Question

И так,"любимый" матан,будьте добры,помогите"Найти промежутки возрастания и убывания функции" y=-1/3x^3+x^2-1 и "Исследовать на выпуклость кривую" y=x^3-9x^2-24x-12

Answer 1

$f(x)=- \frac{1}{3}x^3+x^2-1 \ \ => \ \ f'(x)=-x^2+x \\ x(1-x)>0 \ \ <=> \ \ x \in (0,1) \\ => \ \ f'(x)>0 \ \ <=> \ \ x\in(0,1)$
Итого: $f$ возрастает на $(0,1)$ и убывает на $|R\setminus(0,1)$

$f(x)=x^3-9x^2-24x-12 \ \ => \ \ f'(x)=3x^2-18x-24 \ \ => \\ => \ \ f''(x)=6x-18=6(x-3) \\ 6(x-3)>0 \ \ <=> \ \ x>3 \\ f''(x)>0 \ \ <=> \ \ x>3$
Итого: выпуклое множество на $x>3$

Answer 2

Finding Intervals of Increase and Decrease for the Function y = -1/3x^3 + x^2 - 1

To find the intervals of increase and decrease for the function y = -1/3x^3 + x^2 - 1, we need to determine where the function is increasing and where it is decreasing.

To do this, we can take the derivative of the function and analyze its sign changes.

The derivative of y = -1/3x^3 + x^2 - 1 is given by:

y' = d/dx (-1/3x^3 + x^2 - 1)

Taking the derivative, we get:

y' = -x^2 + 2x

To find the critical points, we set y' equal to zero and solve for x:

-x^2 + 2x = 0

Factoring out x, we get:

x(-x + 2) = 0

This equation is satisfied when x = 0 or x = 2.

Now, we can create a sign chart to determine the intervals of increase and decrease:

| Interval | (-∞, 0) | (0, 2) | (2, ∞) | |-----------|---------|--------|--------| | y' | - | + | - | | y | Decreasing | Increasing | Decreasing |

From the sign chart, we can see that the function is decreasing on the interval (-∞, 0) and (2, ∞), and it is increasing on the interval (0, 2).

Therefore, the function y = -1/3x^3 + x^2 - 1 has intervals of increase on (0, 2) and intervals of decrease on (-∞, 0) and (2, ∞).

Analyzing the Convexity of the Curve y = x^3 - 9x^2 - 24x - 12

To analyze the convexity of the curve y = x^3 - 9x^2 - 24x - 12, we need to determine where the curve is concave up and where it is concave down.

To do this, we can take the second derivative of the function and analyze its sign changes.

The second derivative of y = x^3 - 9x^2 - 24x - 12 is given by:

y'' = d^2/dx^2 (x^3 - 9x^2 - 24x - 12)

Taking the second derivative, we get:

y'' = 6x - 18

To find the inflection points, we set y'' equal to zero and solve for x:

6x - 18 = 0

Solving for x, we get:

x = 3

Now, we can create a sign chart to determine the intervals of concavity:

| Interval | (-∞, 3) | (3, ∞) | |-----------|---------|--------| | y'' | - | + | | y | Concave Down | Concave Up |

From the sign chart, we can see that the curve is concave down on the interval (-∞, 3) and concave up on the interval (3, ∞).

Therefore, the curve y = x^3 - 9x^2 - 24x - 12 is concave down on (-∞, 3) and concave up on (3, ∞).