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Question 804336: I understood everything until I got further on in the homework and now I do not get it! You do not have to solve the whole problem, I am very capable of doing that, but may you please at least help me set them up and explain them too so I can learn how to do it? Thank you so much!
Let f(x)=x^2 -1 and g(x)= 1- 2x. Find the indicated values for
A. g(2g(1))
B. g(1/f(3))
C. g(a+1)-g(a)
D. f(a+1)-f(a)
E. f(x+h) - f(x)/h
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
The one thing you have to remember is that given some ) , ) is the same thing with  replaced by 1. ) is the same thing with  replaced by  . So if \ =\ x^2\ -\ 3x\ +\ 2) , \ =\ (2)^2\ -\ 3(2)\ +\ 2) , \ =\ a^2\ -\ 3a\ +\ 2) , and \ =\ (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 \ =\ x^2\ -\ 1) , evaluate \ -\ f(x)}{h})
Start by defining  =\ (x\ +\ h)^2\ -\ 1)
Now plug in everything you know:
^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
Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it
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