Skillnad mellan versioner av "2.3 Lösning 5a"
Taifun (Diskussion | bidrag) m |
Taifun (Diskussion | bidrag) m |
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| − | <math> \begin{array}{rcl} f(25+h) & = & 4\,(25+h)^2 - 380\,(25+h) + 9000 & = \\ | + | :<math> \begin{array}{rcl} f(25+h) & = & 4\,(25+h)^2 - 380\,(25+h) + 9000 & = \\ |
| − | + | & = & 4\,(625+50\,h+h^2) - 9500 - 380\,h + 9000 & = \\ | |
| − | + | & = & 2500 + 200\,h + 4\,h^2 - 500 - 380\,h & = \\ | |
| − | + | & = & 4\,h^2 - 180\,h + 2000 | |
| − | + | \end{array}</math> | |
<math> f(25) = 4\cdot 25^2 - 380\cdot 25 + 9\,000 = 4\cdot 625 - 9500 + 9000 = 2000 </math> | <math> f(25) = 4\cdot 25^2 - 380\cdot 25 + 9\,000 = 4\cdot 625 - 9500 + 9000 = 2000 </math> | ||
| − | :<math> {\Delta y \over \Delta x} = {f(25+h) - f(25) \over h} = {4\,h^2 - 180\,h + 2000 -2000 \over h} = | + | :<math> \begin{array}{rcl} {\Delta y \over \Delta x} & = & {f(25+h) - f(25) \over h} = {4\,h^2 - 180\,h + 2000 -2000 \over h} & = \\ |
| − | + | & = & {4\,h^2 - 180\,h \over h} = {h\cdot (4\,h - 180) \over h} = 4\,h - 180 & = \\ | |
| + | \end{array}</math> | ||
Versionen från 30 september 2014 kl. 14.22
\[ \begin{array}{rcl} f(25+h) & = & 4\,(25+h)^2 - 380\,(25+h) + 9000 & = \\ & = & 4\,(625+50\,h+h^2) - 9500 - 380\,h + 9000 & = \\ & = & 2500 + 200\,h + 4\,h^2 - 500 - 380\,h & = \\ & = & 4\,h^2 - 180\,h + 2000 \end{array}\]
\( f(25) = 4\cdot 25^2 - 380\cdot 25 + 9\,000 = 4\cdot 625 - 9500 + 9000 = 2000 \)
\[ \begin{array}{rcl} {\Delta y \over \Delta x} & = & {f(25+h) - f(25) \over h} = {4\,h^2 - 180\,h + 2000 -2000 \over h} & = \\
& = & {4\,h^2 - 180\,h \over h} = {h\cdot (4\,h - 180) \over h} = 4\,h - 180 & = \\
\end{array}\]
\( f\,'(25) \; = \; \lim_{h \to 0} \; (4\,h - 180) \; = \; - 180 \)
Dvs vid tiden \( x = 25\, \) sjunker oljans volym med \( 180\, \) liter per minut.
