Skillnad mellan versioner av "2.2 Lösning 8a"

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::<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>
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::<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> \Delta y = f(x + h) \, - \, f(x) = (a + h)^2 - a^2 = a^2 + 2\,a\,h + h^2 - a^2 = 2\,a\,h + h^2 </math>
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::<math> \Delta y = f(x + h) \, - \, f(x) = 2\,h^2 + 4\,x\,h - 5\,h </math>
  
 
::<math> \Delta x \, = \, x + h \, - \, x \, = \, h </math>
 
::<math> \Delta x \, = \, x + h \, - \, x \, = \, h </math>
  
 
::<math> {\Delta y \over \Delta x} = {2\,a\,h + h^2 \over h} = {h\,(2\,a + h) \over h} = 2\,a + h </math>
 
::<math> {\Delta y \over \Delta x} = {2\,a\,h + h^2 \over h} = {h\,(2\,a + h) \over h} = 2\,a + h </math>

Versionen från 17 september 2014 kl. 14.10

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 \( y = f(x) = 2\,x^2 - 5\,x + 32 \):

\[ f(x + h) = 2\,(x+h)^2 - 5\,(x+h) + 32 = 2\,(x^2 + 2\,x\,h + h^2) - 5\,x - 5\,h + 32 = \]
\[ = 2\,x^2 + 4\,x\,h + 2\,h^2 - 5\,x - 5\,h + 32 = 2\,h^2 + 4\,x\,h - 5\,h + (2\,x^2 - 5\,x + 32) \]
\[ \Delta y = f(x + h) \, - \, f(x) = 2\,h^2 + 4\,x\,h - 5\,h \]
\[ \Delta x \, = \, x + h \, - \, x \, = \, h \]
\[ {\Delta y \over \Delta x} = {2\,a\,h + h^2 \over h} = {h\,(2\,a + h) \over h} = 2\,a + h \]