Упр.37.1 ГДЗ Мерзляк 10 класс Базовый уровень (Алгебра)

1) f(x)=x^2+3x, x_0=-1; 5) f(x)=cos(x), x_0=?;
2) f(x)=1/x, x_0=1/2; 6) f(x)=tg(x-?/4), x_0=?/2;
3) f(x)=4vx-3, x_0=9; 7) f(x)=x/(x+1), x_0=-2;
4) f(x)=sin(x), x_0=0; 8) f(x)=v(2x+5), x_0=2.

1. $$f(x)=x^2+3x,\quad x_0=-1$$

   $$f'(x)=2x+3,\quad f'(-1)=2\cdot(-1)+3=1$$

   $$f(-1)=(-1)^2+3\cdot(-1)=1-3=-2$$

   Уравнение касательной:

   $$y=f'(x_0)(x-x_0)+f(x_0)$$

   $$y=1(x+1)-2=x-1$$

2. $$f(x)=\frac{1}{x},\quad x_0=\frac12$$

   $$f'(x)=-\frac{1}{x^2},\quad f’\!\left(\frac12\right)=-\frac{1}{\left(\frac12\right)^2}=-4$$

   $$f\!\left(\frac12\right)=\frac{1}{1/2}=2$$

   $$y=-4\left(x-\frac12\right)+2=-4x+4$$

3. $$f(x)=4\sqrt{x}-3,\quad x_0=9$$

   $$f'(x)=4\cdot\frac{1}{2\sqrt{x}}=\frac{2}{\sqrt{x}},\quad f'(9)=\frac{2}{3}$$

   $$f(9)=4\sqrt{9}-3=12-3=9$$

   $$y=\frac23(x-9)+9=\frac23x+3$$

4. $$f(x)=\sin x,\quad x_0=0$$

   $$f'(x)=\cos x,\quad f'(0)=1,\quad f(0)=0$$

   $$y=1(x-0)+0=x$$

5. $$f(x)=\cos x,\quad x_0=\pi$$

   $$f'(x)=-\sin x,\quad f'(\pi)=0,\quad f(\pi)=-1$$

   $$y=0\cdot(x-\pi)-1=-1$$

6. $$f(x)=\tg\left(x-\frac{\pi}{4}\right),\quad x_0=\frac{\pi}{2}$$

   $$f'(x)=\frac{1}{\cos^2\left(x-\frac{\pi}{4}\right)}$$

   $$f’\!\left(\frac{\pi}{2}\right)=\frac{1}{\cos^2\left(\frac{\pi}{4}\right)}=\frac{1}{\left(\frac{\sqrt2}{2}\right)^2}=2$$

   $$f\!\left(\frac{\pi}{2}\right)=\tg\frac{\pi}{4}=1$$

   $$y=2\left(x-\frac{\pi}{2}\right)+1=2x-\pi+1$$

7. $$f(x)=\frac{x}{x+1},\quad x_0=-2$$

   $$f'(x)=\frac{(x+1)-x}{(x+1)^2}=\frac{1}{(x+1)^2}$$

   $$f'(-2)=\frac{1}{(-2+1)^2}=1,\quad f(-2)=\frac{-2}{-1}=2$$

   $$y=1(x+2)+2=x+4$$

8. $$f(x)=\sqrt{2x+5},\quad x_0=2$$

   $$f'(x)=\frac{1}{\sqrt{2x+5}},\quad f'(2)=\frac{1}{\sqrt9}=\frac13$$

   $$f(2)=\sqrt{9}=3$$

   $$y=\frac13(x-2)+3=\frac13x+\frac73$$

Ответ

1. $$y=x-1$$
2. $$y=-4x+4$$
3. $$y=\frac23x+3$$
4. $$y=x$$
5. $$y=-1$$
6. $$y=2x-\pi+1$$
7. $$y=x+4$$
8. $$y=\frac13x+\frac73$$