3.3 Lösning 8a
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: \( \, f''(x) \, = \, 24\,x^2 + 42\,x + 10 \)
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\( \underline{x_1 = 0} \, \): |
\( \; \) | \[ \qquad \quad f''(0) \, = \, 10 > 0 \quad \Longrightarrow \quad x_1 = 0 \quad {\rm lokalt\;minimum.} \] |
| \( \underline{x_1 = 0} \, \):
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\( \; \) |
\( f''(0) \, = \, 10 > 0 \quad \Longrightarrow \quad x_1 = 0 \quad {\rm lokalt\;minimum.} \) \( f''(-0,625) = 24\cdot(-0,625)^2 + 42\cdot(-0,625) + 10 = -6,876 < 0 \) \( \Longrightarrow \quad x_2 = -0,625 \quad {\rm lokalt\;maximum.} \) \( f''(-2) = 24\cdot(-2)^2 + 42\cdot(-2) + 10 = 22 > 0 \) \( \Longrightarrow \quad x_2 = -2 \quad {\rm lokalt\;minimum.} \) |
\( f''(0) \neq 0 \; {\rm ,} \; f''(-0,625) \neq 0 \; {\rm ,} \; f''(-2) \neq 0 \; \Longrightarrow \; f(x) \, {\rm har\;inga\;terasspunkter.} \)
Lokala maximi- och minimipunkternas \( y\)-koordinater:
\( f(x) \, = \, x^3 - 12\,x^2 + 45\,x - 44 \)
\( f(3) \, = \, 3^3 - 12\cdot 3^2 + 45\cdot 3 - 44 = 10 \; \Longrightarrow \quad (3, 10) \quad {\rm är\;lokal\;maximipunkt.} \)
\( f(5) \, = \, 5^3 - 12\cdot 5^2 + 45\cdot 5 - 44 = 6 \quad \Longrightarrow \quad\; (5, 6) \quad {\rm är\;lokal\;minimipunkt.} \)
