Skillnad mellan versioner av "2.3 Lösning 6a"

Nuvarande version från 10 oktober 2014 kl. 12.33

\[ f\,(-3 + h) = (-3 + h)^2 = (h - 3)^2 = h^2 - 6\,h + 9 \]

\( f(-3) = (-3)^2 = 9 \)

\[ {f(-3+h) - f(-3) \over h} = {h^2 - 6\,h + 9 - 9 \over h} = {h^2 - 6\,h \over h} = {h\,(h - 6) \over h} = h - 6\]

\[ f\,'(-3) \; = \; \lim_{h \to 0} \; (h - 6) \; = \; - 6 \]

@@ Rad 1: / Rad 1: @@
-:<math> \begin{array}{rcl} f(25+h) & = & 4\,(25+h)^2 - 380\,(25+h) + 9000          & = \\
+:<math> f\,(-3 + h) = (-3 + h)^2 = (h - 3)^2 = h^2 - 6\,h + 9 </math>
-                                   & = & 4\,(625+50\,h+h^2) - 9500 - 380\,h + 9000 & = \\
+<math> f(-3) = (-3)^2 = 9 </math>
-                                   & = & 2500 + 200\,h + 4\,h^2 - 500 - 380\,h     & = \\
+:<math> {f(-3+h) - f(-3) \over h} = {h^2 - 6\,h + 9 - 9 \over h} = {h^2 - 6\,h \over h} = {h\,(h - 6) \over h} = h - 6</math>
-                                   & = &  4\,h^2 - 180\,h + 2000
+:<math> f\,'(-3) \; = \; \lim_{h \to 0} \; (h - 6) \; = \; - 6 </math>
-        \end{array}</math>
-<math> f(25) = 4\cdot 25^2 - 380\cdot 25 + 9\,000 = 4\cdot 625 - 9500 + 9000 = 2000 </math>
-:<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>
-:<math> f\,'(25) \; = \; \lim_{h \to 0} \; (4\,h - 180) \; = \; - 180 </math>
-<Big>Dvs vid tiden <math> x = 25\, </math> sjunker oljans volym med <math> 180\, </math> liter per minut.</Big>