My Math Forum Difference quotient

 Algebra Pre-Algebra and Basic Algebra Math Forum

April 12th, 2017, 12:33 PM   #1
Senior Member

Joined: Feb 2016
From: seattle

Posts: 346
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
Attached Images
 20170412_132659.jpg (92.7 KB, 23 views) 20170412_132610.jpg (87.3 KB, 9 views)

 April 12th, 2017, 12:49 PM #2 Senior Member     Joined: Sep 2015 From: CA Posts: 1,207 Thanks: 615 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 Thanks from greg1313 and GIjoefan1976
 April 12th, 2017, 12:57 PM #3 Senior Member   Joined: May 2016 From: USA Posts: 595 Thanks: 251 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 u-substitutions become essential later, but I think they clarify from the beginning. $\therefore u = x + h \implies f(x + h) = f(u) = u^2 = (x + h)^2 = x^2 + 2hx + h^2.$ $\therefore \dfrac{f(x + h) - f(x)}{h} = \dfrac{x^2 + 2hx + h^2 - x^2}{h} =$ $\dfrac{2hx + h^2}{h} = \dfrac{\cancel h(2x + h)}{\cancel h} = 2x + h.$ Thanks from greg1313, romsek and GIjoefan1976
 April 12th, 2017, 01:31 PM #4 Senior Member   Joined: Feb 2016 From: seattle Posts: 346 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: 595 Thanks: 251 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 u-substitution 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: 346 Thanks: 10 (b.) I ended up with (-4h+34)
April 12th, 2017, 04:20 PM   #7
Senior Member

Joined: May 2016
From: USA

Posts: 595
Thanks: 251

Quote:
 Originally Posted by GIjoefan1976 I ended up with (-4h+34)

April 12th, 2017, 04:26 PM   #8
Senior Member

Joined: Feb 2016
From: seattle

Posts: 346
Thanks: 10

(b.)

Quote:
 Originally Posted by JeffM1 No. That is not close. Please show your work.
sorry wrong one hopefully this is closer?
Attached Images
 WIN_20170412_17_24_27_Pro.jpg (87.4 KB, 16 views)

 April 12th, 2017, 05:22 PM #9 Senior Member   Joined: May 2016 From: USA Posts: 595 Thanks: 251 Right on target. Way to go.
April 12th, 2017, 05:26 PM   #10
Senior Member

Joined: Feb 2016
From: seattle

Posts: 346
Thanks: 10

Quote:
 Originally Posted by JeffM1 Right on target. Way to go.
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.

 Tags difference, qoutient, quotient

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post kayna84 Pre-Calculus 2 August 6th, 2015 07:45 AM mophiejoe Calculus 2 February 12th, 2015 09:03 PM missingblood Algebra 6 January 24th, 2012 04:44 PM questioner1 Calculus 2 May 23rd, 2010 03:06 PM missingblood Calculus 3 January 1st, 1970 12:00 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top