Question 654188
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Since you can't "solve" anything here, I presume you mean that you want to simplify the Difference Quotient for the given function.  Please -- read your textbook and use the proper language to communicate with us.


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


So


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


So


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{4\ -\ x^2\ -\ 2h\ -\ h^2\ -\ 4\ +\ x^2}{h}\ =]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{-2h\ -\ h^2}{h}\ =\ -2\ -\ h]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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