Упр.860 ГДЗ Алимов 10-11 класс (Алгебра)
1) f(x)=x2+x+1,x0=1;
2) f(x)=x-3x,x0=2;
3) f(x)=1/x,x0=3;
4) f(x)=1/x,x0=-2;
5) f(x)=sinx,x0=пи/4;
6) f(x)=ex,x0=0;
7) f(x)=lnx,x0=1;
8) f(x)= корень x,x0=1.
$$f(x)=x^2+x+1,\quad x_0=1$$
$$f'(x)=2x+1,\quad f'(1)=3,\quad f(1)=3$$
$$y=f(x_0)+f'(x_0)(x-x_0)=3+3(x-1)=3x$$
$$f(x)=x-3x^2,\quad x_0=2$$
$$f'(x)=1-6x,\quad f'(2)=-11,\quad f(2)=-10$$
$$y=-10-11(x-2)=12-11x$$
$$f(x)=\frac{1}{x},\quad x_0=3$$
$$f'(x)=-\frac{1}{x^2},\quad f'(3)=-\frac{1}{9},\quad f(3)=\frac{1}{3}$$
$$y=\frac{1}{3}-\frac{1}{9}(x-3)=\frac{2}{3}-\frac{1}{9}x$$
$$f(x)=\frac{1}{x},\quad x_0=-2$$
$$f'(x)=-\frac{1}{x^2},\quad f'(-2)=-\frac{1}{4},\quad f(-2)=-\frac{1}{2}$$
$$y=-\frac{1}{2}-\frac{1}{4}(x+2)=-\frac{1}{4}x-1$$
$$f(x)=\sin x,\quad x_0=\frac{\pi}{4}$$
$$f'(x)=\cos x,\quad f’\!\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2},\quad f\!\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}$$
$$y=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\left(x-\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}x+\frac{\sqrt{2}}{2}-\frac{\pi\sqrt{2}}{8}$$
$$f(x)=e^x,\quad x_0=0$$
$$f'(x)=e^x,\quad f'(0)=1,\quad f(0)=1$$
$$y=1+1(x-0)=x+1$$
$$f(x)=\ln x,\quad x_0=1$$
$$f'(x)=\frac{1}{x},\quad f'(1)=1,\quad f(1)=0$$
$$y=0+1(x-1)=x-1$$
$$f(x)=\sqrt{x},\quad x_0=1$$
$$f'(x)=\frac{1}{2\sqrt{x}},\quad f'(1)=\frac{1}{2},\quad f(1)=1$$
$$y=1+\frac{1}{2}(x-1)=\frac{1}{2}x+\frac{1}{2}$$
Ответ
- $$y=3x$$
- $$y=12-11x$$
- $$y=\frac{2}{3}-\frac{1}{9}x$$
- $$y=-\frac{1}{4}x-1$$
- $$y=\frac{\sqrt{2}}{2}x+\frac{\sqrt{2}}{2}-\frac{\pi\sqrt{2}}{8}$$
- $$y=x+1$$
- $$y=x-1$$
- $$y=\frac{1}{2}x+\frac{1}{2}$$