Question

Ma puteți ajuta cu Exercițiul 1. a si b, va rog..

Answer 1

$\displaystyle 1.~f:\mathbb{R}\rightarrow \mathbb{R},~f(x)=\frac{x-2}{x^2+5} \\\\ a)f'(x)=\frac{(5-x)(x+1)}{\Big(x^2+5\Big)^2} ,~x\in\mathbb{R}\\\\ f'(x)=\Bigg(\frac{x-2}{x^2+5}\Bigg)'=\frac{(x-2)'\Big(x^2+5\Big)-\Big(x^2+5\Big)'(x-2)}{\Big(x^2+5\Big)^2} =\\\\=\frac{(x'-2')\Big(x^2+5\Big)-\Big(\Big(x^2\Big)'+5'\Big)(x-2)}{\Big(x^2+5\Big)^2} =\\\\=\frac{1 \cdot \Big(x^2+5\Big)-2x(x-2)}{\Big(x^2+5\Big)^2}=\frac{x^2+5-2x^2+4x}{\Big(x^2+5\Big)^2}=\frac{-x^2+4x+5}{\Big(x^2+5\Big)^2} =$

$\displaystyle =\frac{-x^2-x+5x+5}{\Big(x^2+5\Big)^2}=\frac{-x(x+1)+5(x+1)}{\Big(x^2+5\Big)^2} =\frac{(5-x)(x+1)}{\Big(x^2+5\Big)^2}$

$\displaystyle b)~ \lim_{x\to \infty} \frac{x-2}{x^2+5} =\lim_{x\to \infty} \frac{(x-2)'}{\Big(x^2+5\Big)'}=\lim_{x\to \infty}\frac{1}{2x} =\frac{1}{2 \cdot \infty} =\frac{1}{\infty} =0\Rightarrow y=0$