Subiectul III

Se consideră funcția $f:\mathbb{R}\setminus \{-1\}\to\mathbb{R}$ , $f(x)=\frac{x^2-x-1}{x+1}$ .

Calculați , $x\in\mathbb{R}\setminus \{-1\}$ .
Calculați $\lim_{x\to+\infty}\frac{f(x)\cdot \ln x}{x^2-x-1}$ .
Determinați ecuația asimptotei oblice spre $+\infty$ la graficul funcției .

Calculăm . Fie $x\in\mathbb{R}\setminus \{-1\}$ .

$\begin{align*} {f}'(x)&={\Big(\frac{x^2-x-1}{x+1}\Big)}'\\\\ &=\frac{{(x^2-x-1)}'(x+1)-(x^2-x-1){(x+1)}'}{(x+1)^2}\\\\ &=\frac{(2x-1-0)(x+1)-(x^2-x-1)(1+0)}{(x+1)^2}\\\\ &=\frac{(2x-1)(x+1)-(x^2-x-1)\cdot 1}{(x+1)^2}\\\\ &=\frac{2x\cdot x+ 2x\cdot 1-1\cdot x-1\cdot 1-x^2+x+1}{(x+1)^2}\\\\ &=\frac{x^2+2x}{(x+1)^2} \end{align*}$

$\begin{align*} \Rightarrow {f}'(x)=\frac{x^2+2x}{(x+1)^2}, \forall x\in\mathbb{R}\setminus\{-1\} \end{align*}$ .

Calculăm $\lim_{x\to+\infty}\frac{f(x)\cdot \ln x}{x^2-x-1}$ .

$\begin{align*} &\lim_{x\to+\infty}\frac{f(x)\cdot \ln x}{x^2-x-1}\\\\ &=\lim_{x\to+\infty}\frac{\displaystyle\frac{x^2-x-1}{x+1}\cdot \ln x}{x^2-x-1}\\\\ &=\lim_{x\to+\infty}\frac{\ln x}{x+1} \end{align*}$

Dar $\begin{align*} \frac{\ln x}{x+1}\underset{x\to+\infty}{\longrightarrow}\frac{\infty}{\infty} \end{align*}$ .

Deci avem un caz de nedeterminare.

Aplicând regulile lui l'Hospital, obținem:

$\begin{align*} &\lim_{x\to+\infty}\frac{f(x)\cdot \ln x}{x^2-x-1}\\\\ &=\lim_{x\to+\infty}\frac{\ln x}{x+1}\\\\ &=\lim_{x\to+\infty}\frac{{(\ln x)}'}{{(x+1)}'}\\\\ &=\lim_{x\to+\infty}\frac{\displaystyle\frac{1}{x}}{1+0}\\\\ &=\lim_{x\to+\infty}\frac{\displaystyle\frac{1}{x}}{1}\\\\ &=\lim_{x\to+\infty}\frac{1}{x}\\\\ &=0 \end{align*}$

$\begin{align*} \Rightarrow \lim_{x\to+\infty}\frac{f(x)\cdot \ln x}{x^2-x-1}=0 \end{align*}$ .

Ecuația asimptotei oblice spre $+\infty$ la graficul funcției este .

$\begin{align*} m&=\lim_{x\to+\infty}\frac{f(x)}{x}\\\\ &=\lim_{x\to+\infty}\frac{x^2-x-1}{x(x+1)}\\\\ &=\lim_{x\to+\infty}\frac{x^2-x-1}{x^2+x}\\ \\&=\lim_{x\to+\infty}\frac{x^2\Big(1-\displaystyle\frac{1}{x}-\displaystyle\frac{1}{x^2}\Big)}{x^2\Big(1+\displaystyle\frac{1}{x}\Big)} \end{align*}$

Dar $\begin{align*} \frac{1}{x}\underset{x\to+\infty}{\longrightarrow}0 \end{align*}$ și $\begin{align*} \frac{1}{x^2}\underset{x\to+\infty}{\longrightarrow}0 \end{align*}$ .

Rezultă că $\begin{align*}m=1 \end{align*}$ .

$\begin{align*} n&=\lim_{x\to+\infty}\big(f(x)-mx\big)\\\\ &=\lim_{x\to+\infty}\Big(\frac{x^2-x-1}{x+1}-x\Big)\\\\ &=\lim_{x\to+\infty}\frac{x^2-x-1-x(x+1)}{x+1}\\\\ &=\lim_{x\to+\infty}\frac{x^2-x-1-x^2-x}{x+1}\\\\ &=\lim_{x\to+\infty}\frac{-2x-1}{x+1}\\\\ &=\lim_{x\to+\infty}\frac{x\Big(-2-\displaystyle\frac{1}{x}\Big)}{x\Big(1+\displaystyle\frac{1}{x}\Big)} \end{align*}$

Dar $\begin{align*} \frac{1}{x}\underset{x\to+\infty}{\longrightarrow}0 \end{align*}$ .

Rezultă că $\begin{align*} n=-2 \end{align*}$ .

Deci, ecuația asimptotei oblice spre $\begin{align*} +\infty \end{align*}$ la graficul funcției $\begin{align*} f \end{align*}$ este $\begin{align*} y=x-2 \end{align*}$ .

Se consideră funcția $f:(0,+\infty)\to\mathbb{R}$ , $f(x)=e^x\cdot \sqrt{x+1}$ .

Determinați primitivele funcției $g:(0,+\infty)\to\mathbb{R}$ , $g(x)=\frac{f(x)}{\sqrt{x+1}}$ .
Calculați $\int^2_1 \sqrt{x+1}\cdot f(x)\mathrm{d}x$ .
Calculați aria suprafeței determinate de graficul funcției $h:(0,+\infty)\to\mathbb{R}$ , $h(x)=e^{-x}\cdot f(x)$ , axa și dreptele de ecuații și .

Determinăm primitivele funcției .

$g(x)=\frac{e^x\cdot \sqrt{x+1}}{\sqrt{x+1}}\Leftrightarrow$

$\Leftrightarrow g(x)=e^x$

$\begin{align*} \Rightarrow \int g(x)\mathrm{d}x=\int e^x\mathrm{d}x \end{align*}$

$\begin{align*} \Leftrightarrow \int g(x)\mathrm{d}x=e^x+C, C\in\mathbb{R} \end{align*}$ .

Calculăm $\int^2_1 \sqrt{x+1}\cdot f(x)\mathrm{d}x$ .

$\begin{align*} &\int^2_1 \sqrt{x+1}\cdot f(x)\mathrm{d}x \\\\ &=\int^2_1 \sqrt{x+1}\cdot e^x\cdot \sqrt{x+1}\mathrm{d}x\\\\ &=\int^2_1 e^x( \sqrt{x+1})^2 \mathrm{d}x\\\\ &=\int^2_1 e^x( x+1) \mathrm{d}x\\\\ &=(x+1)e^x\Big|_1^2-\int^2_1 e^x{( x+1)}' \mathrm{d}x\\\\ &=(2+1)e^2-(1+1)e^1-\int^2_1 e^x\cdot 1 \mathrm{d}x\\\\ &=3e^2-2e-\int^2_1 e^x \mathrm{d}x\\\\ &=3e^2-2e-e^x\Big|_1^2\\\\ &=3e^2-2e-(e^2-e^1)\\\\ &=3e^2-2e-e^2+e\\\\ &=2e^2-e \end{align*}$

$\begin{align*} \Rightarrow \int^2_1 \sqrt{x+1}\cdot f(x)\mathrm{d}x =2e^2-e. \end{align*}$

Calculăm aria cerută în enunț.

$\begin{align*} A&= \int_{2}^{3}|h(x)|\mathrm{d}x\\\\ &= \int_{2}^{3}|e^{-x}\cdot f(x)|\mathrm{d}x\\ \\&= \int_{2}^{3}|e^{-x}\cdot e^x\cdot \sqrt{x+1}|\mathrm{d}x\\\\ &= \int_{2}^{3}|e^{-x+x}\cdot \sqrt{x+1}|\mathrm{d}x\\\\ &= \int_{2}^{3}|e^{0}\cdot \sqrt{x+1}|\mathrm{d}x\\\\ &= \int_{2}^{3}|1\cdot \sqrt{x+1}|\mathrm{d}x\\\\ &= \int_{2}^{3}| \sqrt{x+1}|\mathrm{d}x\\\\ &= \int_{2}^{3} \sqrt{x+1}\mathrm{d}x\\\\ &= \int_{2}^{3}(x+1)^{\frac{1}{2}}\mathrm{d}x\\\\ &=\frac{(x+1)^{\frac{1}{2}+1}}{\displaystyle\frac{1}{2}+1}\Big|_2^3\\\\ &=\frac{(x+1)^{\frac{3}{2}}}{\displaystyle\frac{3}{2}}\Big|_2^3\\\\ &=\frac{2}{3}(x+1)\sqrt{x+1}\Big|_2^3\\\\ &=\frac{2}{3}\Big[(3+1)\sqrt{3+1}-(2+1)\sqrt{2+1}\Big]\\\\ &=\frac{2}{3}\big(4\sqrt{4}-3\sqrt{3}\big)\\\\ &=\frac{2}{3}(4\cdot 2-3\sqrt{3})\\\\ &=\frac{2}{3}(8-3\sqrt{3}) \end{align*}$

$\begin{align*} \Rightarrow A=\frac{2}{3}(8-3\sqrt{3}). \end{align*}$