Ответы
Ответ дал:
0
1) y = arctg x; x = 0
y(0) = arctg 0 = 0
y ' (x) = 1/(1 + x^2); y ' (0) = 1/(1 + 0) = 1
Касательная
f(x) = y(0) + y ' (0)*(x - x0) = 0 + 1(x - 0) = x
f(x) = x
2)![y= sqrt[3]{x-1} ; x=1 y= sqrt[3]{x-1} ; x=1](https://tex.z-dn.net/?f=y%3D+sqrt%5B3%5D%7Bx-1%7D+%3B+x%3D1+)
![y(1)= sqrt[3]{1-1}=0 y(1)= sqrt[3]{1-1}=0](https://tex.z-dn.net/?f=y%281%29%3D+sqrt%5B3%5D%7B1-1%7D%3D0+)
- не определено.
Касательная в точке x = 1 не существует.
3)![y=ln frac{x^2-2x+1}{x^2+x+1} ;x=0 y=ln frac{x^2-2x+1}{x^2+x+1} ;x=0](https://tex.z-dn.net/?f=y%3Dln+frac%7Bx%5E2-2x%2B1%7D%7Bx%5E2%2Bx%2B1%7D+%3Bx%3D0)
![y(0)=ln frac{0-2*0+1}{0+0+1} =ln(1)=0 y(0)=ln frac{0-2*0+1}{0+0+1} =ln(1)=0](https://tex.z-dn.net/?f=y%280%29%3Dln+frac%7B0-2%2A0%2B1%7D%7B0%2B0%2B1%7D+%3Dln%281%29%3D0)
![y'(x)= frac{x^2+x+1}{x^2-2x+1}frac{(2x-2)(x^2+x+1)-(x^2-2x+1)(2x+1)}{(x^2+x+1)^2} = \ = frac{(2x-2)(x^2+x+1)-(x^2-2x+1)(2x+1)}{(x^2-2x+1)(x^2+x+1)}= frac{2x^3-2-2x^3+3x^2-1}{(x^2-2x+1)(x^2+x+1)}= \ = frac{3x^2-3}{(x^2-2x+1)(x^2+x+1)} y'(x)= frac{x^2+x+1}{x^2-2x+1}frac{(2x-2)(x^2+x+1)-(x^2-2x+1)(2x+1)}{(x^2+x+1)^2} = \ = frac{(2x-2)(x^2+x+1)-(x^2-2x+1)(2x+1)}{(x^2-2x+1)(x^2+x+1)}= frac{2x^3-2-2x^3+3x^2-1}{(x^2-2x+1)(x^2+x+1)}= \ = frac{3x^2-3}{(x^2-2x+1)(x^2+x+1)}](https://tex.z-dn.net/?f=y%27%28x%29%3D+frac%7Bx%5E2%2Bx%2B1%7D%7Bx%5E2-2x%2B1%7Dfrac%7B%282x-2%29%28x%5E2%2Bx%2B1%29-%28x%5E2-2x%2B1%29%282x%2B1%29%7D%7B%28x%5E2%2Bx%2B1%29%5E2%7D+%3D+%5C+%3D+frac%7B%282x-2%29%28x%5E2%2Bx%2B1%29-%28x%5E2-2x%2B1%29%282x%2B1%29%7D%7B%28x%5E2-2x%2B1%29%28x%5E2%2Bx%2B1%29%7D%3D+frac%7B2x%5E3-2-2x%5E3%2B3x%5E2-1%7D%7B%28x%5E2-2x%2B1%29%28x%5E2%2Bx%2B1%29%7D%3D+%5C+%3D+frac%7B3x%5E2-3%7D%7B%28x%5E2-2x%2B1%29%28x%5E2%2Bx%2B1%29%7D++)
![y'(0)= frac{3*0-3}{(0-2*0+1)(0+0+1)}= frac{-3}{1} =-3 y'(0)= frac{3*0-3}{(0-2*0+1)(0+0+1)}= frac{-3}{1} =-3](https://tex.z-dn.net/?f=y%27%280%29%3D+frac%7B3%2A0-3%7D%7B%280-2%2A0%2B1%29%280%2B0%2B1%29%7D%3D+frac%7B-3%7D%7B1%7D+%3D-3)
y(0) = arctg 0 = 0
y ' (x) = 1/(1 + x^2); y ' (0) = 1/(1 + 0) = 1
Касательная
f(x) = y(0) + y ' (0)*(x - x0) = 0 + 1(x - 0) = x
f(x) = x
2)
Касательная в точке x = 1 не существует.
3)
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