Question

F(x)=7x^3-2x-3 в точку x0=2 написать уравнение касательной к графику функции

Answer 1

Уравнение касательной у=а*(х-х0)+в
а -  производная в точке х0. Производная 21x^2-2. В х0=2 это 84-2=82
Значит а=84. Значение функции в х0 равно 7*8-4-3=49.
Ответ : у=82(х-2)+49 или , что то же самое ,  у=82х-115

Answer 2

To find the equation of the tangent line to the graph of the function $F(x) = 7x^3 - 2x - 3$ at the point $x_0 = 2$, we need to follow these steps:

1. Find the slope of the tangent line at $x = 2$, which is the value of the derivative of $F(x)$ at $x = 2$.
2. Find the $y$-coordinate of the point of tangency, which is $F(2)$.
3. Use the point-slope form of the equation of a line to write the equation of the tangent line.

Step 1: Find the derivative of $F(x)$: $F'(x) = \frac{d}{dx}(7x^3 - 2x - 3)$

To find the derivative, we differentiate each term with respect to $x$: $F'(x) = 3 \cdot 7x^2 - 2 \cdot 1 = 21x^2 - 2$

Step 2: Find the $y$-coordinate of the point of tangency: $F(2) = 7 \cdot 2^3 - 2 \cdot 2 - 3 = 56 - 4 - 3 = 49$

Step 3: Write the equation of the tangent line using the point-slope form: The point-slope form of the equation of a line is $y - y_0 = m \cdot (x - x_0)$, where $(x_0, y_0)$ is a point on the line, and $m$ is the slope of the line.

Now, we substitute the values $x_0 = 2$, $y_0 = 49$, and $m = F'(2) = 21 \cdot 2^2 - 2 = 84 - 2 = 82$ into the point-slope form: $y - 49 = 82 \cdot (x - 2)$

Simplify the equation: $y = 82x - 164 + 49$ $y = 82x - 115$

So, the equation of the tangent line to the graph of the function $F(x) = 7x^3 - 2x - 3$ at the point $x = 2$ is $y = 82x - 115$.