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April 12th, 2017, 12:33 PM   #1
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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?

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April 12th, 2017, 12:49 PM   #2
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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
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April 12th, 2017, 12:57 PM   #3
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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.

$\therefore u = x + h \implies f(x + h) = f(u) = u^2 = (x + h)^2 = x^2 + 2hx + h^2.$

$\therefore \dfrac{f(x + h) - f(x)}{h} = \dfrac{x^2 + 2hx + h^2 - x^2}{h} =$

$\dfrac{2hx + h^2}{h} = \dfrac{\cancel h(2x + h)}{\cancel h} = 2x + h.$
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April 12th, 2017, 01:31 PM   #4
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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.....
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April 12th, 2017, 03:05 PM   #5
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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.
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April 12th, 2017, 04:17 PM   #6
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(b.)

I ended up with (-4h+34)
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April 12th, 2017, 04:20 PM   #7
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Quote:
Originally Posted by GIjoefan1976 View Post
I ended up with (-4h+34)
No. That is not close. Please show your work.
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April 12th, 2017, 04:26 PM   #8
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(b.)

Quote:
Originally Posted by JeffM1 View Post
No. That is not close. Please show your work.
sorry wrong one hopefully this is closer?
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April 12th, 2017, 05:22 PM   #9
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Right on target. Way to go.
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April 12th, 2017, 05:26 PM   #10
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Quote:
Originally Posted by JeffM1 View Post
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.
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