Question

Докажите, что функция y=5x²+7 возрастает на промежутке [0;+∞).

Answer 1

Функция монотонная, значит нет критических точек. На промежутке 0 + бес. в том числе. Возьмем, например,  0 и 1 и сравним значения f(x). f(0) < f(1) => на всем промежутке 0 + бес. это выполняется.

Answer 2

Т,к коэффициент K-положительный то функция будет возвратстающей 

Answer 3

Proof that the function y = 5x² + 7 is increasing on the interval [0, ∞)

To prove that the function y = 5x² + 7 is increasing on the interval [0, ∞), we need to show that the derivative of the function is positive for all values of x in that interval.

Let's find the derivative of the function y = 5x² + 7:

dy/dx = d/dx (5x² + 7)
      = 10x

Now, we need to show that the derivative dy/dx is positive for all x in the interval [0, ∞).

Let's consider two arbitrary values of x in the interval [0, ∞), say x₁ and x₂, where x₁ < x₂.

If dy/dx > 0 for all x in the interval [0, ∞), then dy/dx evaluated at x₁ should be less than dy/dx evaluated at x₂.

Let's evaluate dy/dx at x₁ and x₂:

dy/dx at x₁ = 10x₁
dy/dx at x₂ = 10x₂

Since x₁ < x₂, it follows that 10x₁ < 10x₂. Therefore, dy/dx at x₁ is less than dy/dx at x₂.

This shows that the derivative dy/dx is positive for all x in the interval [0, ∞), which means that the function y = 5x² + 7 is increasing on that interval.

Therefore, we have proved that the function y = 5x² + 7 is increasing on the interval [0, ∞).