Екстремне вредности и превојне тачке функција 2

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Пр.3)   Одредити монотоност, екстремне вредности, конвексност и

            превојне тачке функције $y = \frac{{4x - {x^2} - 4}}{{x - 1}}$.

Пр.4)   Испитати ток функције $y = \left( {x - 1} \right){e^{2x}}$.

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

$Df:x - 1 \ne 0$

$x \ne 1$

$Df:\mathbb{R}\backslash \left\{ 1 \right\}$

$y' = \frac{{{{\left( {4x - {x^2} - 4} \right)}^\prime }\left( {x - 1} \right) - {{\left( {x - 1} \right)}^\prime }\left( {4x - {x^2} - 4} \right)}}{{{{\left( {x - 1} \right)}^2}}} = \frac{{\left( {4 - 2x} \right)\left( {x - 1} \right) - \left( {4x - {x^2} - 4} \right)}}{{{{\left( {x - 1} \right)}^2}}} = $

$ = \frac{{4x - 4 - 2{x^2} + 2x - 4x + {x^2} + 4}}{{{{\left( {x - 1} \right)}^2}}} = \frac{{ - {x^2} + 2x}}{{{{\left( {x - 1} \right)}^2}}}$

\[\frac{{ - {x^2} + 2x}}{{{{\left( {x - 1} \right)}^2}}} = 0\]

\[\begin{array}{*{20}{c}}
{ - {x^2} + 2x = 0}&{}& \wedge  &{}& {{{\left( {x - 1} \right)}^2} \ne 0} \\ 
{x\left( { - x + 2} \right) = 0}&{}&{}&{x \ne 1} \\ 
{\begin{array}{*{20}{c}}
{x = 0}&  \vee  &{}&{ - x + 2 = 0} \\ 
{}&{}&{}&{x = 2} 
\end{array}}&{}&{}&{} 
\end{array}\]

${T_{\min }}\left( {0;4} \right)$

${T_{\max }}\left( {2;0} \right)$

$y'' = \frac{{{{\left( { - {x^2} + 2x} \right)}^\prime }{{\left( {x - 1} \right)}^2} - {{\left( {{{\left( {x - 1} \right)}^2}} \right)}^\prime }\left( { - {x^2} + 2x} \right)}}{{{{\left( {x - 1} \right)}^4}}} = \frac{{\left( { - 2x + 2} \right){{\left( {x - 1} \right)}^2} - 2\left( {x - 1} \right)\left( { - {x^2} + 2x} \right)}}{{{{\left( {x - 1} \right)}^4}}} = $

$ = \frac{{\left( {x - 1} \right)\left( {\left( { - 2x + 2} \right)\left( {x - 1} \right) - 2\left( { - {x^2} + 2x} \right)} \right)}}{{{{\left( {x - 1} \right)}^4}}} = \frac{{ - 2{x^2} + 2x + 2x - 2 + 2{x^2} - 4x}}{{{{\left( {x - 1} \right)}^3}}} = \frac{{ - 2}}{{{{\left( {x - 1} \right)}^3}}}$

\[\begin{gathered}
\frac{{ - 2}}{{{{\left( {x - 1} \right)}^3}}} = 0 \hfill \\
\begin{array}{*{20}{c}}
{ - 2 \ne 0}&{}&{}&{{{\left( {x - 1} \right)}^3} \ne 0} \\ 
{}&{}&{}&{x \ne 1} 
\end{array} \hfill \\ 
\end{gathered} \]

$1 \notin Df \Rightarrow $ нема превојне тачке

Пр.4) $y = \left( {x - 1} \right){e^{2x}}$

$y' = {\left( {x - 1} \right)^\prime }{e^{2x}} + \left( {x - 1} \right){\left( {{e^{2x}}} \right)^\prime } = {e^{2x}} + \left( {x - 1} \right) \cdot {e^{2x}}{\left( {2x} \right)^\prime } = $

$= {e^{2x}} + 2\left( {x - 1} \right) \cdot {e^{2x}} = {e^{2x}} \cdot \left( {1 + 2x - 2} \right) = {e^{2x}} \cdot \left( {2x - 1} \right)$

${T_{\min }}\left( {\frac{1}{2}; - \frac{e}{2}} \right)$

$y'' = {\left( {2x - 1} \right)^\prime } \cdot {e^{2x}} + {\left( {{e^{2x}}} \right)^\prime } \cdot \left( {2x - 1} \right) = {e^{2x}} \cdot 2 + {e^{2x}} \cdot 2 \cdot \left( {2x - 1} \right) = $

$ = {e^{2x}}\left( {2 + 4x - 2} \right) = 4x \cdot {e^{2x}}$

$4x \cdot {e^{2x}} = 0$

$x = 0$

$P\left( {0; - 1} \right)$