Skillnad mellan versioner av "2.2 Lösning 8a"
Från Mathonline
Taifun (Diskussion | bidrag) m |
Taifun (Diskussion | bidrag) m |
||
| Rad 7: | Rad 7: | ||
::<math> f(x + h) = 2\,(x+h)^2 - 5\,(x+h) + 32 = 2\,(x^2 + 2\,x\,h + h^2) - 5\,x - 5\,h + 32 = </math> | ::<math> f(x + h) = 2\,(x+h)^2 - 5\,(x+h) + 32 = 2\,(x^2 + 2\,x\,h + h^2) - 5\,x - 5\,h + 32 = </math> | ||
| − | ::<math> = 2\,x^2 + 4\,x\,h + 2\,h^2 - 5\,x - 5\,h + 32 = 2\,h^2 + 4\,x\,h - 5\,h + (2\,x^2 - 5\,x + 32) </math> | + | ::<math> = {\color{Red} {2\,x^2}} + 4\,x\,h + 2\,h^2 {\color{Red} {- 5\,x}} - 5\,h + {\color{Red} {32}} = 2\,h^2 + 4\,x\,h - 5\,h + ({\color{Red} {2\,x^2}} {\color{Red} {- 5\,x}} + {\color{Red} {32}}) </math> |
::<math> \Delta y = f(x + h) \, - \, f(x) = 2\,h^2 + 4\,x\,h - 5\,h </math> | ::<math> \Delta y = f(x + h) \, - \, f(x) = 2\,h^2 + 4\,x\,h - 5\,h </math> | ||
::<math> {\Delta y \over \Delta x} = {2\,h^2 + 4\,x\,h - 5\,h \over h} = {h\,(2\,h + 4\,x - 5) \over h} = 2\,h + 4\,x - 5 </math> | ::<math> {\Delta y \over \Delta x} = {2\,h^2 + 4\,x\,h - 5\,h \over h} = {h\,(2\,h + 4\,x - 5) \over h} = 2\,h + 4\,x - 5 </math> | ||
Versionen från 3 november 2014 kl. 13.36
Definitionen till ändringskvot i intervallet mellan \( x \, \) och \( x+h \, \):
- \[ {\Delta y \over \Delta x} \; = \; {f(x + h) \, - \, f(x) \over h} \]
Tillämpad på vårt exempel \( y = f(x) = 2\,x^2 - 5\,x + 32 \):
- \[ f(x + h) = 2\,(x+h)^2 - 5\,(x+h) + 32 = 2\,(x^2 + 2\,x\,h + h^2) - 5\,x - 5\,h + 32 = \]
- \[ = {\color{Red} {2\,x^2}} + 4\,x\,h + 2\,h^2 {\color{Red} {- 5\,x}} - 5\,h + {\color{Red} {32}} = 2\,h^2 + 4\,x\,h - 5\,h + ({\color{Red} {2\,x^2}} {\color{Red} {- 5\,x}} + {\color{Red} {32}}) \]
- \[ \Delta y = f(x + h) \, - \, f(x) = 2\,h^2 + 4\,x\,h - 5\,h \]
- \[ {\Delta y \over \Delta x} = {2\,h^2 + 4\,x\,h - 5\,h \over h} = {h\,(2\,h + 4\,x - 5) \over h} = 2\,h + 4\,x - 5 \]
