Question

Даю 49+ балов, только решите реще! Пж!

Answer 1

Ответ:

Объяснение:

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Answer 2

$1)\; \; f(x)=sinx+cosx\; \; ,\; \; x_0=\pi \\\\f(x_0)=f(\pi )=sin\pi +cos\pi =0-1=-1\\\\f'(x)=cosx-sinx\\\\f'(x_0)=f'(\pi )=cos\pi -sin\pi =-1-0=-1\\\\y=f(x_0)+f'(x_0)(x-x_0)\\\\y=-1-(x-\pi )\\\\\underline {y=-x+(\pi -1)}\\\\\\2)\; \; f(x)=\frac{x-1}{x+1}\; \; ,\; \; x_0=-2\\\\f'(x)=\frac{x+1-(x-1)}{(x+1)^2}=\frac{2}{(x+1)^2}\\\\f'(-2)=\frac{2}{(-2+1)^2}=\frac{2}{1}=2$

$3)\; \; f(x)=\frac{1}{4}\, x^4- \frac{1}{3}\, x^3-3x^2+2\\\\f'(x)=x^3-x^2-6x\geq 0\\\\a)\; \; x(x^2-x-6)\geq 0\\\\x(x-3)(x+2)\geq 0\; \; \; ---[-2\, ]+++[\, 0\, ]---[\, 3\, ]+++\\\\\underline {x\in [-2,0\, ]\cup [\, 3,+\infty )}\\\\b)\; \; x(x-3)(x+2)<0\; \; \Rightarrow \; \; \underline {x\in (-\infty ,-2)\cup (0,3)}$