Функције – нуле функције 1

Текст задатака објашњених у видео лекцији:

Одредити нуле функције:

пр.1)  $y={x^3}-7{x^2} +10x$

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

пр.3)  $y= \frac{\sqrt {{x^2} - 4x + 3} }{{x - 1}}$

пр.1)  $y={x^3}-7{x^2} +10x$

$Df:\mathbb{R}$

нуле: $y = 0$

${x^3} - 7{x^2} + 10x = 0$

$x\left( {{x^2} - 7x + 10} \right) = 0$

\[\begin{array}{*{20}{c}}
{x = 0}& \cup &{{x^2} - 7x + 10 = 0} \\
{}&{}&{{x_{1,2}} = \frac{{7 \pm \sqrt {49 - 40} }}{2}} \\
{}&{}&{{x_{1,2}} = \frac{{7 \pm 3}}{2}} \\
{}&{}&{\begin{array}{*{20}{c}}
{{x_1} = 5}&{{x_2} = 2}
\end{array}}
\end{array}\]

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

\[\begin{gathered}
Df:{x^2} - 1 \ne 0 \hfill \\
{x^2} \ne 1 \hfill \\
x \ne \pm 1 \hfill \\
Df:\mathbb{R}\backslash \left\{ { - 1;1} \right\} \hfill \\
\hfill \\
y = 0 \hfill \\
\frac{{{x^2} - 4x + 4}}{{{x^2} - 1}} = 0 \hfill \\
{x^2} - 4x + 4 = 0 \hfill \\
{x_{1,2}} = \frac{{4 \pm \sqrt {16 - 16} }}{2} \hfill \\
x = 2 \hfill \\
\end{gathered} \]

пр.3)  $y= \frac{\sqrt {{x^2} - 4x + 3} }{{x - 1}}$

\[\begin{array}{*{20}{c}}
{Df:}&{x - 1 \ne 0}& \cap &{{x^2} - 4x + 3 \geqslant 0} \\
{}&{x \ne 1}&{}&{{x^2} - 4x + 3 = 0} \\
{}&{}&{}&{{x_{1,2}} = \frac{{4 \pm \sqrt {16 - 12} }}{2}} \\
{}&{}&{}&{\begin{array}{*{20}{c}}
{{x_1} = 1}&{{x_2} = 3}
\end{array}}
\end{array}\]

\[\begin{gathered}
x \in \left( { - \infty ;1} \right] \cup \left[ {3; + \infty } \right) \hfill \\
\hfill \\
Df:x \in \left( { - \infty ;1} \right) \cup \left[ {3; + \infty } \right) \hfill \\
\end{gathered} \]

нуле: $y = 0$

\[\begin{gathered}
\frac{{\sqrt {{x^2} - 4x + 3} }}{{x - 1}} = 0 \hfill \\
\sqrt {{x^2} - 4x + 3} = 0 \hfill \\
{x^2} - 4x + 3 = 0 \hfill \\
\begin{array}{*{20}{c}}
{{x_1} = 1}&{{x_2} = 3} 
\end{array} \hfill \\ 
\end{gathered} \]