Функције – знак функције 1

Задаци

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)

$$\begin{aligned}&f\left( x \right) = \frac{{2{x^2} + x - 1}}{{{x^2} - 4}} \\ &Df:{x^2} - 4 \ne 0 \\ &\left( {x - 2} \right)\left( {x + 2} \right) \ne 0 \\ &\begin{array}{cccccccccccccccccccc} {x - 2 \ne 0}&{}&{x + 2 \ne 0} \\ {x \ne 2}&{}&{x \ne - 2} \end{array} \\ &Df:\mathbb{R}\backslash \left\{ { - 2;2} \right\}\end{aligned}$$

нуле:

$$\begin{aligned}&y = 0 \\ &\frac{{2{x^2} + x - 1}}{{{x^2} - 4}} = 0 \\ &2{x^2} + x - 1 = 0 \\ &{x_{1,2}} = \frac{{ - 1 \pm \sqrt {1 + 8} }}{4} = \frac{{ - 1 \pm 3}}{4} \\ &{x_1} = - 1 \in Df \\ &{x_2} = \frac{1}{2} \in Df\end{aligned}$$

знак:

$$\begin{aligned}&y > 0,x \in \left( { - \infty ; - 2} \right) \cup \left( { - 1;\frac{1}{2}} \right) \cup \left( {2; + \infty } \right) \\ &y > 0,x \in \left( { - 2; - 1} \right) \cup \left( {\frac{1}{2};2} \right)\end{aligned}$$

пр.6)

$$\begin{aligned}&Df:\sqrt {2x + 4} \ne 0 \cap 2x + 4 \geqslant 0 \\ &2x + 4 \ne 0 \\ &\begin{array}{cccccccccccccccccccc} {2x \ne - 4}&{}&{2x \geqslant - 4} \\ {x \ne - 2}&{}&{x \geqslant - 2} \end{array} \\ &Df:x \in \left( { - 2; + \infty } \right)\end{aligned}$$

нуле:

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

знак:

$$\begin{aligned}&y > 0,x \in \left( { - \frac{2}{3};2} \right) \\ &y > 0,x \in \left( { - 2; - \frac{2}{3}} \right) \cup \left( {2; + \infty } \right)\end{aligned}$$