Skillnad mellan versioner av "3.3 Lösning 8a"
Taifun (Diskussion | bidrag) m |
Taifun (Diskussion | bidrag) m |
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| Rad 1: | Rad 1: | ||
| − | + | För att kunna derivera utvecklas <math> \, f(x) \, </math> till ett polynom: | |
| − | + | :<math> f(x) \, = \, x^2 \, (x + 1) \, (2\,x + 5) + 1 \, = \, (x^3 + x^2) \, (2\,x + 5) + 1 \, = </math> | |
| − | + | ::<math> \quad = \, 2\,x^4 + 5\,x^3 + 2\,x^3 + 5\,x^2 + 1 \, = \, 2\,x^4 + 7\,x^3 + 5\,x^2 + 1 </math> | |
| − | + | Vi deriverar <math> \, f(x) \, </math> två gånger: | |
| + | |||
| + | :<math>\begin{array}{rcl} f(x) & = & 2\,x^4 + 7\,x^3 + 5\,x^2 + 1 \\ | ||
| + | f'(x) & = & 8\,x^3 + 21\,x^2 + 10\,x \\ | ||
| + | f''(x) & = & 24\,x^2 + 42\,x + 10 | ||
| + | \end{array}</math> | ||
| + | |||
| + | Derivatans nollställen: | ||
| + | |||
| + | :<math>\begin{array}{rcl} 8\,x^3 + 21\,x^2 + 10\,x & = & 0 \\ | ||
| + | x\,(8\,x^2 + 21\,x + 10) & = & 0 \\ | ||
| + | x_1 & = & 0 \\ | ||
| + | 8\,x^2 + 21\,x + 10 & = & 0 \\ | ||
| + | x^2 + \frac{21}{8}\,x + \frac{10}{8} & = & 0 \\ | ||
| + | x^2 + 2,625\,x + 1,25 & = & 0 \\ | ||
| + | x_{2,3} & = & -1,3125 \pm \sqrt{1,7227 - 1,25} \\ | ||
| + | x_{2,3} & = & -1,3125 \pm 0,6875 \\ | ||
| + | x_2 & = & - 0,625 \\ | ||
| + | x_3 & = & - 2 | ||
| + | \end{array}</math> | ||
| + | |||
| + | Vi Sätter in derivatans nollställen i andraderivatan <math> \, f''(x) \, = \, 24\,x^2 + 42\,x + 10 </math> | ||
| + | <table> | ||
| + | <tr> | ||
| + | <td><math> \underline{x_1 = 0} \, </math><span style="color:black">:</span> | ||
| + | |||
| + | <math> \underline{x_2 = -0,625} \, </math><span style="color:black">:</span> | ||
| + | |||
| + | |||
| + | |||
| + | <math> \underline{x_3 = -2} \, </math><span style="color:black">:</span> | ||
| + | |||
| + | |||
| + | |||
| + | </td> | ||
| + | <td><math> \; </math></td> | ||
| + | <td> | ||
| + | <math> f''(0) \, = \, 10 > 0 \quad \Longrightarrow \quad x_1 = 0 \quad {\rm lokalt\;minimum.} </math> | ||
| + | |||
| + | <math> f''(-0,625) = 24\cdot(-0,625)^2 + 42\cdot(-0,625) + 10 = -6,9 < 0 </math> | ||
| + | |||
| + | <math> \Longrightarrow \quad x_2 = -0,625 \quad {\rm lokalt\;maximum.} </math> | ||
| + | |||
| + | <math> f''(-2) = 24\cdot(-2)^2 + 42\cdot(-2) + 10 = 22 > 0 </math> | ||
| + | |||
| + | <math> \Longrightarrow \quad x_3 = -2 \quad {\rm lokalt\;minimum.} </math> | ||
| + | |||
| + | </td> | ||
| + | </tr> | ||
| + | </table> | ||
| + | |||
| + | <math> f''(0) \neq 0 \; {\rm ,} \; f''(-0,625) \neq 0 \; {\rm ,} \; f''(-2) \neq 0 \; \Longrightarrow \; f(x) \, {\rm har\;inga\;terasspunkter.} </math> | ||
| + | |||
| + | |||
| + | Maximi- och minimipunkternas koordinater<span style="color:black">:</span> | ||
| + | |||
| + | <math> f(x) \, = \, 2\,x^4 + 7\,x^3 + 5\,x^2 + 1 </math> | ||
| + | |||
| + | <math> f(0) \, = \, 1 \; \Longrightarrow \quad (0, 1) \quad {\rm är\;lokal\;minimipunkt.} </math> | ||
| + | |||
| + | <math> f(-0,625) \, = \, 2\cdot(-0,625)^4 + 7\cdot(-0,625)^3 + 5\cdot(-0,625)^2 + 1 = 1,549 </math> | ||
| + | |||
| + | ::::<math> \Longrightarrow \quad\; (-0,625; 1,549) \quad {\rm är\;lokal\;maximipunkt.} </math> | ||
| + | |||
| + | <math> f(-2) \, = \, 2\cdot(-2)^4 + 7\cdot(-2)^3 + 5\cdot(-2)^2 + 1 = -3 </math> | ||
| + | |||
| + | ::::<math> \Longrightarrow \quad\; (-2, -3) \quad {\rm är\;lokal\;minimipunkt.} </math> | ||
Nuvarande version från 14 februari 2016 kl. 12.54
För att kunna derivera utvecklas \( \, f(x) \, \) till ett polynom:
\[ f(x) \, = \, x^2 \, (x + 1) \, (2\,x + 5) + 1 \, = \, (x^3 + x^2) \, (2\,x + 5) + 1 \, = \]
- \[ \quad = \, 2\,x^4 + 5\,x^3 + 2\,x^3 + 5\,x^2 + 1 \, = \, 2\,x^4 + 7\,x^3 + 5\,x^2 + 1 \]
Vi deriverar \( \, f(x) \, \) två gånger:
\[\begin{array}{rcl} f(x) & = & 2\,x^4 + 7\,x^3 + 5\,x^2 + 1 \\ f'(x) & = & 8\,x^3 + 21\,x^2 + 10\,x \\ f''(x) & = & 24\,x^2 + 42\,x + 10 \end{array}\]
Derivatans nollställen:
\[\begin{array}{rcl} 8\,x^3 + 21\,x^2 + 10\,x & = & 0 \\ x\,(8\,x^2 + 21\,x + 10) & = & 0 \\ x_1 & = & 0 \\ 8\,x^2 + 21\,x + 10 & = & 0 \\ x^2 + \frac{21}{8}\,x + \frac{10}{8} & = & 0 \\ x^2 + 2,625\,x + 1,25 & = & 0 \\ x_{2,3} & = & -1,3125 \pm \sqrt{1,7227 - 1,25} \\ x_{2,3} & = & -1,3125 \pm 0,6875 \\ x_2 & = & - 0,625 \\ x_3 & = & - 2 \end{array}\]
Vi Sätter in derivatans nollställen i andraderivatan \( \, f''(x) \, = \, 24\,x^2 + 42\,x + 10 \)
| \( \underline{x_1 = 0} \, \):
\( \underline{x_2 = -0,625} \, \):
\( \underline{x_3 = -2} \, \):
|
\( \; \) |
\( f''(0) \, = \, 10 > 0 \quad \Longrightarrow \quad x_1 = 0 \quad {\rm lokalt\;minimum.} \) \( f''(-0,625) = 24\cdot(-0,625)^2 + 42\cdot(-0,625) + 10 = -6,9 < 0 \) \( \Longrightarrow \quad x_2 = -0,625 \quad {\rm lokalt\;maximum.} \) \( f''(-2) = 24\cdot(-2)^2 + 42\cdot(-2) + 10 = 22 > 0 \) \( \Longrightarrow \quad x_3 = -2 \quad {\rm lokalt\;minimum.} \) |
\( f''(0) \neq 0 \; {\rm ,} \; f''(-0,625) \neq 0 \; {\rm ,} \; f''(-2) \neq 0 \; \Longrightarrow \; f(x) \, {\rm har\;inga\;terasspunkter.} \)
Maximi- och minimipunkternas koordinater:
\( f(x) \, = \, 2\,x^4 + 7\,x^3 + 5\,x^2 + 1 \)
\( f(0) \, = \, 1 \; \Longrightarrow \quad (0, 1) \quad {\rm är\;lokal\;minimipunkt.} \)
\( f(-0,625) \, = \, 2\cdot(-0,625)^4 + 7\cdot(-0,625)^3 + 5\cdot(-0,625)^2 + 1 = 1,549 \)
- \[ \Longrightarrow \quad\; (-0,625; 1,549) \quad {\rm är\;lokal\;maximipunkt.} \]
\( f(-2) \, = \, 2\cdot(-2)^4 + 7\cdot(-2)^3 + 5\cdot(-2)^2 + 1 = -3 \)
- \[ \Longrightarrow \quad\; (-2, -3) \quad {\rm är\;lokal\;minimipunkt.} \]
