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\documentclass{article} \usepackage[utf8]{inputenc} \usepackage[serbian]{babel} \usepackage{tikz, xcolor, float} \usepackage{amsmath, amssymb, amsthm} \usepackage{caption, subcaption} \usepackage{multicol} \usepackage[inline]{enumitem}

\newtheorem{zad}{Zadatak}

\renewcommand{\figurename}{Slika} \newcommand{\Placeholder}[2][(10, 10)] { \begin{tikzpicture} \draw[help lines] (0, 0) grid #1; \pgfgetlastxy{\x}{\y} \node[rotate=-25, align=center] at (\x / 2, \y / 2) {{\Huge Placeholder} \ #2}; \end{tikzpicture} }

\newcommand{\placeholder}[2][(10, 10)] { \begin{tikzpicture} \draw[help lines] (0, 0) grid #1; \pgfgetlastxy{\x}{\y} \node[rotate=-25, align=center] at (\x / 2, \y / 2) {{\Huge Placeholder} \ #2}; \end{tikzpicture} }

\begin{document} \begin{zad} \end{zad} $x \rightarrow x^{2} - 2x + 5$ i $x \rightarrow 3x$

$x \rightarrow x^{2} - 2x + 5 \rightarrow 3x$ pi\v se se kra\' ce $3(x^{2}-2x+5)$

$X \rightarrow 3x \rightarrow x^{2} - 2x + 5$ pi\v se se kra\' ce $(3x)^{2} - 2 * 3x + 5$

\begin{zad}
\end{zad}

$f : x \rightarrow x^{2}, h: x \rightarrow 2x, g: x \rightarrow (x+5)$

$(h \circ g)x = h(g(x)) = h(x+5) = 2(x+5) ; h \circ g : x \rightarrow 2(x+5)$

$(g \circ h)x = g(h(x)) = g(2x) = 2x+5 ; g \circ h: x \rightarrow 2x+5$\\

\begin{align*}
     (h \circ f \circ g \circ h \circ f \circ g)x &=  h(f \circ(g(h(f(g(x)))))) \\
     &=  h(f \circ(g(h(f(x+5^{2}))))) \\
     &=  h(f \circ(g(h(2(x+5)^{2}))))) \\
     &=  h(f (g \circ 2(x^{2} + 5)^{2} + 5) \\
     &=  h(f ([2(x^{2} + 5)^{2} + 5]^{2}) \\
     &=  2[2(x^{2} + 5)^{2} + 5]^{2} \\
    h \circ f \circ g \circ h \circ f \circ g : x  &\rightarrow  2[2(x^{2} + 5)^{2} + 5]^{2} \\
\end{align*}

\begin{zad}    
\end{zad}

Zato \v sto polazni skup ima vi\v sestruki element, \ sto pokazuje sagitalna \v sena gde uz elementa 5 polaznog skupa polaze vi\v se strelice, tj $f(x)$ nije samo singlton ili prazan skup.

${(5,y) , y \in R}$

\begin{figure}[H]
    \center
    \Placeholder[(3, 3)]{Ovde stoji opis ovog crteza}
    \caption{Ovo je figura}
\end{figure}

\begin{zad}
\end{zad}

$k: x \rightarrow 2x^{2}-7$ izra\v cunaj $k(-2)$ i $k(- \dfrac{1}{10})$,

$k(-2) = 2 * (-2)^{2} - 7 = 2 * 4 - 7 = 8 - 7 = 1$

$k(- \dfrac{1}{10} = 2(- \dfrac{1}{10})^{2} - 7 = 2 * \dfrac{1}{100} - 7 = \dfrac{1}{50} - \dfrac{350}{50} = - \dfrac{349}{50}$

\begin{zad}
\end{zad}

$f : x \rightarrow x^{4}, x \in {1,2,3,4} f^{-1} = ?$

Odre\dj uje se skup - cilj $f x^{4} \in {1,16,81,256}$ a to je polazni skup inverzne funkcije $f^{-1}$.

Kako je $f : y = x^{4}$ i kada $x$ i $y$ uzajamno zamene mesta, onda je $x = y^{4} \Rightarrow  y = \sqrt[4]{1} = \pm 1 jer je (+1)^{4} = 1$ i $(-1)^{4} = 1$ ; $\sqrt[4]{16} = \pm 2 ;$ ...

Relacija $f^{-1}$ napisana ekstenzivno je:

$\int_{}^{-1} = {(1,1),(1,-1),(16,2),(16,-2),(81,3),(256,4),(256,-4)}$

o\v cigledno da $f^{-1}$ nije funckija.

\begin{zad}
\end{zad}

$g(x) \rightarrow x^{3} - 3x, [-1,1], x_{1} = -1, x_{2} = 1$

$x_{2} - x_{1} = 1 - (- 1) = 2 > 0; g(x_{1}) = g(-1) = (-1)^{3} - 3 * (-1) = 2 > 0$

$g(x_{2}) = g(1) = 1^{3} - 3 * 1 = -2 < 0$

Kako je $x_{2} - x_{1} > 0$ i $g(x_{2}) - g(x_{1}) < 0 \Rightarrow \dfrac{g(x_{2} - g(x_{1}}{x_{2} - x_{1}} < 0$

Zna\v ci funckija $g$ opada.

$g(x) \rightarrow x^{3} - 3x, [1,\infty], x_{1} = 1, x^{2} = 2$,

$x_{1} - x^{2} = 1 > 0 , g(x_{1}) = g(1) = - 2 ; g(x_{2}) = g(2) = 2 > 0$

$g(x_{1}) - g(x^{2}) = 4 > 0$

Kako je $x_{2} - x_{1} > 0$ i $g(x_{2}) - g(x_{1}) > 0$, sledi

$\dfrac{g(x_{2} - g(x_{1}}{x_{2} - x_{1}} > 0$ zna\v ci funkcija $g$ raste.

\begin{zad}

\end{zad}

$f : x \rightarrow ax^{2} (y=ax^{2}, f(x) = ax^{2})$

$a > 0, x_{1} = 3, x_{2} = 5, 3,5 \in R^{+}, y = ax^{2}$

$y_{1} = ax^{2} = a * 3^{2} = 9a, y_{2} = a5^{2} = 25a$

$ \dfrac{y_{2} - y_{1}}{x_{2} - x_{1}} = \dfrac{25a - 9a}{5 - 3}  = \dfrac{16a}{2} = 8a$ funckija raste.

Primena koli\v cnika $ \dfrac{y_{2} - y_{1}}{x_{2} - x_{1}} $

\end{document}

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Note : ovo je samo pola fajla. Nije ceo gotov.

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