Question 184252
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ f(x) = 4x^2 + 2 ]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ f(x + h) = 4(x + h)^2 + 2 ]


So:


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


John
*[tex \Large e^{i\pi} + 1 = 0]
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