Subiectul III

Se consideră funcția $f:(0,+\infty)\to\mathbb{R},\ f(x)=x-\ln{x}$ .
Calculați $f'(x),\ x\in(0,+\infty)$ .
Determinați ecuația tangentei la graficul funcției în punctul de abscisă , situat pe graficul funcției .
Demonstrați că $x\geq \ln{x}+1$ , pentru orice $x\in(0,+\infty)$ .

Calculăm .

$\begin{align*} f'(x) &=\big(x-\ln{x}\big)'\\ &=x'-(\ln{x})'\\ &=1-\frac{1}{x} \end{align*}$

$\begin{align*} \Leftrightarrow f'(x) &=1-\frac{1}{x},\ \forall\ x\in(0,+\infty).\end{align*}$

Determinăm ecuația tangentei la graficul funcției în punctul de abscisă folosind următoarea formulă:

$\begin{align*} y-f(1)=f'(1)(x-1) \end{align*}$

$\begin{align*} f(1)&=1-\ln{1}\\ &=1-0\\ &=1 \end{align*}$

$\begin{align*} \Leftrightarrow f(1)&=1. \end{align*}$

$\begin{align*} f'(1)&=\Big(f(1)\Big)'\\ &=1'\\ &=0 \end{align*}$

$\begin{align*} \Leftrightarrow f'(1)&=0. \end{align*}$

$\begin{align*} &\Rightarrow y-f(1)=f'(1)(x-1) \\ &\Leftrightarrow y-1=0\cdot(x-1)\\ &\Leftrightarrow y-1=0\\ &\Leftrightarrow y=1 \end{align*}$

Ecuația tangentei la graficul funcției în punctul de abscisă x_0=1 este $\begin{align*} y=1 \end{align*}$ .

Aflăm pe din .

$\begin{align*} &f'(x)=0 \end{align*}$

$\begin{align*} &\Leftrightarrow 1-\frac{1}{x}=0 \\\\ &\Leftrightarrow 1=\frac{1}{x}\\\\ &\Leftrightarrow 1\cdot x=1\\\\ &\Leftrightarrow x=1\end{align*}$

Calculăm $\begin{align*} f'(1) \end{align*}$ :

$\begin{align*} f'(1)&=1-\frac{1}{1}\\ &=1-1\\ &=0\\ \end{align*}$

$\begin{align*} \Leftrightarrow f'(1)&=0. \end{align*}$

$\begin{align*} f'(x)&=1-\frac{1}{x}\\\\ &=\frac{x}{x}-\frac{1}{x}\\\\ &=\frac{x-1}{x}\\ \end{align*}$

$\begin{align*}\Leftrightarrow f'(x)&=\frac{x-1}{x}. \end{align*}$

Facem tabelul de variație:

$\begin{align*} &\Rightarrow f'(x)<0, \ \forall \ x\in(0,1) \end{align*}$

$\Rightarrow f'(x)>0, \ \forall \ x\in(1,+\infty)$

$\begin{align*} &\Rightarrow f(x)\geq f(1) \Leftrightarrow x-\ln{x}\geq 1\end{align*}$

$\Leftrightarrow x\geq 1+\ln{x}, \ \forall \ x\in(0,+\infty).$

Se consideră funcția $f:\mathbb{R}\to\mathbb{R},\ f(x)=x(x+1)(x-1)$ .
Arătați că $\int_{2}^{3} \displaystyle\frac{f(x)}{x(x-1)}\ \mathrm{dx}=\frac{7}{2}.$
Determinați primitiva $F:\mathbb{R}\to\mathbb{R}$ a funcției știind că
Arătați că $\int_{2}^{e} \displaystyle\frac{f(x)\cdot \ln{x}}{x^2-1}\ \mathrm{dx}=\frac{e^2}{4}-2\ln{2}+1.$

Calculăm $\int_{2}^{3} \displaystyle\frac{f(x)}{x(x-1)}\ \mathrm{dx}.$

$\begin{align*} \int_{2}^{3} \displaystyle\frac{f(x)}{x(x-1)}\ \mathrm{dx}&=\int_{2}^{3} \displaystyle\frac{x(x+1)(x-1)}{x(x-1)}\ \mathrm{dx}\\\\ &=\int_{2}^{3} (x+1)\ \mathrm{dx}\\\\ &=\Big(\frac{x^2}{x}+x\Big)\Big|_2^3\\ \\&=\Big(\frac{3^2}{2}+3\Big)-\Big(\frac{2^2}{2}+2\Big)\\\\ &=\frac{9}{2}+3-\frac{4}{2}-2\\\\ &=\frac{9-4}{2}+1\\\\ &=\frac{5}{2}+\frac{2}{2}\\\\ &=\frac{5+2}{2}\\\\ &=\frac{7}{2} \end{align*}$

$\Rightarrow \int_{2}^{3} \displaystyle\frac{f(x)}{x(x-1)}\ \mathrm{dx}=\frac{7}{2}.$

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

$\begin{align*} f(x)&=x(x+1)(x-1)\\ &=x(x^1-1)\\ &=x\cdot x^2-x\cdot x\\ &=x^3-x \end{align*}$

Primitiva $\begin{align*} F \end{align*}$ a funcției o calculăm astfel:

$\begin{align*} \int f(x)\ \mathrm{dx} &= \int (x^3-x)\ \mathrm{dx}\\ &=\frac{x^4}{4}-\frac{x^2}{2}+c \end{align*}$

$\begin{align*} \Leftrightarrow F&=\frac{x^4}{4}-\frac{x^2}{2}+c,\ c\in\mathbb{R}. \end{align*}$

Dar

$\begin{align*}& F(1)=-1 \end{align*}$

$\begin{align*}&\Leftrightarrow \frac{1^4}{4}-\frac{1^2}{2}+c=-1\\\\ &\Leftrightarrow \frac{1}{4}-\frac{1}{2}+c=-1\\\\ &\Leftrightarrow c=-1-\frac{1}{4}+\frac{1}{2}\\\\ &\Leftrightarrow c=-\frac{4}{4}-\frac{1}{4}+\frac{2}{4}\\\\ &\Leftrightarrow c=\displaystyle\frac{-4-1+2}{4}\\\\ &\Leftrightarrow c=-\frac{3}{4} \end{align*}$

$\begin{align*} \Rightarrow F(x)=\frac{x^4}{4}-\frac{x^2}{2}-\frac{3}{4}. \end{align*}$

Calculăm $\int_{2}^{e} \displaystyle\frac{f(x)\cdot \ln{x}}{x^2-1}\ \mathrm{dx}.$

$\begin{align*} \int_{2}^{e} \displaystyle\frac{f(x)\cdot \ln{x}}{x^2-1}\ \mathrm{dx} &= \int_{2}^{e} \displaystyle\frac{x(x+1)(x-1)\cdot \ln{x}}{x^2-1}\ \mathrm{dx} \\\\ &= \int_{2}^{e} \displaystyle\frac{x(x^2-1)\cdot \ln{x}}{x^2-1}\ \mathrm{dx}\\ \\&=\int_2^e x\cdot \ln{x} \ \mathrm{dx}\\ \\&=\frac{x^2}{2}\cdot\ln{x}\Big|_2^e-\int_2^e \frac{1}{x}\cdot\frac{x^2}{2}\ \mathrm{dx}\\\\ &=\Big(\frac{e^2}{2}\cdot\ln{e}-\frac{2^2}{2}\ln{2}\Big)-\frac{1}{2}\int_2^e x\ \mathrm{dx}\\\\ &=\frac{e^2}{2}\cdot1-\frac{4}{2}\ln{2}-\frac{1}{2}\cdot\frac{x^2}{2}\Big|_2^e\\\\ &=\frac{e^2}{2}-2\ln{2}-\frac{1}{2}\Big(\frac{e^2}{2}-\frac{2^2}{2}\Big)\\\\ &=\frac{e^2}{2}-2\ln{2}-\frac{1}{2}\cdot\frac{e^2}{2}+\frac{1}{2}\cdot\frac{4}{2}\\\\ &=\frac{2\cdot e^2}{4}-2\ln{2}-\frac{e^2}{4}+1\\ \\&=\frac{e^2}{4}-2\ln{2}+1\\ \end{align*}$

$\Leftrightarrow \int_{2}^{e} \displaystyle\frac{f(x)\cdot \ln{x}}{x^2-1}\ \mathrm{dx}=\frac{e^2}{4}-2\ln{2}+1.$