April 12th, 2017, 12:33 PM  #1 
Senior Member Joined: Feb 2016 From: seattle Posts: 377 Thanks: 10  Difference quotient
I am guessing I have wrote these wrong as won't the  sign make it where everthing gets subtracted or am I supposed to change the sign after I do something or do something before I change the sign? Thanks 
April 12th, 2017, 12:49 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 1,944 Thanks: 1011 
You are treating all of these as if $f(x+h) = f(x) + h$ This is incorrect In (1) for example $f(x+h) = (x+h)^2 =x^2 + 2xh + h^2$ so $\dfrac{f(x+h)f(x)}{h} = \dfrac{2xh+h^2}{h} = 2x + h$ see if you can apply this to the rest of them 
April 12th, 2017, 12:57 PM  #3 
Senior Member Joined: May 2016 From: USA Posts: 1,030 Thanks: 420 
Your problem here is that you really have not yet grasped function notation. f(x) = some formula involving x means that f(u) is the same formula with u replacing x in the formula. $f(x) = x^2 \implies f(u) = u^2.$ That is easy. The confusion arise when you are given f(g(x)). Set g(x) = u, and the confusion dissipates. These usubstitutions become essential later, but I think they clarify from the beginning. $\therefore u = x + h \implies f(x + h) = f(u) = u^2 = (x + h)^2 = x^2 + 2hx + h^2.$ $\therefore \dfrac{f(x + h)  f(x)}{h} = \dfrac{x^2 + 2hx + h^2  x^2}{h} =$ $\dfrac{2hx + h^2}{h} = \dfrac{\cancel h(2x + h)}{\cancel h} = 2x + h.$ 
April 12th, 2017, 01:31 PM  #4 
Senior Member Joined: Feb 2016 From: seattle Posts: 377 Thanks: 10 
ahhh, thanks you both! for sure I do still have to learn how to read these and how to say it correctly I think it's partly the = sign as I see it as saying x is to be x2 but it is really saying.....

April 12th, 2017, 03:05 PM  #5 
Senior Member Joined: May 2016 From: USA Posts: 1,030 Thanks: 420 
A bit of vocabulary may help. You know I presume what an expression is. It is a numeral, a pronumeral, or a combination of numerals, pronumerals or both joined by operators and grouping symbols. A function has an argument. In f(x), the argument is x. The function represents a rule, a set of instructions on what to do with the argument. In most cases for elementary algebra and calculus, the rule is defined by an expression so, for, $f(x) = x^2  2x + 1$, we have x as the argument and $x^2  2x + 1$ as the expression defining the rule. With me so far? We defined the rule using a simple pronumeral as the argument, x in this case, but any expression can be the argument. Whatever you see as the argument of that function you replace that entire expression wherever x is in the expression defining the rule. When the argument's expression contains the pronumeral used in the defining rule, beginning students may lose the thread. So I am going to explain an example in two different ways. $ f(x) = x^2  2x + 1 \implies f(3x  4) = what?$ The usubstitution way. $u = (3x  4) \implies f(3x  4) = f(u) = u^2  2u + 1 =$ $(3x  4)^2  2(3x  4) + 1 = 9x^2  24x + 16  6x + 8 + 1 = 9x^2  30x + 25.$ That method prevents confusion but adds steps. The quick way is $f(3x  4) = (3x  4)^2  2(3x  4) + 1 =$ $9x^2  24x + 16  6x + 8 + 1 = 9x^2  30x + 25.$ Once you fell comfortable with the notation, you can skip the u substitution. Now do the remaining problems and show us what you get. 
April 12th, 2017, 04:17 PM  #6 
Senior Member Joined: Feb 2016 From: seattle Posts: 377 Thanks: 10  (b.)
I ended up with (4h+34)

April 12th, 2017, 04:20 PM  #7 
Senior Member Joined: May 2016 From: USA Posts: 1,030 Thanks: 420  
April 12th, 2017, 04:26 PM  #8 
Senior Member Joined: Feb 2016 From: seattle Posts: 377 Thanks: 10  (b.) sorry wrong one hopefully this is closer?

April 12th, 2017, 05:22 PM  #9 
Senior Member Joined: May 2016 From: USA Posts: 1,030 Thanks: 420 
Right on target. Way to go.

April 12th, 2017, 05:26 PM  #10 
Senior Member Joined: Feb 2016 From: seattle Posts: 377 Thanks: 10  Thanks JeffM1 I appreciate your asking me to show what I got for them; I will work out the others to show. I know you all know how much I really do want to know this and be less of a stress....
Last edited by skipjack; April 13th, 2017 at 12:52 AM. 

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