Question

F(x)=-1/3x^3+x^2+3x f'(x)>0

Answer 1

$f(x)=-\frac{1}{3}x^3+x^2+3x\\f'(x)=(-\frac{1}{3}x^3+x^2+3x)'=-x^2+2x+3\\-x^2+2x+3>0\\D=2^2-4\cdot(-1)\cdot(-3)=4+12=16\\\sqrt D=\sqrt{16}=4\\x_1=\frac{-2-4}{2\cdot(-1)}=\frac{-6}{-2}=3\\x_2=\frac{-2+4}{2\cdot(-1)}=\frac{2}{-2}=-1$

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 -       -1      +      3    -

Ответ :

$x\in(-1;3)$

Answer 2

To determine where the function F(x) is increasing, we need to find the derivative of F(x) and then determine where it is greater than zero.

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

F'(x) = (-1)x^2 + 2x + 3

Now we need to find the values of x where F'(x) is greater than zero:

F'(x) > 0

(-1)x^2 + 2x + 3 > 0

To solve for x, we can use the quadratic formula:

x = (-b ± sqrt(b^2 - 4ac)) / 2a

where a = -1, b = 2, and c = 3.

Plugging in the values, we get:

x = (-2 ± sqrt(2^2 - 4(-1)(3))) / 2(-1)

x = (-2 ± sqrt(16)) / (-2)

x = (-2 ± 4) / (-2)

x = -1 or x = 3

So the function F(x) is increasing on the intervals (-∞, -1) and (3, ∞).