Question 311326
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ x^2\ +\ 3x\ +\ 1]


So, *[tex \Large f(a)] is 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(a)\ =\ a^2\ +\ 3a\ +\ 1]


and *[tex \Large f(a\ +\ h)] is


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(a\ +\ h)\ =\ (a\ +\ h)^2\ +\ 3(a\ +\ h)\ +\ 1]


so


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{f(a\ +\ h)\ -\ f(a)}{h}\ =\ \frac{\left((a\ +\ h)^2\ +\ 3(a\ +\ h)\ +\ 1\right)\ -\ \left(a^2\ +\ 3a\ +\ 1\right)}{h}]


Expanding the binomials:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{f(a\ +\ h)\ -\ f(a)}{h}\ =\ \frac{\left(a^2\ +\ 2ah\ +\ h^2\ +\ 3a\ +\ 3h\ +\ 1\right)\ -\ \left(a^2\ +\ 3a\ +\ 1\right)}{h}]


Removing parentheses and collecting like terms:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{f(a\ +\ h)\ -\ f(a)}{h}\ =\ \frac{2ah\ +\ h^2\ +\ 3h}{h}]


Then eliminating the factor of *[tex \Large h] common to both numerator and denominator:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{f(a\ +\ h)\ -\ f(a)}{h}\ =\ 2a\ +\ h\ +\ 3]


And that's all there is to it.


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