Subiectul III

Se consideră funcția $f:\mathbb{R}\to\mathbb{R}$ , $f(x)=\frac{x^2-3}{x^2+3}$ .

Calculați $\lim_{x\to+\infty}f(x)$ .
Arătați că ${f}'(x)=\frac{12x}{(x^2+3)^2}$ , $x\in\mathbb{R}$ .
Arătați că funcția este convexă pe intervalul .

Calculăm $\lim_{x\to+\infty}f(x)$ .

$\begin{align*} \lim_{x\to+\infty}f(x)&=\lim_{x\to+\infty}\frac{x^2-3}{x^2+3}\\\\ &=\lim_{x\to+\infty}\frac{x^2\Big(1-\displaystyle\frac{3}{x^2}\Big)}{x^2\Big(1+\displaystyle\frac{3}{x^2}\Big)} \end{align*}$

$\begin{align*} \frac{3}{x^2}\underset{x\to+\infty}{\longrightarrow}0 \end{align*}$

$\begin{align*} &\Rightarrow \lim_{x\to+\infty}f(x)=\frac{1-0}{1+0}\\\\ &\Leftrightarrow \lim_{x\to+\infty}f(x)=1. \end{align*}$

Calculăm $\begin{align*} {f}'(x) \end{align*}$ .

$\begin{align*} {f}'(x)&={\Big(\frac{x^2-3}{x^2+3}\Big)}'\\\\ &=\frac{{(x^2-3)}'(x^2+3)-{(x^2+3)}'(x^2-3)}{(x^2+3)^2}\\\\ &=\frac{(2x-0)(x^2+3)-(2x+0)(x^2-3)}{(x^2+3)^2}\\\\ &=\frac{2x(x^2+3)-2x(x^2-3)}{(x^2+3)^2}\\\\ &=\frac{2x^3+6x-2x^3+6x}{(x^2+3)^2}\\\\ &=\frac{12x}{(x^2+3)^2} \end{align*}$

Rezultă că:

$\begin{align*} {f}'(x)=\frac{12x}{(x^2+3)^2} \end{align*}$ .

Calculăm $\begin{align*} {f}''(x) \end{align*}$ .

$\begin{align*} {f}''(x)&={f}'(x)={\Big[\frac{12x}{(x^2+3)^2}\Big]}'\\\\ &=\frac{{12x}'(x^2+3)^2-{[(x^2+3)^2]}'\cdot 12x}{[(x^2+3)^2]^2}\\\\ &=\frac{12(x^2+3)^2-12x\cdot 2(x^2+3){(x^2+3)}'}{(x^2+3)^4}\\ &=\frac{12(x^2+3)^2-24x(x^2+3)(2x+0)}{(x^2+3)^4}\\\\ &=\frac{12(x^2+3)^2-48x^2(x^2+3)}{(x^2+3)^4}\\\\ &=\frac{(x^2+3)(12(x^2+3)-48x^2)}{(x^2+3)^4}\\\\ &=\frac{12(x^2+3)-48x^2}{(x^2+3)^3}\\\\ &=\frac{12x^2+36-48x^2}{(x^2+3)^3}\\\\ &=\frac{36-36x^2}{(x^2+3)^3}\\\\ &=\frac{36(1-x^2)}{(x^2+3)^3}\\ \end{align*}$

Rezultă că:

$\begin{align*} {f}''(x)=\frac{36(1-x^2)}{(x^2+3)^3}\\ \end{align*}$

Pentru $\begin{align*} x\in(-1,1)\Rightarrow 1-x^2>0 \end{align*}$ .

Deci:

$\begin{align*} {f}''>0, \forall x\in(-1,1) \end{align*}$

Rezultă că funcția $\begin{align*} f \end{align*}$ este convexă pe intervalul $\begin{align*} (-1,1) \end{align*}$ .

Se consideră funcția $f:(0, +\infty)\to\mathbb{R}$ , $f(x)=\ln x$ .

Arătați că $\int_{1}^{e}f(x)\cdot {f}'(x)\mathrm{d}x=\frac{1}{2}$ .
Arătați că $\int_{1}^{e}x^3f(x)\mathrm{d}x=\frac{3\mathrm{e}^4+1}{16}$ .
Determinați aria suprafeței plane delimitate de graficul funcției , axa și dreptele de ecuații și $x=\mathrm{e}$ .

Calculăm $\int_{1}^{e}f(x)\cdot {f}'(x)\mathrm{d}x.$

$\begin{align*} \int_{1}^{e}f(x)\cdot {f}'(x)\mathrm{d}x&=\int_{1}^{e}\ln x\cdot {(\ln x)}'\mathrm{d}x\\\\ &=\int_{1}^{e}\ln x \cdot \frac{1}{x}\mathrm{d}x\\\\ &=\frac{1}{2}\int_{1}^{e}2\ln x \cdot \frac{1}{x}\mathrm{d}x\\\\ &=\frac{1}{2}\int_{1}^{e} {[\ln^2 x]}'\mathrm{d}x\\\\ &=\frac{1}{2} \ln^2 x\Big|_1^{\mathrm{e}}\\\\ &=\frac{1}{2}\big(\ln ^2 \mathrm{e}- \ln^2 1\big)\\\\ &=\frac{1}{2}(1^2-0)\\\\ &=\frac{1}{2}\cdot1\\\\ &=\frac{1}{2} \end{align*}$

Rezultă că:

$\begin{align*} \int_{1}^{e}f(x)\cdot {f}'(x)\mathrm{d}x=\frac{1}{2}. \end{align*}$

Calculăm $\int_{1}^{e}x^3f(x)\mathrm{d}x.$

$\begin{align*} \int_{1}^{e}x^3f(x)\mathrm{d}x&=\int_{1}^{e}x^3\ln x\mathrm{d}x\\\\ &=\int_{1}^{e}{\Big(\frac{x^4}{4}\Big)}'\ln x\mathrm{d}x\\\\ &=\frac{x^4}{4}\ln x\Big|_1^{\mathrm{e}}-\int_{1}^{e}\Big(\frac{x^4}{4}\Big){(\ln x)}'\mathrm{d}x \\\\ &=\frac{\mathrm{e}^4}{4}\ln \mathrm{e}-\frac{1^4}{4}\ln 1-\int_{1}^{e}\frac{x^4}{4}\cdot \frac{1}{x}\mathrm{d}x \\\\ &=\frac{\mathrm{e}^4}{4}\cdot 1-\frac{1}{4}\cdot 0-\int_{1}^{e}\frac{x^3}{4}\mathrm{d}x \\\\ &=\frac{\mathrm{e}^4}{4}- 0-\frac{1}{4}\cdot\int_{1}^{e}x^3\mathrm{d}x \\\\ &=\frac{\mathrm{e}^4}{4}-\frac{1}{4} \cdot\frac{x^4}{4}\Big|_{1}^{\mathrm{e}}\\\\ &=\frac{\mathrm{e}^4}{4}-\frac{1}{4} \Big(\frac{\mathrm{e}^4}{4}-\frac{1^4}{4}\Big)\\\\ &=\frac{\mathrm{e}^4}{4}-\frac{\mathrm{e}^4}{16}+\frac{1}{16}\\\\ &=\frac{4\mathrm{e}^4-\mathrm{e}^4+1}{16}\\\\ &=\frac{3\mathrm{e}^4+1}{16} \end{align*}$

Am obținut că:

$\begin{align*} \int_{1}^{e}x^3f(x)\mathrm{d}x=\frac{3\mathrm{e}^4+1}{16}. \end{align*}$

Aflăm aria suprafeței plane delimitate de graficul funcției , axa și dreptele de ecuații și $x=\mathrm{e}$ .

$\begin{align*} \mathcal{A}&=\int_{1}^{\mathrm{e}}|f(x)|\mathrm{d}x\\\\ &=\int_{1}^{\mathrm{e}}\ln x\mathrm{d}x\\ \\&=\int_{1}^{\mathrm{e}}1\cdot \ln x\mathrm{d}x\\\\ &=\int_{1}^{\mathrm{e}}{x}'\cdot \ln x\mathrm{d}x\\\\ &=x\ln x\Big|_{1}^{\mathrm{e}}-\int_{1}^{\mathrm{e}}x\cdot{ (\ln x)}'\mathrm{d}x\\\\ &=(\mathrm{e}\cdot \ln \mathrm{e}-1\cdot \ln 1)-\int_{1}^{\mathrm{e}}x\cdot\frac{1}{x}\mathrm{d}x\\\\ &=(\mathrm{e}\cdot 1-1\cdot 0)-\int_{1}^{\mathrm{e}}1\mathrm{d}x\\\\ &=(\mathrm{e}-0)-x\Big|_1^{\mathrm{e}}\\\\ &=\mathrm{e}-(\mathrm{e}-1)\\\\ &=\mathrm{e}-\mathrm{e}+1\\\\ &=1 \end{align*}$

În concluzie, aria suprafeței căutate este:

$\begin{align*} \mathcal{A}=1. \end{align*}$