Question

F'(x)>0 если f(x)= cosx - x/2

Answer 1

Решение
F'(x)>0 если f(x)= cosx - x/2
F`(x) = - sunx - 1/2
- sunx - 1/2 > 0
- sinx > 1/2
sinx < - 1/2
- π - arcsin(-1/2) + 2πn < x < arcsin(-1/2) + 2πn, n ∈ Z
- π - (- π/6) + 2πn < x < - π/6 + 2πn, n ∈ Z
- 5π/6 + 2πn < x < - π/6  + 2πn, n ∈ Z

Answer 2

To find the derivative of the function f(x) = cos(x) - x/2, we can differentiate each term separately using the rules of differentiation. Let's calculate it step by step:

1. Differentiating cos(x) with respect to x: The derivative of cos(x) is -sin(x).

2. Differentiating -x/2 with respect to x: The derivative of -x/2 is -1/2.

3. Combining the derivatives: Since the derivative is a linear operator, we can add the derivatives of each term to find the derivative of the entire function: f'(x) = -sin(x) - 1/2.

Now, to determine when f'(x) > 0, we need to find the values of x for which -sin(x) - 1/2 > 0. Let's solve this inequality:

-sin(x) - 1/2 > 0

Adding sin(x) to both sides:

-sin(x) + sin(x) - 1/2 > sin(x)

-sin(x) + sin(x) - 1/2 > 0

-1/2 > sin(x)

Since the range of the sine function is [-1, 1], the inequality -1/2 > sin(x) is always true. Therefore, for any value of x, the derivative f'(x) = -sin(x) - 1/2 is less than zero (negative), not greater than zero (positive).

In conclusion, f'(x) is always negative for the function f(x) = cos(x) - x/2, indicating that the function is decreasing for all values of x.