Упр.194 ГДЗ Колмогоров 10-11 класс (Алгебра)

Найдём производные по определению.

1. $$f(x)=x^2-3x$$

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

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

   $$\frac{\Delta f}{\Delta x}=2x+\Delta x-3$$

   При $$\Delta x\to 0$$ получаем $$f'(x)=2x-3$$.

   Тогда $$f'(-1)=2\cdot(-1)-3=-5,$$ $$f'(2)=2\cdot 2-3=1.$$

2. $$f(x)=2x^3$$

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

   $$\frac{\Delta f}{\Delta x}=6x^2+6x\Delta x+2(\Delta x)^2$$

   При $$\Delta x\to 0$$ получаем $$f'(x)=6x^2$$.

   Тогда $$f'(0)=6\cdot 0^2=0,$$ $$f'(1)=6\cdot 1^2=6.$$

3. $$f(x)=\frac{1}{x}$$

   $$\frac{\Delta f}{\Delta x}=\frac{1}{\Delta x}\left(\frac{1}{x+\Delta x}-\frac{1}{x}\right)=-\frac{1}{x(x+\Delta x)}$$

   При $$\Delta x\to 0$$ получаем $$f'(x)=-\frac{1}{x^2}.$$

   Тогда $$f'(-2)=-\frac{1}{(-2)^2}=-\frac14,$$ $$f'(1)=-\frac{1}{1^2}=-1.$$

4. $$f(x)=4-x^2$$

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

   $$\frac{\Delta f}{\Delta x}=-2x-\Delta x$$

   При $$\Delta x\to 0$$ получаем $$f'(x)=-2x.$$

   Тогда $$f'(3)=-2\cdot 3=-6,$$ $$f'(0)=-2\cdot 0=0.$$

Ответ

а) $$f'(-1)=-5,$$ $$f'(2)=1$$; б) $$f'(0)=0,$$ $$f'(1)=6$$; в) $$f'(-2)=-\frac14,$$ $$f'(1)=-1$$; г) $$f'(3)=-6,$$ $$f'(0)=0$$.