Функције – знак функције 1

Текст задатака објашњених у видео лекцији:

пр.5)  $y=\frac{{2{x^2}+x -1}}{{{x^2} - 4}}$

пр.6)  $y=\frac{{4+4x-3{x^2}}}{\sqrt {2x + 4}}$

пр.5)

\[\begin{gathered}
f\left( x \right) = \frac{{2{x^2} + x - 1}}{{{x^2} - 4}} \hfill \\
Df:{x^2} - 4 \ne 0 \hfill \\
\left( {x - 2} \right)\left( {x + 2} \right) \ne 0 \hfill \\
\begin{array}{*{20}{c}}
{x - 2 \ne 0}&{}&{x + 2 \ne 0} \\
{x \ne 2}&{}&{x \ne - 2}
\end{array} \hfill \\
Df:\mathbb{R}\backslash \left\{ { - 2;2} \right\} \hfill \\
\end{gathered} \]

нуле:

\[\begin{gathered}
y = 0 \hfill \\
\frac{{2{x^2} + x - 1}}{{{x^2} - 4}} = 0 \hfill \\
2{x^2} + x - 1 = 0 \hfill \\
{x_{1,2}} = \frac{{ - 1 \pm \sqrt {1 + 8} }}{4} = \frac{{ - 1 \pm 3}}{4} \hfill \\
{x_1} = - 1 \in Df \hfill \\
{x_2} = \frac{1}{2} \in Df \hfill \\
\end{gathered} \]

знак:

\[\begin{gathered}
y > 0,x \in \left( { - \infty ; - 2} \right) \cup \left( { - 1;\frac{1}{2}} \right) \cup \left( {2; + \infty } \right) \hfill \\
y > 0,x \in \left( { - 2; - 1} \right) \cup \left( {\frac{1}{2};2} \right) \hfill \\
\end{gathered} \]

пр.6)

\[\begin{gathered}
Df:\sqrt {2x + 4} \ne 0 \cap 2x + 4 \geqslant 0 \hfill \\
2x + 4 \ne 0 \hfill \\
\begin{array}{*{20}{c}}
{2x \ne - 4}&{}&{2x \geqslant - 4} \\
{x \ne - 2}&{}&{x \geqslant - 2}
\end{array} \hfill \\
Df:x \in \left( { - 2; + \infty } \right) \hfill \\
\end{gathered} \]

нуле:

\[\begin{gathered}
y = 0 \hfill \\
\frac{{4 + 4x - 3{x^2}}}{{\sqrt {2x + 4} }} = 0 \hfill \\
4 + 4x - 3{x^2} = 0 \hfill \\
- 3{x^2} + 4x + 4 = 0 \hfill \\
{x_{1,2}} = \frac{{ - 4 \pm \sqrt {16 + 48} }}{{ - 6}} = \frac{{ - 4 \pm 8}}{{ - 6}} \hfill \\
{x_1} = 2 \in Df \hfill \\
{x_2} = - \frac{2}{3} \in Df \hfill \\
\end{gathered} \]

знак:

\[\begin{gathered}
y > 0,x \in \left( { - \frac{2}{3};2} \right) \hfill \\
y > 0,x \in \left( { - 2; - \frac{2}{3}} \right) \cup \left( {2; + \infty } \right) \hfill \\
\end{gathered} \]