Question

Монотонность функции y=√(x-5).

Answer 1

Множество значений функции E(y)=[0;+∞). Область определения функции D(y) = [5;+∞). Функция y = √(x-5) возрастает на промежутке [0;+∞).

Answer 2

To determine the monotonicity of the function $y = \sqrt{x-5}$, we need to analyze its first derivative with respect to $x$. If the first derivative is positive, the function is increasing. If the first derivative is negative, the function is decreasing. If the first derivative is zero, it indicates critical points where the function might change its monotonicity.

Let's find the first derivative of the function $y = \sqrt{x-5}$:

$y = \sqrt{x-5} = (x-5)^{\frac{1}{2}}$

Using the power rule for differentiation, the first derivative is:

$y' = \frac{1}{2}(x-5)^{\frac{1}{2}-1} \cdot \frac{d}{dx}(x-5) = \frac{1}{2}(x-5)^{-\frac{1}{2}}$

To investigate monotonicity, we need to analyze the sign of the first derivative. For $x-5 > 0$, which means $x > 5$, the first derivative is always positive. This indicates that the function is monotonically increasing in the interval $x > 5$.

In summary, the function $y = \sqrt{x-5}$ is monotonically increasing for $x > 5$.