Question

F(x)= \frac{ 1 }{ 2 } x ^ { 2 } -3x ​

Answer 1

Ответ:

При х ∈ (- ∞ ; 3] f(x) убывает;

при х ∈ [3; + ∞) f(x) возрастает.

у min = - 4,5.

Пошаговое объяснение:

1) D(f) = R

f(x) = 1/2•x² - 3x - квадратичная, графиком является парабола.

а = 1/2, а > 0, ветви параболы направлены вверх, своего наименьшего значения функция достигает в вершине параболы.

2) х вершины = -b/(2a) = 3/(2•1/2) = 3.

При х ∈ (- ∞ ; 3] f(x) убывает;

при х ∈ [3; + ∞) f(x) возрастает.

х min = 3, y min = f(3) = 1/2•9 - 9 = - 4,5.

Answer 2

The function $F(x) = \frac{1}{2}x^2 - 3x$ is a quadratic function. It is in the form $ax^2 + bx + c$, where:

• $a = \frac{1}{2}$
• $b = -3$
• (c = 0 (since there is no constant term)\

This function represents a parabola that opens upwards because the coefficient $a$ is positive. To find the vertex of the parabola, you can use the formula:

$$x_{\text{vertex}} = -\frac{b}{2a}$$

In this case:

$$x_{\text{vertex}} = -\frac{-3}{2\left(\frac{1}{2}\right)} = 3$$

Now that you know the x-coordinate of the vertex is 3, you can find the corresponding y-coordinate by plugging this value into the function:

$$F(3) = \frac{1}{2}(3^2) - 3(3) = \frac{9}{2} - 9 = -\frac{9}{2}$$

So, the vertex of the parabola is at (3, -9/2), and the axis of symmetry is the vertical line $x = 3$. The parabola opens upward, and it crosses the x-axis at the points where $F(x) = 0$, which are the roots of the quadratic equation:

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

You can solve for $x$ by factoring or using the quadratic formula. In this case, you can factor out $x$ from the equation:

$$x\left(\frac{1}{2}x - 3\right) = 0$$

This equation is satisfied when either $x = 0$ or $\frac{1}{2}x - 3 = 0$. So, the roots are $x = 0$ and $x = 6$.

Therefore, the graph of the function $F(x) = \frac{1}{2}x^2 - 3x$ is a parabola that opens upward, with a vertex at (3, -9/2), and it crosses the x-axis at $x = 0$ and (x = 6).