Question

с алгебры, буду очень благодарен

Answer 1

${}\ \ \ \ \boxed{\ y=f(x)+f'(x_0)(x-x_0)\ }\\\\1)\ \ f(x)=x^2-4\ \ x_0=2\\\\f(2)=0\\\\f'(x)=2x\ \ ,\ \ f'(2)=4\\\\y=0+4(x-2)\ \ ,\ \ \ \boxed{y=4x-8\ }\\\\\\2)\ \ f(x)=\dfrac{1}{3}\, x^3-2x^2\ \ ,\ \ x_0=3\\\\f(3)=9-18=-9\\\\f'(x)=x^2-4x\ \ ,\ \ f'(3)=9-12=-3\\\\y=-9-3(x-3)\ \ ,\ \ \ \ y=-9-3x+9\ \ ,\ \ \ \boxed{\ y=-3x\ }\\\\\\3)\ \ f(x)=\dfrac{x+1}{x^2+1}\ \ ,\ \ \ x_0=1\\\\f(1)=\dfrac{2}{2}=1\\\\f'(x)=\dfrac{x^2+1-2x(x+1)}{(x^2+1)^2}=\dfrac{-x^2-2x+1}{(x^2+1)^2}\ \ ,\ \ f'(1)=\dfrac{-2}{4}=-\dfrac{1}{2}$

$y=1-\dfrac{1}{2}\cdot (x-1)\ \ ,\ \ \ \boxed{\ y=-\dfrac{1}{2}\, x+\dfrac{3}{2}\ }\\\\\\4)\ \ f(x)=-x^2+3x\\\\OX:\ \ y=0\ \ \ \to \ \ \ -x^2+3x=0\ \ ,\ \ -x(x-3)=0\ \ ,\ \ x_1=0\ ,\ \ x_2=3\\\\f'(x)=-2x+3\\\\a)\ \ f(0)=0\ \ ,\ \ f'(0)=3\\\\y=0+3(x-0)\ \ ,\ \ \ \boxed{\ y=3x\ }\\\\b)\ \ f(3)=-9+9=0\ \ ,\ \ \ f'(3)=-2\cdot 3+3=-3\\\\y=0-3(x-3)\ \ ,\ \ \ \ \boxed{\ y=-3x+9\ }$