Упр.180 ГДЗ Колмогоров 10-11 класс (Алгебра)

а) $$\Delta f=f(x_0+\Delta x)-f(x_0)=1-3(x_0+\Delta x)^2-(1-3x_0^2)$$

$$=1-3x_0^2-6x_0\Delta x-3(\Delta x)^2-1+3x_0^2=-6x_0\Delta x-3(\Delta x)^2$$

$$\Delta f=-3\Delta x(2x_0+\Delta x)$$

б) $$\Delta f=f(x_0+\Delta x)-f(x_0)=a(x_0+\Delta x)+b-(ax_0+b)=a\Delta x$$

в) $$\Delta f=f(x_0+\Delta x)-f(x_0)=2(x_0+\Delta x)^2-2x_0^2$$

$$=2(x_0^2+2x_0\Delta x+(\Delta x)^2)-2x_0^2=4x_0\Delta x+2(\Delta x)^2$$

г) $$\Delta f=f(x_0+\Delta x)-f(x_0)=\frac{1}{x_0+\Delta x}-\frac{1}{x_0}$$

$$=\frac{x_0-(x_0+\Delta x)}{x_0(x_0+\Delta x)}=-\frac{\Delta x}{x_0(x_0+\Delta x)}$$

Ответ

а) $$\Delta f=-3\Delta x(2x_0+\Delta x)$$

б) $$\Delta f=a\Delta x$$

в) $$\Delta f=4x_0\Delta x+2(\Delta x)^2$$

г) $$\Delta f=-\dfrac{\Delta x}{x_0(x_0+\Delta x)}$$