Функције - асимптоте функција 4
Домен, нуле, знак и асимптоте функције.
Задаци
Текстови задатака објашњених у видео лекцији.
Одредити асимптоту графика функције:
пр.9) $f(x) = \frac{{6x - {x^2} - 9}}{{x - 2}}$
$Df:x - 2 \ne 0$
$Df:\mathbb{R}\backslash \left\{ 2 \right\}$
нула функције:
$f\left( x \right) = 0$
$\frac{{6x - {x^2} - 9}}{{x - 2}} = 0$
$6x - {x^2} - 9 = 0$
$ - {x^2} + 6x - 9 = 0$
${x_{1/2}} = \frac{{ - 6 \pm \sqrt {36 - 36} }}{{ - 2}}$
$x = 3$
знак функције:
$f\left( x \right) < x \in \left( {2;3} \right) \cup \left( {3; + \infty } \right)$
$f\left( x \right) > x \in \left( { - \infty ,2} \right)$
асимптоте:
$Df:\mathbb{R}\backslash \left\{ 2 \right\}$
B.A.:
$\mathop {\lim }\limits_{x \to {2^ - }} \frac{{6x - {x^2} - 9}}{{x - 2}} = \frac{{12 - 4 - 9}}{{{2^ - } - 2}} = \frac{{ - 1}}{{{0^ - }}} = + \infty $
$\mathop {\lim }\limits_{x \to {2^ + }} \frac{{6x - {x^2} - 9}}{{x - 2}} = \frac{{12 - 4 - 9}}{{{2^ + } - 2}} = \frac{{ - 1}}{{{0^ + }}} = - \infty $
X.A.:
$\mathop {\lim }\limits_{x \to \pm \infty } \frac{{6x - {x^2} - 9}}{{x - 2}} = \mathop {\lim }\limits_{x \to \pm \infty } \frac{{\frac{{6x}}{{{x^2}}} - \frac{{{x^2}}}{{{x^2}}} - \frac{9}{{{x^2}}}}}{{\frac{x}{{{x^2}}} - \frac{2}{{{x^2}}}}} = \frac{{0 - 1 - 0}}{{0 - 0}} = \frac{{ - 1}}{0} = - \infty $ нема X.A.
К.А.: десна
$y = kx + n$
\[\begin{gathered}
k = \mathop {\lim }\limits_{x \to \infty } \frac{{f\left( x \right)}}{x} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{6x - {x^2} - 9}}{{x - 2}}}}{{\frac{x}{1}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{6x - {x^2} - 9}}{{x\left( {x - 2} \right)}} = \mathop {\lim }\limits_{x \to \infty } \frac{{6x - {x^2} - 9}}{{{x^2} - 2x}} = \hfill \\
= \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{6x}}{{{x^2}}} - \frac{{{x^2}}}{{{x^2}}} - \frac{9}{{{x^2}}}}}{{\frac{{{x^2}}}{{{x^2}}} - \frac{{2x}}{{{x^2}}}}} = \frac{{ - 1}}{1} = - 1 \hfill \\
\end{gathered} \]
$k = - 1$
\[\begin{gathered}
n = \mathop {\lim }\limits_{x \to \infty } \left( {f\left( x \right) - kx} \right) = \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{6x - {x^2} - 9}}{{x - 2}} - \left( { - 1} \right)x} \right) = \mathop {\lim }\limits_{x \to \infty } \frac{{6x - {x^2} - 9 + x\left( {x - 2} \right)}}{{x - 2}} = \hfill \\
= \mathop {\lim }\limits_{x \to \infty } \frac{{4x - 9}}{{x - 2}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{4x}}{x} - \frac{9}{x}}}{{\frac{x}{x} - \frac{2}{x}}} = \frac{4}{1} = 4 \hfill \\
\end{gathered} \]
$n = 4$
Десна К.А. $y = - x + 4$
лева
$y = kx + n$
$k = \mathop {\lim }\limits_{x \to - \infty } \frac{{f\left( x \right)}}{x} = \mathop {\lim }\limits_{x \to - \infty } \frac{{6x - {x^2} - 9}}{{x\left( {x - 2} \right)}} = \mathop {\lim }\limits_{t \to \infty } \frac{{ - 6t - {t^2} - 9}}{{{t^2} + 2t}} = \mathop {\lim }\limits_{t \to \infty } \frac{{\frac{{ - 6t}}{{{t^2}}} - \frac{{{t^2}}}{{{t^2}}} - \frac{9}{{{t^2}}}}}{{1 + \frac{{2t}}{{{t^2}}}}} = - 1$
$x = - t$
$x \to - \infty $
$t \to \infty $
$k = - 1$
\[\begin{gathered}
n = \mathop {\lim }\limits_{x \to - \infty } \left( {f\left( x \right) - kx} \right) = \mathop {\lim }\limits_{x \to - \infty } \left( {\frac{{6x - {x^2} - 9}}{{x - 2}} - \left( { - 1} \right)x} \right) = \hfill \\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{6x - {x^2} - 9 + x\left( {x - 2} \right)}}{{x - 2}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{4x - 9}}{{x - 2}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{4x}}{x} - \frac{9}{x}}}{{\frac{x}{x} - \frac{2}{x}}} = \frac{4}{1} = 4 \hfill \\
\end{gathered} \]
$n = 4$
Лева К.А. $y = - x + 4$