Question

F(x)=ctgx-cosx f'(x)=0​

Answer 1

ctgx=-1/sin^2x

cosx=-sinx

подстовляем значение в выражение

-1/sin^2x-(-sinx)=-1/sin^2x+sinx. это и будет ответом

Answer 2

To find the critical points of the function $f(x) = \cot(x) - \cos(x)$, we first need to calculate its derivative, $f'(x)$, and then solve for the values of $x$ where $f'(x) = 0$. The derivative of $f(x)$ is found using the rules of differentiation:

$f(x) = \cot(x) - \cos(x)$

To find $f'(x)$, we'll differentiate each term separately:

1. Differentiate $\cot(x)$:

Using the chain rule, $\frac{d}{dx}(\cot(x)) = -\csc^2(x)\cdot \frac{d}{dx}(x) = -\csc^2(x)$.

2. Differentiate $-\cos(x)$:

The derivative of $-\cos(x)$ is $\sin(x)$ since the derivative of $\cos(x)$ is $-\sin(x)$.

Now, we can find $f'(x)$:

$f'(x) = -\csc^2(x) - \sin(x)$

To find the critical points, set $f'(x) = 0$ and solve for $x$:

$-\csc^2(x) - \sin(x) = 0$

Let's simplify this equation:

$-\csc^2(x) - \sin(x) = 0$

Divide through by $-1$ to make it easier to work with:

$\csc^2(x) + \sin(x) = 0$

Now, you can try to solve this equation for $x$. Unfortunately, there isn't a simple algebraic solution for $x$ in this case. You would typically need to use numerical methods or a graphing calculator/computer software to approximate the values of $x$ where $f'(x) = 0$. These values of $x$ will be the critical points of the function $f(x) = \cot(x) - \cos(x)$.