Question

Решите неравенство f’(x)>0: 1) f(x)=cos^2 x/2-sin^2 x/2;

Answer 1

Ответ:

Объяснение:

Сначала упроси функцию,затем найдём производную ,потом решим неравенство

Answer 2

To solve the inequality $f'(x) > 0$ for $f(x) = \frac{\cos^2(x/2) - \sin^2(x/2)}{2}$, we need to find the derivative $f'(x)$ and determine the intervals where it is positive.

Let's find the derivative first:

$f(x) = \frac{\cos^2(x/2) - \sin^2(x/2)}{2}$

Using the chain rule and basic trigonometric derivatives, we get:

$f'(x) = -\frac{1}{2}\sin(x)\cos(x)$

Now, we want to find where $f'(x) > 0$:

$-\frac{1}{2}\sin(x)\cos(x) > 0$

This inequality is true when either $\sin(x) < 0$ and $\cos(x) > 0$ or $\sin(x) > 0$ and $\cos(x) < 0$. These conditions are satisfied in the second and fourth quadrants.

In the second quadrant, $\sin(x) < 0$ and $\cos(x) > 0$, so the inequality holds for $\pi < x < \frac{3\pi}{2}$.

In the fourth quadrant, $\sin(x) > 0$ and $\cos(x) < 0$, so the inequality holds for $\frac{3\pi}{2} < x < 2\pi$.

Therefore, the solution to $f'(x) > 0$ is:

$x \in (\pi, \frac{3\pi}{2}) \cup (\frac{3\pi}{2}, 2\pi)$