Question

Помогите пж решить задачу!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 1

$\displaystyle\\f(x)=x^2+1\ \ \ \ x_0=1\\\\y=f'(x_0)(x-x_0)+f(x_0)\\\\f(x_0)=1^2+1=2\\\\f'(x)=2x\\\\f'(x_0)=2*1=2\\\\y=2(x-1)+2=2x-2+2=2x\\\\\\\boxed{y=2x}$

Answer 2

Ответ:

$y=x^2+1\ \ ,\ \ a=1\\\\y(a)=y(1)=1^2+1=2\\\\y'(x)=2x\ \ ,\ \ y'(a)=y'(1)=2\cdot 1=2\\\\kasatelnaya:\ \ \ \underline {y=y(a)+y'(a)(x-a)}\\\\y=2+2(x-1)\\\\y=2+2x-2\\\\\underline {y=2x\ }$