Question

Выпишите уравнение нормали к графику функции в точке с абсциссой а)х₀=1 ; б)х₀=-2 ; в)х₀=0
1) f(x)=12ˣ
2) f(x)=sinх

Answer 1

Уравнение нормали:

$y_n=f(x_0)-dfrac{1}{f'(x_0)} (x-x_0)$

1)

$f(x)=12^x\f'(x)=12^xln12$

а)

$x_0=1\f(x_0)=f(1)=12^1=12\f'(x_0)=f'(1)=12^1ln12=12ln12$

$y_n=12-dfrac{1}{12ln12} (x-1)=12-dfrac{x}{12ln12} +dfrac{1}{12ln12}\\y_n=-dfrac{x}{12ln12} +dfrac{144ln12+1}{12ln12}$

б)

$x_0=-2\f(x_0)=f(1)=12^{-2}=dfrac{1}{144} \\f'(x_0)=f'(-2)=12^{-2}ln12=dfrac{1}{144} ln12$

$y_n=dfrac{1}{144}- dfrac{144}{ln12} (x+2)=dfrac{1}{144}- dfrac{144}{ln12} x- dfrac{288}{ln12} \\y_n=- dfrac{144}{ln12} x+dfrac{1}{144}- dfrac{288}{ln12}$

в)

$x_0=0\f(x_0)=f(1)=12^0=1 \f'(x_0)=f'(-2)=12^0ln12=ln12$

$y_n=1-dfrac{1}{ln12} (x-0)\\y_n=-dfrac{x}{ln12} +1$

2)

$f(x)=sin x\f'(x)=cos x$

а)

$x_0=1\f(x_0)=f(1)=sin1\f'(x_0)=f'(1)=cos1$

$y_n=sin1-dfrac{1}{cos1} (x-1)=sin1-dfrac{x}{cos1} +dfrac{1}{cos1} \\y_n=-dfrac{x}{cos1} +sin1+dfrac{1}{cos1}$

б)

$x_0=-2\f(x_0)=f(-2)=sin(-2)=-sin2\f'(x_0)=f'(-2)=cos(-2)=cos2$

$y_n=-sin2-dfrac{1}{cos2} (x+2)=-sin2-dfrac{x}{cos2} -dfrac{2}{cos2} \\y_n=-dfrac{x}{cos2} -sin2-dfrac{2}{cos2}$

в)

$x_0=0\f(x_0)=f(0)=sin0=0\f'(x_0)=f'(0)=cos0=1$

$y_n=0-dfrac{1}{1} (x-0) \\y_n=-x$

Answer 2

еще раз спасибо