Question 804336
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The one thing you have to remember is that given some *[tex \Large f(x)], *[tex \Large f(1)] is the same thing with *[tex \Large x] replaced by 1.  *[tex \Large f(a)] is the same thing with *[tex \LARGE x] replaced by *[tex \Large a].  So if *[tex \Large f(x)\ =\ x^2\ -\ 3x\ +\ 2], *[tex \Large f(2)\ =\ (2)^2\ -\ 3(2)\ +\ 2], *[tex \Large f(a)\ =\ a^2\ -\ 3a\ +\ 2], and *[tex \Large f(gobbledegook)\ =\ (gobbledegook)^2\ -\ 3(gobbledegook)\ +\ 2]


For part A, start by evaluating g(1).  Then multiply that value by 2.  Then evalute g at that value.  g(1) is 1 - 2(1) = -1. 2 times -1 is -2.  Then g(-2) is 1 - 2(-2) = 1 - (-4) = 5.


Now, let's do part E.


Given *[tex \LARGE f(x)\ =\ x^2\ -\ 1], evaluate *[tex \LARGE \frac{f(x\ +\ h)\ -\ f(x)}{h}]


Start by defining *[tex \LARGE f(x\ +\ h) =\ (x\ +\ h)^2\ -\ 1]


Now plug in everything you know:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{(x\ +\ h)^2\ -\ 1\ -\ (x^2\ -\ 1)}{h}]


Now all you have to do is expand the binomial and collect like terms.  If you are careful about your signs, all of the terms you have left after collecting like terms will have at least one factor of h.  You can then divide through by h, getting that pesky thing out of your denominator -- always the goal when working out a difference quotient.


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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