Является ли убывающая функция y = x^2 + √x? да нет

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The function given is y = x2 + √x.

To determine if a function is decreasing, we need to look at its derivative. If the derivative is negative over an interval, the function is decreasing on that interval.

First, let's find the derivative of the function:

• The derivative of x2 is 2x.
• The derivative of √x (which is x1/2) is (1/2)x-1/2, or 1 / (2√x).

So, the derivative of y with respect to x (dy/dx) is:

dy/dx = 2x + 1 / (2√x)

Now, let's analyze the sign of the derivative:

• For the function to be defined, x must be greater than or equal to 0 (due to the square root).
• Also, for the derivative term 1 / (2√x) to be defined, x must be strictly greater than 0.

If x > 0:

• 2x is positive.
• 1 / (2√x) is also positive.

Therefore, dy/dx = 2x + 1 / (2√x) will always be positive for any x > 0.

Since the derivative is always positive, the function is always increasing (or at least non-decreasing) on its domain where it is defined.

A decreasing function has a derivative that is negative.

In this case, the derivative is always positive for x > 0.

Ответ: нет