Question

Bună seara! Aş dori să ştiu cum se rezolvă Subiectul III, exerciţiul 1 cu subpunctele: a, b şi c. De asemenea, dacă s-ar putea cu o explicaţie amănunţită. Mulţumesc!

Answer 1

$\displaystyle \mathtt{1.~f:\mathbb{R}\rightarrow \mathbb{R},~f(x)=2x^3-9x^2+12x+1}\\ \\ \mathtt{a)~f'(x)=6(x-1)(x-2),~x \in\mathbb{R}}\\ \\ \mathtt{f'(x)=\left(2x^3-9x^2+12x+1\right)'=\left(2x^3\right)'-\left(9x^2\right)'+(12x)'+1'=}\\ \\ \mathtt{=2\left(x^3\right)'-9\left(x^2\right)'+12x'+0=2\cdot3x^2-9\cdot2x+12\cdot1=}\\ \\ \mathtt{=6x^2-18x+12=6\left(x^2-3x+2\right)=6(x-1)(x-2)}$

$\displaystyle \mathtt{b)~ \lim_{x \to +\infty} \frac{2x^3-f(x)}{f'(x)}}\\ \\ \mathtt{ \lim_{x \to +\infty} \frac{2x^3-\left(2x^3-9x^2+12x+1\right)}{6x^2-18x+12}= \lim_{x \to +\infty}\frac{2x^3-2x^3+9x^2-12x-1}{6x^2-18x+12}=}\\ \\ \mathtt{= \lim_{x \to +\infty}\frac{9x^2-12x-1}{6x^2-18x+12}=\lim_{x \to +\infty}\frac{x^2\left(9- \frac{12}{x} - \frac{1}{x^2} \right)}{x^2\left(6- \frac{18}{x}+ \frac{12}{x^2}\right)}=}$
$\displaystyle \mathtt{=\lim_{x \to +\infty} \frac{9- \frac{12}{x} - \frac{1}{x^2} }{6- \frac{18}{x}+ \frac{12}{x^2}}= \frac{9-0-0}{6-0+0}= \frac{9}{6}= \frac{3}{2} }$

$\displaystyle \mathtt{c)~ \frac{y-f(a)}{x-a}=f'(a)}\\ \\ \mathtt{ \frac{y-f(1)}{x-1}=f'(1) }\\ \\ \mathtt{f(1)=2\cdot1^3-9\cdot1^2+12\cdot1+1=2-9+12+1=6}\\ \\ \mathtt{f'(x)=\left(2x^3-9x^2+12x+1\right)'=6(x-1)(x-2)}\\ \\ \mathtt{f'(1)=6(1-1)(1-2)=6\cdot0\cdot(-1)=0}\\ \\ \mathtt{ \frac{y-6}{x-1}=0 }\\ \\ \mathtt{y-6=0}\\ \\ \mathtt{y=6}$