Question

1) f(x)=1-1/x^4 f штрих (x) - ?2) f(x)=sinx F(-П)= -1 F(x)=?3) f(x)=x^3+2 M (2;15) ∈ F(x) F(x)=?

Answer 1

Ответ:

$1)f'(x)=\dfrac{4}{x^5}\\2)F(x)=-\cos x-2\\3)F(x)=\dfrac{x^4}{4}+2x+7$

Объяснение:

$1)f(x)=1-\frac{1}{x^4}\Rightarrow f'(x)=(1-\frac{1}{x^4})'=-(\frac{1}{x^4})'=- (-4)\cdot x^{-5}=\frac{4}{x^5}$

$2)f(x)=\sin x\Rightarrow F(x)=-\cos x+C\\F(-\pi)=1+C=-1\Rightarrow C=-2\Rightarrow F(x)=-\cos x-2.$

$3)f(x)=x^3+2\Rightarrow F(x)=\dfrac{1}{4}x^4+2x+C\\M(2;15)\in F(x)\Rightarrow 15=\frac{1}{4}\cdot 2^4+2\cdot 2+C\\15=4+4+C\Rightarrow C=7\Rightarrow F(x)=\dfrac{x^4}{4}+2x+7$

Answer 2

Let's go through each of the given functions and their respective questions:

1. $f(x) = 1 - \frac{1}{x^4}$

To find the derivative $f'(x)$ of the function $f(x)$, we'll use the power rule and the chain rule:

$f'(x) = \frac{4}{x^5}$

2. $f(x) = \sin(x)$

The question asks for the value of $F(x)$ when $F(-\pi) = -1$. Given that $F(x)$ is not explicitly defined, we can't determine the exact form of $F(x)$ with the provided information.

3. $f(x) = x^3 + 2$

The question states that the point $(2, 15)$ is in the function $F(x)$. However, it's not clear how $F(x)$ is related to $f(x)$ in this context. If $F(x)$ is meant to be the integral of $f(x)$, we can find it as follows:

$F(x) = \int f(x) \, dx = \int (x^3 + 2) \, dx = \frac{1}{4}x^4 + 2x + C$

Now, using the given point $(2, 15)$, we can solve for $C$:

$15 = F(2) = \frac{1}{4}(2^4) + 2(2) + C$ $15 = 4 + 4 + C$ $C = 7$

So, the function $F(x)$ in this context is:

$F(x) = \frac{1}{4}x^4 + 2x + 7$