Упр.1543 ГДЗ Алимов 10-11 класс (Алгебра)

1) f(x)=x3-x2/2 + x, x0=1/3;
2) f(x)=lnx/x, x0=1;
3) f(x) = x^-3 — 2/x2 + 3x, x0=3;
4) f(x) = cosx/sinx, x0=пи/4.

1. $$f(x)=x^3-\frac{x^2}{2}+x,\quad x_0=\frac13$$

   $$f'(x)=3x^2-x+1$$

   $$f’\!\left(\frac13\right)=3\cdot\left(\frac13\right)^2-\frac13+1=\frac13-\frac13+1=1$$

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

   $$f'(x)=\frac{(\ln x)’\cdot x-\ln x\cdot x’}{x^2}=\frac{\frac1x\cdot x-\ln x}{x^2}=\frac{1-\ln x}{x^2}$$

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

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

   $$f'(x)=-3x^{-4}+4x^{-3}+3=-\frac{3}{x^4}+\frac{4}{x^3}+3$$

   $$f'(3)=-\frac{3}{3^4}+\frac{4}{3^3}+3=-\frac{3}{81}+\frac{4}{27}+3=\frac{1}{27}+\frac{4}{27}+3=3+\frac{5}{27}=3\frac{5}{27}$$

4. $$f(x)=\frac{\cos x}{\sin x},\quad x_0=\frac{\pi}{4}$$

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

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

Ответ

1) $$1$$; 2) $$1$$; 3) $$3\frac{5}{27}$$; 4) $$-2$$.