Упр.229 ГДЗ Колягин Ткачёва 11 класс (Алгебра)
1) f(х) = х2 + e^-x;
2) f(x) = cos х;
3) f(x) = корень х +1 + еx/2;
4) f(x) = х2 + 3х + 2/2x+1;
5) f(x) = ln(2x+1)+ 3/x+1;
6) f(x) = 2/3*(x + 3) корень x + 3.
$$f(x)=x^2+e^{-x}, \quad x_0=0$$
$$f'(x)=2x-e^{-x}$$
$$f'(0)=0-1=-1$$
Угол наклона касательной к оси $$Ox$$: $$\alpha=\arctg(-1)=-\frac{\pi}{4}$$.
Искомый угол между осью $$Oy$$ и касательной:
$$\beta=\frac{\pi}{2}-\alpha=\frac{\pi}{2}+\frac{\pi}{4}=\frac{3\pi}{4}$$
$$f(x)=\cos x, \quad x_0=0$$
$$f'(x)=-\sin x$$
$$f'(0)=-\sin 0=0$$
$$\alpha=\arctg 0=0$$
$$\beta=\frac{\pi}{2}-0=\frac{\pi}{2}$$
$$f(x)=\sqrt{x+1}+e^{x/2}, \quad x_0=0$$
$$f'(x)=\frac{1}{2\sqrt{x+1}}+\frac{1}{2}e^{x/2}$$
$$f'(0)=\frac{1}{2\sqrt{1}}+\frac{1}{2}e^0=1$$
$$\alpha=\arctg 1=\frac{\pi}{4}$$
$$\beta=\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}$$
$$f(x)=x^2+3x+\frac{2}{2x+1}, \quad x_0=0$$
$$f'(x)=2x+3-\frac{4}{(2x+1)^2}$$
$$f'(0)=0+3-4=-1$$
$$\alpha=\arctg(-1)=-\frac{\pi}{4}$$
$$\beta=\frac{\pi}{2}-\alpha=\frac{\pi}{2}+\frac{\pi}{4}=\frac{3\pi}{4}$$
$$f(x)=\ln(2x+1)+\frac{3}{x+1}, \quad x_0=0$$
$$f'(x)=\frac{2}{2x+1}-\frac{3}{(x+1)^2}$$
$$f'(0)=2-3=-1$$
$$\alpha=\arctg(-1)=-\frac{\pi}{4}$$
$$\beta=\frac{\pi}{2}-\alpha=\frac{3\pi}{4}$$
$$f(x)=\frac{2}{3}(x+3)\sqrt{x+3}, \quad x_0=0$$
$$f(x)=\frac{2}{3}(x+3)^{3/2}$$
$$f'(x)=\frac{2}{3}\cdot\frac{3}{2}(x+3)^{1/2}=\sqrt{x+3}$$
$$f'(0)=\sqrt{3}$$
$$\alpha=\arctg\sqrt{3}=\frac{\pi}{3}$$
$$\beta=\frac{\pi}{2}-\frac{\pi}{3}=\frac{\pi}{6}$$
Ответ
1) $$\frac{3\pi}{4}$$; 2) $$\frac{\pi}{2}$$; 3) $$\frac{\pi}{4}$$; 4) $$\frac{3\pi}{4}$$; 5) $$\frac{3\pi}{4}$$; 6) $$\frac{\pi}{6}$$.
