Помогите пожалуйста 60 балов. Created by kemelbek7. algebra-ru

Помогите пожалуйста 60 балов...

kemelbek7 @kemelbek7

August 2022 1 9 Report

Помогите пожалуйста 60 балов

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NNNLLL54

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$1)\; \; y=x^3+2x^2-4x-3\; \; ,\; \; x_0=-2\\\\y'(x)=3x^2+4x-4\; \; ,\; \; y'(-2)=0\\\\y(-2)=-8+8+8-3=5\\\\kasatelnaya:\; \; y-y(x_0)=y'(x_0)\cdot (x-x_0)\\\\y-5=0(x+2)\; \; \to \; \; \; \boxed {\; y=5\; }\\\\normal\perp kasatelnoj\; \; \to \; \; \; normal:\; \; \boxed {\; x=-2\; }$

$2)\; \; y=x^2-5x+4\; \; ,\; \; x_0=-1\\\\y'=2x-5\; \; ,\; \; \; y'(-1)=-2-5=-7\\\\y(-1)=1+5+4=10\\\\kasatelmnaya:\; \; y-10=-7\, (x+1)\; \; ,\; \; \boxed {\; y=-7x+3\; }\\\\normal:\; \; y-y(x_0)=-\frac{1}{y'(x_0)}\cdot (x-x_0)\\\\y-10=\frac{1}{7}\, (x+1)\; \; ,\; \; \boxed {\; y=\frac{1}{7}\, x+10\frac{1}{7}\; }$

$3)\; \; y=\sqrt{x}\; \; ,\; \; x_0=4\\\\y'=\frac{1}{2\sqrt{x}}\; \; ,\; \; y'(4)=\frac{1}{4}\\\\y(4)=\sqrt4=2\\\\kasanelmnaya:\; \; y-2=\frac{1}{4}\, (x-4)\; \; ,\; \; \boxed {\; y=\frac{1}{4}\, x+1\; }\\\\normal:\; \; y-2=-4(x-4)\; \; ,\; \; \boxed {\; y=-4x+18\; }$

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