Question

va roggg, dau coroana, macar un exercițiu, va implorrrr​

Answer 1

Explicație pas cu pas:

a)

$f^{\prime}(x) = \Big(\dfrac{2x}{3x + 1} \Big)^{\prime} = \dfrac{(2x)^{\prime}(3x + 1) - 2x(3x + 1)^{\prime}}{(3x + 1)^{2}} = \\ = \dfrac{2(3x + 1) - 2x \cdot 3}{(3x + 1)^{2}} = \dfrac{6x + 2 - 6x}{(3x + 1)^{2}} = \bf \dfrac{2}{(3x + 1)^{2}}$

b)

$\lim _{x \rightarrow +\infty}f(x) = \lim _{x \rightarrow +\infty} \dfrac{2x}{3x + 1} = \lim _{x \rightarrow +\infty} \dfrac{2}{3 + \dfrac{1}{x}} = \\ = \dfrac{2}{3 + \dfrac{1}{\infty}} = \dfrac{2}{3 + 0} = \dfrac{2}{3}$

ecuația asimptotei orizontale la +∞:

$\implies \bf y = \dfrac{2}{3}$

c)

$f^{"}(x) = \bigg(\dfrac{2}{(3x + 1)^{2}} \bigg)^{\prime} = 2\bigg(\dfrac{1}{(3x + 1)^{2}} \bigg)^{\prime} = \\ = 2\bigg( - \dfrac{2(3x + 1)^{\prime}}{(3x + 1)^{3}} \bigg) = - \dfrac{4 \cdot 3}{(3x + 1)^{3}} = \bf - \dfrac{12}{(3x + 1)^{3}}$

3x + 1 > 0 => f"(x) < 0 => f(x) este concavă pe domeniul de definiție

Answer 2

$\displaystyle2.~~f:(0,+\infty)\rightarrow\mathbb{R},~f(x)=x^2+lnx-1\\\\ a)~ \int\limits^4_1 (f(x)-lnx+1)dx=21\\\\\int\limits^4_1 (f(x)-lnx+1)dx=\int\limits^4_1 \Big(x^2+lnx-1-lnx+1\Big)dx=\int\limits^4_1x^2dx=\\\\ =\cfrac{x^3}{3} \Bigg|_1^4=\cfrac{4^3}{3} -\cfrac{1^3}{3} =\cfrac{64-1}{3}=\cfrac{63}{3} =21$

$\displaystyle b)~\int\limits^4_2\cfrac{x}{f(x)-lnx} dx=\cfrac{1}{2}ln5 \\\\ \int\limits^4_2\cfrac{x}{f(x)-lnx} dx=\int\limits^4_2 \cfrac{x}{x^2+lnx-1-lnx} dx=\int\limits^4_2\cfrac{x}{x^2-1}dx=\\\\=\int\limits^4_2 \cfrac{x}{(x-1)(x+1)}dx=\cfrac{1}{2} \int\limits^4_2 \cfrac{1}{x-1} dx+\cfrac{1}{2} \int\limits^4_2\cfrac{1}{x+1} dx=\\\\=\cfrac{1}{2} \cdot ln|x-1|\Bigg|_2^4+\cfrac{1}{2} \cdot ln|x+1|\Bigg|_2^4=$

$=\cfrac{1}{2} \cdot(ln|4-1|-ln|2-1|)+\cfrac{1}{2} \cdot (ln|4+1|-ln|2+1|)=\\\\=\cfrac{1}{2} \cdot (ln3-ln1)+\cfrac{1}{2} \cdot(ln5-ln3)=\cfrac{1}{2} \cdot ln3+\cfrac{1}{2} \cdot ln\cfrac{5}{3} =\\\\=\cfrac{1}{2}\Bigg(ln3+ln\cfrac{5}{3} \Bigg)=\cfrac{1}{2} ln\Bigg(\not3 \cdot \cfrac{5}{\not3} \Bigg)=\cfrac{1}{2}ln5$

$\displaystyle c)~\int\limits^a_1\cfrac{f(x)}{x^2} dx=\cfrac{a-lna}{a} ,~a\in(1,+\infty)\\\\\int\limits^a_1\cfrac{f(x)}{x^2} dx=\int\limits^a_1\cfrac{x^2+lnx-1}{x^2} dx=\int\limits^a_1 \cfrac{x^2}{x^2} dx+\int\limits^a_1\cfrac{lnx}{x^2} dx-\int\limits^a_1\cfrac{1}{x^2} dx=\\\\=\int\limits^a_11dx+\int\limits^a_1\cfrac{lnx}{x^2}dx-\int\limits^a_1\cfrac{1}{x^2} dx =x \Bigg|_1^a-\cfrac{lnx-1}{x} \Bigg|_1^a+\cfrac{1}{x}\Bigg|_1^a=$

$\displaystyle=a-1-\Bigg(\cfrac{lna-1}{a} -\cfrac{ln1-1}{1} \Bigg) -\Bigg(\cfrac{1}{a} -\cfrac{1}{1} \Bigg)=a-1-\cfrac{lna-1}{a} -1-\cfrac{1}{a} +1=\\\\=\cfrac{a^2-lna+1-a-1}{a} =\cfrac{a^2-lna-a}{a} \\\\ \int\limits^a_1\cfrac{f(x)}{x^2} dx=\cfrac{a-lna}{a} \Rightarrow \cfrac{a^2-lna-a}{a} =\cfrac{a-lna}{a} \Rightarrow a^2-lna-a=a-lna\Rightarrow \\\\ \Rightarrow a^2-lna-a-a+lna=0\Rightarrow a^2-2a=0\Rightarrow a(a-2)=0\Rightarrow a=2$