Skillnad mellan versioner av "3.4 Lösning 4a"
Från Mathonline
Taifun (Diskussion | bidrag) m |
Taifun (Diskussion | bidrag) m |
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| Rad 16: | Rad 16: | ||
\end{array}</math> | \end{array}</math> | ||
| − | <math> {\color{White} x} \ | + | <math> {\color{White} x} \quad \underline{x_1 = 1} \, </math><span style="color:black">:</span> |
::<math> f''(x) \, = \, -2\,x \, + \, 4 </math> | ::<math> f''(x) \, = \, -2\,x \, + \, 4 </math> | ||
| Rad 22: | Rad 22: | ||
::<math> f''(1) \, = \, -2\cdot 1 + 4 = 2 > 0 \quad \Longrightarrow \quad x_1 = 1 \quad {\rm lokalt\;minimum.} </math> | ::<math> f''(1) \, = \, -2\cdot 1 + 4 = 2 > 0 \quad \Longrightarrow \quad x_1 = 1 \quad {\rm lokalt\;minimum.} </math> | ||
| − | <math> {\color{White} x} \ | + | <math> {\color{White} x} \quad \underline{x_2 = 3} \, </math><span style="color:black">:</span> |
| − | ::<math> f''(3) \, = \, -2\cdot 3 + 4 = | + | ::<math> f''(3) \, = \, -2\cdot 3 + 4 = -2 < 0 \quad \Longrightarrow \quad x_2 = 5 \quad {\rm lokalt\;maximum.} </math> |
| − | ::<math> f''( | + | ::<math> f''(1) \neq 0 \quad {\rm och} \quad f''(3) \neq 0 \quad \Longrightarrow \quad f(x) \, {\rm har\;inga\;terasspunkter.} </math> |
| − | + | ::<math> f(x) \, = \, -\,{x^3 \over 3} \, + \, 2\,x^2 \, - \, 3\,x \, + \, 1 </math> | |
| − | ::<math> f( | + | ::<math> f(1) \, = \, -\,{1^3 \over 3} \, + \, 2\cdot 1^2 \, - \, 3\cdot 1 \, + \, 1 = 10 \quad \Longrightarrow \quad (3, 10) \quad {\rm är\;lokal\;maximipunkt.} </math> |
| − | + | ||
| − | + | ||
::<math> f(5) \, = \, 5^3 - 12\cdot 5^2 + 45\cdot 5 - 44 = 6 \quad \Longrightarrow \quad (5, 6) \quad {\rm är\;lokal\;minimipunkt.} </math> | ::<math> f(5) \, = \, 5^3 - 12\cdot 5^2 + 45\cdot 5 - 44 = 6 \quad \Longrightarrow \quad (5, 6) \quad {\rm är\;lokal\;minimipunkt.} </math> | ||
Versionen från 22 januari 2015 kl. 10.02
- \[\begin{array}{rcl} f(x)&=&-\,{x^3 \over 3} \, + \, 2\,x^2 \, - \, 3\,x \, + \, 1 \\ f'(x)&=&-\,x^2 \, + \, 4\,x \, - \, 3 \\ f''(x)&=&-2\,x \, + \, 4 \end{array}\]
- \[\begin{array}{rcl} -\,x^2 \, + \, 4\,x \, - \, 3 & = & 0 \\ x^2 \, - \, 4\,x \, + \, 3 & = & 0 \\ \end{array}\]
- \[ \begin{array}{rcl} {\rm Vieta:} \quad x_1 \cdot x_2 & = & 3 \\ x_1 + x_2 & = & -(-4) = 4 \\ &\Downarrow& \\ x_1 & = & 1 \\ x_2 & = & 3 \end{array}\]
\( {\color{White} x} \quad \underline{x_1 = 1} \, \):
- \[ f''(x) \, = \, -2\,x \, + \, 4 \]
- \[ f''(1) \, = \, -2\cdot 1 + 4 = 2 > 0 \quad \Longrightarrow \quad x_1 = 1 \quad {\rm lokalt\;minimum.} \]
\( {\color{White} x} \quad \underline{x_2 = 3} \, \):
- \[ f''(3) \, = \, -2\cdot 3 + 4 = -2 < 0 \quad \Longrightarrow \quad x_2 = 5 \quad {\rm lokalt\;maximum.} \]
- \[ f''(1) \neq 0 \quad {\rm och} \quad f''(3) \neq 0 \quad \Longrightarrow \quad f(x) \, {\rm har\;inga\;terasspunkter.} \]
- \[ f(x) \, = \, -\,{x^3 \over 3} \, + \, 2\,x^2 \, - \, 3\,x \, + \, 1 \]
- \[ f(1) \, = \, -\,{1^3 \over 3} \, + \, 2\cdot 1^2 \, - \, 3\cdot 1 \, + \, 1 = 10 \quad \Longrightarrow \quad (3, 10) \quad {\rm är\;lokal\;maximipunkt.} \]
- \[ f(5) \, = \, 5^3 - 12\cdot 5^2 + 45\cdot 5 - 44 = 6 \quad \Longrightarrow \quad (5, 6) \quad {\rm är\;lokal\;minimipunkt.} \]
