Question

Stie cineva sa rezolve subiectul 3? si cat se poate de detaliat multumesc oricum !

Answer 1

multe dar nu prea grele..exceptand 2c) oleaca mai subtila...

Answer 2

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

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

$\displaystyle \mathtt{c)~ \frac{y-f(a)}{x-a}=f'(a) }\\ \\ \mathtt{ \frac{y-f(1)}{x-1}=f'(1) }\\ \\ \mathtt{f(1)=2 \cdot 1^3-9 \cdot 1^2+12\cdot1+1=2 \cdot 1-9\cdot1+12+1=}\\ \\ \mathtt{=2-9+12+1=-7+12+1=5+1=6}\\ \\ \mathtt{f'(x)=\left(2x^3-9x^2+12x+1\right)'=6(x-1)(x-2)}\\ \\ \mathtt{f'(1)=6(1-1)(1-2)=6 \cdot 0 \cdot(-1)=0 \cdot (-1)=0}\\ \\ \mathtt{ \frac{y-6}{x-1} =0}\\ \\ \mathtt{y-6=0}\\ \\ \mathtt{y=0+6}\\ \\ \mathtt{y=6}$

$\displaystyle \mathtt{2.~~~f:\mathbb{R}\rightarrow\mathbb{R},~f(x)=x^2-2x}\\ \\ \mathtt{a)~\int\limits_{-1}^1\left(f(x)+2x\right)dx= \frac{2}{3}}\\ \\ \mathtt{\int\limits_{-1}^1\left(x^2-2x+2x\right)dx=\int\limits_{-1}^1x^2dx= \frac{x^3}{3} \Bigg|_{-1}^1= \frac{1^3}{3}- \frac{(-1)^3}{3}=}\\ \\ \mathtt{= \frac{1}{3}- \frac{-1}{3}= \frac{1}{3} + \frac{1}{3}= \frac{1+1}{3}= \frac{2}{3}}$

$\displaystyle \mathtt{b)~\int\limits_0^1e^x\left(x^2-f(x)\right)dx=\int\limits_0^1e^x\left(x^2-\left(x^2-2x\right)\right)dx=}\\ \\ \mathtt{\int\limits_0^1e^x\left(x^2-x^2+2x\right)dx=\int\limits_0^12xe^xdx=2\int\limits_0^1xe^xdx}$
$\displaystyle \mathtt{\int\limits f(x)g'(x)dx=f(x)g(x)-\int\limits f'(x)g(x)dx}\\ \\ \mathtt{f(x)=x\Rightarrow f'(x)=x'=1}\\ \\ \mathtt{g'(x)=e^x\Rightarrow g(x)=\int\limits e^xdx=e^x}\\ \\ \mathtt{\int\limits xe^xdx=xe^x-\int\limits e^xdx=xe^x-e^x+C=e^x(x-1)+C}\\ \\ \mathtt{2\int\limits_0^1xe^xdx=2\left(e^x(x-1)\right)\Bigg|_0^1=2\left(e^1(1-1)-e^0(0-1)\right)=}\\ \\ \mathtt{=2(e \cdot 0-1 \cdot (-1))=2(0+1)=2 \cdot 1=2}$

$\displaystyle \mathtt{c)~\int\limits_0^1\left(x^2-2x\right)dx=\int\limits_0^1x^2-\int\limits_0^12x=\int\limits_0^1x^2-2\int\limits_0^1x=\frac{x^3}{3} \Bigg|_0^1-2 \frac{x^2}{2} \Bigg|_0^1=}\\ \\ \mathtt{=\left( \frac{1^3}{3}- \frac{0^3}{3}\right)-2\left( \frac{1^2}{2} - \frac{0^2}{2} \right)= \frac{1}{3}-2 \cdot \frac{1}{2}= \frac{1}{3}-1= \frac{1-3}{3}=- \frac{2}{3} }\\ \\ \mathtt{\mathcal{A}=\int\limits_0^1\big|f(x)\big|dx=\left| -\frac{2}{3}\right|= \frac{2}{3}}$