increase of the function. In your answer, indicate the length of the largest of them.
Solution:
In the figure, all intervals on which the derivative is positive are highlighted in color, which means the function itself increases on these intervals.
The length of the largest of them is 6.
Task 9.
The figure shows a graph of the derivative of a function defined on the interval. At what point on the segment does it take on the greatest value?
Solution:
Let's see how the graph behaves on the segment, which is what we are interested in only the sign of the derivative .
The sign of the derivative on is minus, since the graph on this segment is below the axis.
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. A secant MR is drawn through an arbitrary point P(a + ∆x; f(a + ∆x)) of the graph.
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 on 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 the geometric meaning of the 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 the following form:
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).
Let's consider next 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)