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)

$f(x)=12^x\\f'(x)=12^x\ln12$

а)

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

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

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)

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

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

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)

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

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