Question

2..........................

Answer 1

$f:\mathbb{R}\to \mathbb{R},\,\, f(x) = \sqrt[3]{ax^2-x^3}$

- nu are asimptote verticale fiindcă nu sunt puncte în care să nu aibă sens.

- nu are asimptote orizontale pentru că limitele la ±infinit nu sunt finite.

Asimptote oblice:

$y = mx+n$

$m = \lim\limits_{x\to \pm \infty} \dfrac{f(x)}{x} = \lim\limits_{x\to \pm \infty} \dfrac{\sqrt[3]{ax^2-x^3}}{x} =\lim\limits_{x\to \pm \infty} \dfrac{\sqrt[3]{x^3(\frac{a}{x}-1)}}{x} =$

$= \lim\limits_{x\to \pm \infty} \dfrac{x\sqrt[3]{(\frac{a}{x}-1)}}{x} = \lim\limits_{x\to \pm \infty} \sqrt[3]{(\frac{a}{x}-1)} = \sqrt[3]{(0-1)} = -1$

$A(1,1) \in y= mx+n \Rightarrow 1 = m\cdot 1+n \Rightarrow 1 = -1+n \Rightarrow n = 2$

$n = \lim\limits_{x\to\pm \infty}\left(f(x) -mx\right)$

$\Rightarrow \lim\limits_{x\to\pm \infty}\left(f(x) -mx\right) = 2$

$\Rightarrow \lim\limits_{x\to\pm \infty}\left(\sqrt[3]{ax^2-x^3} -(-1)\cdot x\right) = 2$

$\Rightarrow \lim\limits_{x\to\pm \infty}\left(\sqrt[3]{ax^2-x^3} + x\right) = 2$

$\Rightarrow \lim\limits_{x\to\pm \infty}\left(\sqrt[3]{x^3(\frac{a}{x}-1)} + x\right) = 2$

$\Rightarrow \lim\limits_{x\to\pm \infty}\left(x\sqrt[3]{(\frac{a}{x}-1)} + x\right) = 2$

$\Rightarrow \lim\limits_{x\to\pm \infty}x\left(\sqrt[3]{(\frac{a}{x}-1)} + 1\right) = 2$

$\Rightarrow \lim\limits_{x\to\pm \infty}\dfrac{\sqrt[3]{(a\cdot \frac{1}{x}-1)} + 1}{\frac{1}{x}} = 2$

$\text{Notez }\frac{1}{x} = t \Rightarrow t \to \frac{1}{\pm \infty} \Rightarrow t\to 0$

$\Rightarrow \lim\limits_{t\to 0}\dfrac{\sqrt[3]{(at-1)} + 1}{t} = 2$

$\Rightarrow \lim\limits_{t\to 0}\dfrac{(at-1)^{\frac{1}{3}} + 1}{t} = 2$

$\overset{L'H}{\Rightarrow} \lim\limits_{t\to 0}\dfrac{\frac{1}{3}(at-1)^{\frac{1}{3}-1}\cdot (a\cdot 1-0) + 0}{1} = 2$

$\Rightarrow \lim\limits_{t\to 0}\left(\frac{1}{3}(at-1)^{-\frac{2}{3}}\cdot a\right)= 2$

$\Rightarrow \frac{1}{3}(a\cdot 0-1)^{-\frac{2}{3}}\cdot a = 2$

$\Rightarrow \frac{a}{3}= 2$

$\Rightarrow \boxed{a = 6}$