Question

Решите уравнение y'(x)= 0, если y(x)= 3x/x^2+1​

Answer 1

Ответ:

y(x)=3*x/(x^2)+1=3/x+1

y'=-3/x^2

обращается в 0, если x=+oo или -oo

Объяснение:

Answer 2

To find the solution to the equation $y'(x) = 0$ for the given function $y(x) = \frac{3x}{x^2 + 1}$, we need to find the values of $x$ where the derivative of $y(x)$ is equal to zero.

First, let's find the derivative of $y(x)$:

$y(x) = \frac{3x}{x^2 + 1}$

To find $y'(x)$, we can use the quotient rule:

$y'(x) = \frac{d}{dx} \left(\frac{3x}{x^2 + 1}\right)$ $y'(x) = \frac{(x^2 + 1) \cdot 3 - 3x \cdot 2x}{(x^2 + 1)^2}$ $y'(x) = \frac{3x^2 + 3 - 6x^2}{(x^2 + 1)^2}$ $y'(x) = \frac{-3x^2 + 3}{(x^2 + 1)^2}$

Now, we need to find the values of $x$ for which $y'(x) = 0$:

$0 = \frac{-3x^2 + 3}{(x^2 + 1)^2}$

Since the numerator $-3x^2 + 3$ is never equal to zero (it's a non-zero constant), the only way for the fraction to be zero is when the denominator $(x^2 + 1)^2$ is equal to infinity. However, since $x^2 + 1$ is always positive, its square $(x^2 + 1)^2$ is also always positive and can never be equal to zero.

Thus, there are no solutions to the equation $y'(x) = 0$ for the given function $y(x) = \frac{3x}{x^2 + 1}$. In other words, the function $y(x)$ does not have any critical points or turning points. It is a monotonic function on its entire domain.