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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:

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$f\left( x \right) < x \in \left( {2;3} \right) \cup \left( {3; + \infty } \right)$

$f\left( x \right) > x \in \left( { - \infty ,2} \right)$

 

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. $y =  - x + 4$

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