Question

F(x)= x^4/2+3x^3+x+11... надите значение производной в точке x0=-2

Answer 1

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Вот решение
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Answer 2

$f(x)= \frac{x^4}{2}+3x^3+x+11 \\\\ f\prime (x)=(\frac{x^4}{2}+3x^3+x+11)\prime=\frac{1}{2}\cdot4x^{4-1}+3\cdot3x^{3-1}+1+0=\\ =2x^3+9x^2+1\\ \\ f\prime (-2)=2\cdot(-2)^3+9\cdot(-2)^2+1=2\cdot(-8)+9\cdot4+1=\\ =-16+36+1=21$

Ответ: 21

Answer 3

To find the derivative of the function $F(x) = \frac{x^4}{2} + 3x^3 + x + 11$ and then evaluate it at $x_0 = -2$, we can follow these steps:

Step 1: Find the derivative of $F(x)$ with respect to $x$. Step 2: Substitute $x_0 = -2$ into the derivative expression to find the value of the derivative at that point.

Let's start with Step 1:

The derivative of $F(x)$ with respect to $x$ can be found by taking the derivative of each term individually:

$F'(x) = \frac{d}{dx} \left( \frac{x^4}{2} \right) + \frac{d}{dx} (3x^3) + \frac{d}{dx} (x) + \frac{d}{dx} (11)$

Using the power rule of differentiation, we get:

$F'(x) = \frac{1}{2} \cdot 4x^3 + 3 \cdot 3x^2 + 1$

Simplifying further:

$F'(x) = 2x^3 + 9x^2 + 1$

Now, let's move to Step 2:

Substitute $x_0 = -2$ into $F'(x)$:

$F'(-2) = 2(-2)^3 + 9(-2)^2 + 1$

$F'(-2) = 2(-8) + 9(4) + 1$

$F'(-2) = -16 + 36 + 1$

$F'(-2) = 21$

So, the value of the derivative of $F(x)$ at $x_0 = -2$ is 21.