Question

F'(x)= x^2-4Найти xmax

Answer 1

Ответ:

Пошаговое объяснение:

Answer 2

Осталось выяснить, что известно о функции, непрерывна ли она? Так как это функция у = х³/3-4х+с, т.е. она определена всюду, то ее производная везде определена. Тогда х=±2 -  точки области определения являются критическими, и можно выяснять вопрос о точках максимума. Так как по методу интервалов

____-2_______2__

+               -           +

ясно, что своего максимума функция достигает в точке максимума х=-2, поскольку при переходе через нее производная меняет знак с плюса на  минус.

Answer 3

To find the maximum value of the function, we need to first find the critical points. Critical points are the points where the derivative of the function is either zero or undefined.

Given the function $F'(x) = x^2 - 4$, we need to find where $F'(x) = 0$ and check if there are any points where $F'(x)$ is undefined.

$F'(x) = x^2 - 4 = 0$

To find the critical points, we solve for $x$:

$x^2 - 4 = 0$

Adding 4 to both sides:

$x^2 = 4$

Taking the square root of both sides:

$x = \pm 2$

So, we have two critical points: $x = 2$ and $x = -2$.

Now, to determine which one of these points corresponds to a maximum, minimum, or neither, we need to examine the second derivative test or use a sign chart for the first derivative.

Since $F'(x) = x^2 - 4$, the second derivative will be:

$F''(x) = \frac{d}{dx}(x^2 - 4) = 2x$.

Now, we can evaluate the second derivative at each critical point:

At $x = 2$, $F''(2) = 2(2) = 4$.

At $x = -2$, $F''(-2) = 2(-2) = -4$.

The second derivative test states that:

1. If $F''(x) > 0$ at a critical point, then the function has a local minimum at that point.
2. If $F''(x) < 0$ at a critical point, then the function has a local maximum at that point.
3. If $F''(x) = 0$ at a critical point, the test is inconclusive.

So, in this case, at $x = 2$, $F''(2) > 0$, which means the function has a local minimum at $x = 2$. However, at $x = -2$, $F''(-2) < 0$, which means the function has a local maximum at $x = -2$.

Thus, the value of $x$ that corresponds to the maximum value of the function is $x = -2$.