# Difference quotient

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• Jun 5th 2009, 06:17 PM
CaptianHawk1
Difference quotient
Next problem:

For each function f(x), determine this difference quotient in a simplified form.
f
(x + h) − f(x)

h

a.) f(x) = 2x + 1

The problem I have here is that I understand all of the steps except for the first one, which is this:

2(x + h) + 1 − (2x + 1)
h

The second part of the first step ((2x + 1)) makes complete sense.

It's this I can't wrap my head around:

2(x + h) + 1

How does f(x) = 2x + 1 turn this f(x + h) into this
2(x + h) + 1
?
• Jun 5th 2009, 06:25 PM
yeongil
Quote:

Originally Posted by CaptianHawk1
It's this I can't wrap my head around:

2(x + h) + 1

How does f(x) = 2x + 1 turn this f(x + h) into this
2(x + h) + 1
?

The function is f(x) = 2x + 1. When you put in a number for x, like f(4), you're essentially saying, "I want the value of f(x) when x = 4." So to find this value you plug in 4 for x:
f(4) = 2(4) + 1 = 9

But you don't have to just put in numbers. You could put in variables and expressions too. For example, I want to know what f(y) is. Who cares whether I know what y is equal to or not? I plug in y for x:
f(y) = 2y + 1.

In your case, you want to know the value of the function when x = x + h. Plug in x + h every time you see an x:
f(x + h) = 2(x + h) + 1.

I'll give you one more example. What if I want to know the value of the function when x = ab + c?
f(ab + c) = 2(ab + c) + 1.

01
• Jun 6th 2009, 01:56 PM
CaptianHawk1
I get it now. I was doing it in reverse, that is taking the 2x + 1 and trying to plug it into x's in the fraction.

Keep in mind, I'm teaching myself this stuff. (Happy)

Thanks again.