Question

F(x)=7x^2-3x+15 x0=3;-2 найти к1 косательная и к2 косательная​

Answer 1

Производная функции в заданной точке равна тангенсу угла наклона касательной к графику функции в этой точке.

Производная равна -8х + 7 = -9

                                     -8х = -16

                                         х = -16/-8 = 2.

Подставим это значение в уравнение f(x)=3+7x-4x^2  и находим координату по оси у:

у = 3 + 7*2 - 4*2² = 3 + 14 - 16 = 17 - 16 = 1

Ответ: (2; 1)

Answer 2

To find the equations of the tangent lines to the function f(x) = 7x^2 - 3x + 15 at the points x0 = 3 and x0 = -2, we need to calculate the slopes of the tangents at those points and then use the point-slope form to determine the equations.

Let's start by finding the derivative of the function f(x) to obtain the slope of the tangent line at any given point:

f(x) = 7x^2 - 3x + 15

Taking the derivative with respect to x:

f'(x) = d/dx (7x^2 - 3x + 15) = 14x - 3

Now, let's find the slope of the tangent line at x = 3:

f'(3) = 14(3) - 3 = 39

So, the slope of the tangent line at x = 3 is 39.

Using the point-slope form of a linear equation, which is y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope, we can write the equation of the tangent line at x = 3:

y - f(3) = 39(x - 3)

Next, let's find the slope of the tangent line at x = -2:

f'(-2) = 14(-2) - 3 = -31

The slope of the tangent line at x = -2 is -31.

Using the point-slope form, the equation of the tangent line at x = -2 can be written as:

y - f(-2) = -31(x + 2)

Now, we need to calculate the values of f(3) and f(-2) to complete the equations.

For x = 3: f(3) = 7(3)^2 - 3(3) + 15 = 63 - 9 + 15 = 69

For x = -2: f(-2) = 7(-2)^2 - 3(-2) + 15 = 28 + 6 + 15 = 49

Substituting these values into the equations, we get:

Equation of the tangent line at x = 3: y - 69 = 39(x - 3)

Equation of the tangent line at x = -2: y - 49 = -31(x + 2)

These are the equations of the tangent lines to the function f(x) = 7x^2 - 3x + 15 at the points x = 3 and x = -2, respectively.