Question 800239
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ Q(x)\ =\ \frac{f(x\,+\,h)\ -\ f(x)}{h}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ Q_f(x)\ =\ \frac{\left(4(x\,+\,h)^2\ -\ 3(x\,+\,h)\ +\ 1\right)\ -\ \left(4x^2\ -\ 3x\ +\ 1\right)}{h}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ Q_f(x)\ =\ \frac{\left(4(x^2\,+\,2xh\,+\,h^2)\ +\ 3(x\,+\,h)\ +\ 1\right)\ -\ \left(4x^2\ -\ 3x\ +\ 1\right)}{h}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ Q_f(x)\ =\ \frac{8xh\ +\ 4h^2\ +\ 3h}{h}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ Q_f(x)\ =\ 8x\ +\ 4h\ +\ 3]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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