Question

Помогите пожалуйста!!!!!!

Answer 1

       $\boxed{\ y=f(x_0)+f'(x_0)(x-x_0)\ }$

$1)\ \ f(x)=2x^3-3x\ \ ,\ \ \ x_0=1\\\\f(1)=2-3=-3\\\\f'(x)=6x^2-3\ \ ,\ \ \ f'(1)=6-3=3\\\\y=-3+3(x-1)\\\\\boxed{\ y=3x\ }\\\\\\2)\ \ f(x)=\sqrt{4x^2+3x}\ \ ,\ \ \ x_0=-1\\\\f(-1)=\sqrt{4-3}=1\\\\f'(x)=\dfrac{8x+3}{2\sqrt{4x^2+3x}}\ \ ,\ \ \ f'(-1)=\dfrac{-8+3}{2\sqrt{4-3}}=\dfrac{-5}{2}=-2,5\\\\y=1-2,5(x+1)\\\\\boxed{\ y=-2,5x-1,5\ }$

$3)\ \ f(x)=2x^3-5x+2\\\\OY:\ x=0\ \ \to \ \ f(0)=2\ \ \ \Rightarrow \ \ \ x_0=2\\\\f(x_0)=f(2)=2\cdot 8-5\cdot 2+2=8\\\\f'(x)=6x^2-5\ \ ,\ \ \ f'(x_0)=f'(2)=6\cdot 4-5=19\\\\y=8+19(x-2)\\\\\boxed{\ y=19x-30\ }$