Упр.40.1 ГДЗ Мерзляк 10 класс Углубленный уровень (Алгебра)

Задача

1) f(x)=x^2+3x, x_0=-1; 4) f(x)=tg(x-?/4), x_0=?/2;
2) f(x)=4vx-3, x_0=9; 5) f(x)=x/(x+1), x_0=-2;
3) f(x)=sin(x), x_0=0; 6) f(x)=v(2x+5), x_0=2.

Подробный ответ

$$f(x)=x^2+3x,\quad x_0=-1$$
$$f'(x)=2x+3$$
$$f'(-1)=2\cdot(-1)+3=1$$
$$f(-1)=(-1)^2+3\cdot(-1)=1-3=-2$$
Уравнение касательной:
$$y=1(x+1)-2=x-1$$
$$f(x)=4\sqrt{x}-3,\quad x_0=9$$
$$f'(x)=\frac{4}{2\sqrt{x}}=\frac{2}{\sqrt{x}}$$
$$f'(9)=\frac{2}{\sqrt{9}}=\frac{2}{3}$$
$$f(9)=4\sqrt{9}-3=12-3=9$$
$$y=\frac{2}{3}(x-9)+9=\frac{2}{3}x+3$$
$$f(x)=\sin x,\quad x_0=0$$
$$f'(x)=\cos x$$
$$f'(0)=\cos 0=1$$
$$f(0)=\sin 0=0$$
$$y=1(x-0)+0=x$$
$$f(x)=\tg\left(x-\frac{\pi}{4}\right),\quad x_0=\frac{\pi}{2}$$
$$f\left(\frac{\pi}{2}\right)=\tg\left(\frac{\pi}{2}-\frac{\pi}{4}\right)=\tg\frac{\pi}{4}=1$$
$$f'(x)=\frac{1}{\cos^2\left(x-\frac{\pi}{4}\right)}$$
$$f’\left(\frac{\pi}{2}\right)=\frac{1}{\cos^2\frac{\pi}{4}}=2$$
$$y=2\left(x-\frac{\pi}{2}\right)+1=2x-\pi+1$$
$$f(x)=\frac{x}{x+1},\quad x_0=-2$$
$$f'(x)=\frac{(x+1)-x}{(x+1)^2}=\frac{1}{(x+1)^2}$$
$$f'(-2)=\frac{1}{(-2+1)^2}=1$$
$$f(-2)=\frac{-2}{-2+1}=2$$
$$y=1(x+2)+2=x+4$$
$$f(x)=\sqrt{2x+5},\quad x_0=2$$
$$f'(x)=\frac{2}{2\sqrt{2x+5}}=\frac{1}{\sqrt{2x+5}}$$
$$f'(2)=\frac{1}{\sqrt{2\cdot 2+5}}=\frac{1}{\sqrt{9}}=\frac{1}{3}$$
$$f(2)=\sqrt{2\cdot 2+5}=\sqrt{9}=3$$
$$y=\frac{1}{3}(x-2)+3=\frac{1}{3}x+\frac{7}{3}$$