April 12th, 2017, 05:56 PM  #11 
Senior Member Joined: May 2016 From: USA Posts: 1,029 Thanks: 420 
Please stop being apologetic. I doubt anyone here is ever bothered by someone who works at learning. A problem with teaching mathematics is that we got over our hurdles so long ago that often we have difficulty even remembering what they were. This is compounded because we seldom explain that the modern exposition of mathematics has been formulated to address conceptual difficulties that either were not perceived or were glossed over by the very brilliant mathematicians of the past. That is, what we teach is not designed to address the initial student's difficulties, but rather to address issues that have never occurred to the beginning student but have concerned excellent mathematicians. Difference quotients stand at the doorway to calculus. It took close to 200 years for mathematicians to develop a satisfactory notation and rigorous development of calculus. We try to cram the results of that process of centuries into a few weeks. 
April 12th, 2017, 06:08 PM  #12  
Senior Member Joined: Feb 2016 From: Australia Posts: 1,591 Thanks: 546 Math Focus: Yet to find out.  Quote:
 
April 13th, 2017, 01:02 AM  #13 
Global Moderator Joined: Dec 2006 Posts: 18,956 Thanks: 1603 
Mistakes are often very obvious once spotted. In this case, (x + h)³ was expanded correctly, but should have been followed by " x³". 
April 13th, 2017, 04:17 AM  #14  
Senior Member Joined: Feb 2016 From: seattle Posts: 377 Thanks: 10  Quote:
Okay JeffM1, I just get nervous those helping me will forget how much I am appreciative if I don't show my gratitude. I don't want to upset anyone but maybe I will do as you say I really appreciate your words and of course also agree completely with your words, so thanks for saying so!  
April 13th, 2017, 04:23 AM  #15 
Senior Member Joined: Feb 2016 From: seattle Posts: 377 Thanks: 10  Okay thanks for taking a look at this as well. So I am guessing the final answer is correct, but I should have included x^3 in my second step to show it was still part of the....until my 3rd step?

April 13th, 2017, 05:19 AM  #16  
Senior Member Joined: May 2016 From: USA Posts: 1,029 Thanks: 420  Quote:
My fault. I looked at only the first and last lines and so did not notice that you failed to bring the minus x cubed down in the second line and so really should not have canceled the x cubed terms in the third line. Good catch by skipjack.  
April 13th, 2017, 10:50 AM  #17 
Senior Member Joined: Feb 2016 From: seattle Posts: 377 Thanks: 10  skipping to (f.)
Not sure if I have this set up right for a start? thanks 
April 13th, 2017, 11:22 AM  #18 
Senior Member Joined: May 2016 From: USA Posts: 1,029 Thanks: 420  You are still making the same mistake. Try the usubstitution method until you get the hang of it. $f(x) = 3x^2  2x + 4 \implies f(u) = 3u^2  2u + 4$ $Let\ (x + h) = u \implies f(x + h) = f(u) = 3u^2  2u + 4 =$ $3(x + h)^2  2(x + h) + 4 = 3x^2 + 6hx + 3h^2  2x  2h + 4.$ $\therefore \dfrac{f(x + h)  f(x)}{h} = \dfrac{3x^2 + 6hx + 3h^2  2x  2h + 4  (3x^2  2x + 4)}{h} = what?$ f(x) = formula in x means that f(expression) is the formula with every x replaced by the expression. What gets beginning students confused is that when the expression contains x, they can't figure out which x goes where. So the usubstitution method is to get f(u), where every x in the formula is replaced with u. No chance to get mixed up by the x's. Then f(expression) involves replacing every u with the expression. 

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