Question

Помогите пожалуйста с алгеброй

Answer 1

Ответ:

Объяснение:

2.

$f(x)=tg(\frac{x}{3}+\frac{\pi }{6})\ \ \ \ \ (0;y)\ \ \ \ \Rightarrow\ \ \ \ \ x_0=0. \\f(0)=tg(\frac{0}{3}+\frac{\pi }{6})=tg\frac{\pi }{6}=\frac{\sqrt{3} }{3}.\\ f'(x)=(tg(\frac{x}{3}+\frac{\pi }{6}))'=\frac{(\frac{x}{3}+\frac{\pi }{6} )' }{cos^2(\frac{x}{3}+\frac{\pi }{6}) } =\frac{1}{3*cos^2(\frac{x}{3}+\frac{\pi }{6}) } .\\ f'(0)=\frac{1}{3*cos^2(\frac{0}{3}+\frac{\pi }{6} ) } =\frac{1}{3*cos^2(\frac{\pi }{6} ) } = \frac{1}{3*(\frac{\sqrt{3} }{2})^2 } =\frac{1}{3*\frac{3}{4} }=\frac{4}{9}.$

$y_k=\frac{\sqrt{3} }{3}+\frac{4}{9}*(x-0)=\frac{4}{9}x+ \frac{\sqrt{3} }{3} .$

$OTBET:\ y_k=\frac{4x}{9} +\frac{\sqrt{3} }{3} .$

3.

$f(x)=\frac{x+1}{3-x^2} \ \ \ \ \ (x;0)\\\frac{x+1}{3-x^2}=0\\ x+1=0\\x_0=-1.\\f(-1)=\frac{-1+1}{3-(-1)^2} =\frac{0}{2}=0.$

$f'(x)=(\frac{x+1}{3-x^2})'=\frac{(x+1)'*(3-x^2)-(x+1)*(3-x^2)'}{(3-x^2)^2} =\frac{3-x^2-(-2x)*(x+1)}{(3-x^2)^2}=\\ =\frac{3-x^2+2x^2+2x}{(3-x^2)^2} =\frac{x^2+2x+3}{(3-x^2)^2}.\\ f'(-1)=\frac{(-1)^2+2*(-1)+3}{(3-(-1)^2)^2} =\frac{1-2+3}{(3-1)^2}=\frac{2}{2^2} =\frac{2}{4}=0,5.$

$y_k=0+0,5*(x-(-1))=0,5*(x+1)=0,5x+0,5.\\OTBET:\ y_k=0,5x+0,5.$

4.

$f(x)=\frac{x^3}{3}-\frac{9x^2}{8} +x+\sqrt{5} \ \ \ \ y=0,5x+3\ \ \ \ \ \ x_0=?\\f'(x)=(\frac{x^3}{3}-\frac{9x^2}{8} +x+\sqrt{5})'=0,5\\x^2-\frac{9x}{4} +1=\frac{1}{2} \ |*4\\4x^2-9x+4=2\\4x^2-9x+2=0\\D=49\ \ \ \ \sqrt{D}=7\\ x_2=2\ \ \ \ x_2=0,25.\\OTBET:\ x_{0_1}=2,\ \ x_{0_2}=0,25.$