Функције – испитивање функција 2

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Пр.2)   Испитати функцију $f\left( x \right) = \frac{{{x^2}}}{{x - 2}}$.

1. Домен функције                

Df:  $x - 2 \ne 0$

$x \ne 2$

2. Нуле функције

$f\left( x \right) = 0$

$\frac{{{x^2}}}{{x - 2}} = 0$

${x^2} = 0$

$x = 0$

Тачка $A\left( {0;0} \right)$ је нула функције.

3. Пресек са y-осом

$x = 0;f\left( 0 \right) = \frac{{{0^2}}}{{0 - 2}} = 0$

$A\left( {0;0} \right)$

4. Знак функције

$y > 0$    $x \in \left( {2; + \infty } \right)$

$y < 0$    $x \in \left( { - \infty ;0} \right) \cup \left( {0;2} \right)$

5. Парност

$f\left( { - x} \right) = \frac{{{{\left( { - x} \right)}^2}}}{{ - x - 2}} = \frac{{{x^2}}}{{ - x - 2}} \ne f\left( x \right)$

$ - f\left( x \right) =  - \frac{{{x^2}}}{{x - 2}} = \frac{{{x^2}}}{{ - x + 2}}$

Функција није парна, није непарна.

6. Асимптоте

В.А.

$\mathop {\lim }\limits_{x \to {2^ - }} \frac{{{x^2}}}{{x - 2}} = \frac{4}{{{2^ - } - 2}} = \frac{4}{{{0^ - }}} =  - \infty $

$\mathop {\lim }\limits_{x \to {2^ + }} \frac{{{x^2}}}{{x - 2}} = \frac{4}{{{2^ + } - 2}} = \frac{4}{{{0^ + }}} =  + \infty $

Х.А.

$\mathop {\lim }\limits_{x \to \infty } \frac{{{x^2}:{x^2}}}{{\left( {x - 2} \right):{x^2}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{x}{{{x^2}}}}}{{\frac{x}{{{x^2}}} - \frac{2}{{{x^2}}}}} = \frac{1}{{0 - 0}} = \frac{1}{0} = \infty $ нема Х.А.

К.А.

$y = kx + n$

$k = \mathop {\lim }\limits_{x \to \infty } \frac{{f\left( x \right)}}{x} = \mathop {\lim }\limits_{x \to \infty } \frac{{{x^2}:{x^2}}}{{\left( {{x^2} - 2x} \right):{x^2}}} = \frac{1}{{1 - 0}} = 1$

$k = 1$

$n = \mathop {\lim }\limits_{x \to \infty } \left( {f\left( x \right) - kx} \right) = \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{{x^2}}}{{x - 2}} - x} \right) = \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{{x^2} - {x^2} + 2x}}{{x - 2}}} \right) = \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{2x:x}}{{x - 2:x}}} \right) = $

$ = \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{\frac{{2x}}{x}}}{{\frac{x}{x} - \frac{2}{x}}}} \right) = \frac{2}{{1 - 0}} = 2$

$n = 2$

Права $y = x + 2$ је коса асимптота.

7. Монотоност и екстремне вредности

$f\left( x \right) = \frac{{{x^2}}}{{x - 2}}$

$f'\left( x \right) = \frac{{{{\left( {{x^2}} \right)}^\prime }\left( {x - 2} \right) - {{\left( {x - 2} \right)}^\prime }{x^2}}}{{{{\left( {x - 2} \right)}^2}}} = \frac{{2x\left( {x - 2} \right) - {x^2}}}{{{{\left( {x - 2} \right)}^2}}} = \frac{{2{x^2} - 4x - {x^2}}}{{{{\left( {x - 2} \right)}^2}}} = \frac{{{x^2} - 4x}}{{{{\left( {x - 2} \right)}^2}}}$

$\frac{{{x^2} - 4x}}{{{{\left( {x - 2} \right)}^2}}} = 0$

${x^2} - 4x = 0;x - 2 \ne 0$

$x = 0  \vee  x = 4;x \ne 2$

${T_{\max }}\left( {0;0} \right)$

${T_{\min }}\left( {4;8} \right)$

8. Конвексност и превојне тачке

$f\left( x \right) = \frac{{{x^2}}}{{x - 2}}$

$f'\left( x \right) = \frac{{{x^2} - 4x}}{{{{\left( {x - 2} \right)}^2}}}$

$f''\left( x \right) = \frac{{{{\left( {{x^2} - 4x} \right)}^\prime }{{\left( {x - 2} \right)}^2} - {{\left( {{{\left( {x - 2} \right)}^2}} \right)}^\prime }\left( {{x^2} - 4x} \right)}}{{{{\left( {x - 2} \right)}^4}}} = \frac{{\left( {2x - 4} \right){{\left( {x - 2} \right)}^2} - 2\left( {x - 2} \right)\left( {{x^2} - 4x} \right)}}{{{{\left( {x - 2} \right)}^4}}} = $

$ = \frac{{\left( {x - 2} \right)\left( {\left( {2x - 4} \right)\left( {x - 2} \right) - 2\left( {{x^2} - 4x} \right)} \right)}}{{{{\left( {x - 2} \right)}^4}}} = \frac{{2{x^2} - 4x - 4x + 8 - 2{x^2} + 8x}}{{{{\left( {x - 2} \right)}^3}}} = \frac{8}{{{{\left( {x - 2} \right)}^3}}}$

$\frac{8}{{{{\left( {x - 2} \right)}^3}}} = 0$

$x \ne 2$

нема превојне тачке