Question

F (x)=1/2x^2 Xo(Иксноль)=-2

Answer 1

F(x)=х² /2, х0=-2.

Уравнение касательной имеет вид:
y= f(x0) + f’(x0)*(x - x0)
Найдём составляющие: f(x0)=f(-2)=(-2)²/2=4/2=2,
 f’(x)=2х/2=х,                    f’(x0)=f’(-2)=-2,
                                         х-х0=х-(-2)=х+2 и подставим:
y= 2-2(x+2), то есть у=-2х-2 - искомое уравнение касательной.

Answer 2

I'm sorry, but I'm not quite sure what you're asking for.

It appears that you have provided a function F(x) which is defined as 1/2x^2 multiplied by some unknown value Xo (which may be a constant or a function of some variable) and that you have also given a specific value of -2 for the input variable x (i.e. x = -2).

However, there isn't a specific question or request included in your statement. Could you please provide more information or clarify your question?