Question

Найти точку минимума y=x^3-27x+11 (решить самим или проверить решение) y'=3x^2-27 3x^2-27=0 3x^2=27 x^2=27/3 x^2=9 x=+-3 Точка минимума -3

Answer 1

На интервале от (-оо; -3) и (3; оо) производная "+"

на (-3;3) "-". Минимум та точка где производная меняет свой знак с "-" на "+". Значит точка минимума = 3

Answer 2

To find the minimum point of the function $y=x^3-27x+11$, we first need to find the critical points. Critical points occur when the derivative of the function is equal to zero or is undefined.

Let's find the derivative of the function $y=x^3-27x+11$ with respect to $x$:

$y' = \frac{d}{dx}(x^3-27x+11) = 3x^2 - 27$

Now, we set the derivative equal to zero and solve for $x$:

$3x^2 - 27 = 0$

Divide both sides by 3:

$x^2 - 9 = 0$

Now, let's solve for $x$:

$x^2 = 9$

Taking the square root of both sides:

$x = \pm 3$

So, we have two critical points: $x = 3$ and $x = -3$.

To determine whether these points are minima or maxima, we can use the second derivative test. We need to find the second derivative of the function and then plug in the critical points.

The second derivative of $y=x^3-27x+11$ with respect to $x$ is:

$y'' = \frac{d^2}{dx^2}(x^3-27x+11) = \frac{d}{dx}(3x^2 - 27) = 6x$

Now, let's evaluate the second derivative at the critical points:

For $x = 3$:

$y'' = 6 \cdot 3 = 18$

For $x = -3$:

$y'' = 6 \cdot (-3) = -18$

Since the second derivative is positive at $x = 3$ (i.e., $y'' > 0$), it means that the function is concave up at that point, indicating a local minimum. Similarly, since the second derivative is negative at $x = -3$ (i.e., $y'' < 0$), it means that the function is concave down at that point, indicating a local maximum.

Thus, the point $x = -3$ corresponds to a local maximum, and the point $x = 3$ corresponds to a local minimum. To find the corresponding $y$ values, let's plug $x = 3$ and $x = -3$ back into the original function:

For $x = 3$:

$y = 3^3 - 27 \cdot 3 + 11 = 27 - 81 + 11 = -43$

So, the local minimum is at the point $(-3, -43)$.