Evaluate the difference quotient

In summary: I am sorry if it was confusing. I hope it is clear now.In summary, the conversation discusses the concept of difference quotients and how to evaluate them by plugging in values into a function. The goal is to simplify the equation and get rid of the h on the bottom. The participants also discuss techniques such as cross-multiplying to simplify the numerator.
  • #1
vorcil
398
0
and simplify your awnser

f(x)= (X+3) / (x+1)

don't know how to do it,

my attempt

f(x+h)-f(x)
/h

((x+3)/(x+1)+h)-(x+3/(x+1)
/h

bah
how do i start and finish this?
 
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  • #2
This is the very basic thing you should know before you learn derivatives in Calculus. This is just a simple plug-the-value into the function. That means you change x's in function f(x) into x+h to transform the function into f(x) into f(x+h).

Give it a try again. You should be able to do this problem if you know middle school math.
 
  • #3
kNYsJakE said:
This is the very basic thing you should know before you learn derivatives in Calculus. This is just a simple plug-the-value into the function. That means you change x's in function f(x) into x+h to transform the function into f(x) into f(x+h).

Give it a try again. You should be able to do this problem if you know middle school math.

Didn't know they taught difference quotients to 12 year olds...

Don't have to be a total ****
don't post if you're not going to help, no point in criticizing people, I'm here to look for help on difference quotients
...

I know how to plug stuff in, I JUST DONT KNOW HOW TO START IT
do i put
(x+3+h)/(x+1+h) as the first part for (x+h)
or is it
(x+h)/(x+h)+3 for (x+h)
 
  • #4
Of course they don't teach difference quotients to 12 years olds. But, they do teach how to plug in values into functions.

I didn't mean to criticize or anything. I was just saying that you don't have to think of this problem in a hard way, because all you need to do is just plug in the value into the function. You learn how to plug in values into function in middle school. Don't you? That's what I meant.

First part that you have gotten above looks right. That's all you need to do. Put the first part you got into the difference quotients and see what you get. =]

vorcil said:
Didn't know they taught difference quotients to 12 year olds...

Don't have to be a total ****
don't post if you're not going to help, no point in criticizing people, I'm here to look for help on difference quotients
...

I know how to plug stuff in, I JUST DONT KNOW HOW TO START IT
do i put
(x+3+h)/(x+1+h) as the first part for (x+h)
or is it
(x+h)/(x+h)+3 for (x+h)
 
  • #5
I saw that you had posted this somewhere else also. I don't think it is a good idea to ask the same question in two different threads. You are allowed to edit posts, and it would probably have been better if you had just replied to your previous thread instead of starting another one.

The difference quotient is:
[tex] \frac{f(x+h)-f(x)}{h} [/tex]
The goal in evaluating difference quotients is to perform algebra until you get rid of the h on the bottom, i.e. simplify the equation. This is preparing you for ideas in calculus (which can be explained more after the problem at hand is finished if you want).

To find f(x+h), you just substitute in x+h everywhere x appears in f(x). Then
[tex]
\frac{f(x+h)-f(x)}{h} = \frac{ \frac{x+h+3}{x+h+1} - \frac{x+3}{x+1} }{h}
[/tex]
What technique do you know for simplifying the numerator? Can you do it? Hint: cross-multiply.
 
  • #6
n!kofeyn said:
Hint: cross-multiply.
"Cross multiply" is something you can do when you have a proportion: an equation involving two rational expressions. E.g., if a/b = c/d, then ad = bc. In this case the OP has an expression, so the options are limited.

A better hint is to combine the two rational expressions in the numerator so that they have the same denominator, and then go from there.
 
  • #7
Mark44 said:
"Cross multiply" is something you can do when you have a proportion: an equation involving two rational expressions. E.g., if a/b = c/d, then ad = bc. In this case the OP has an expression, so the options are limited.

A better hint is to combine the two rational expressions in the numerator so that they have the same denominator, and then go from there.

Cross-multiply can also mean
[tex]
\frac{a}{b}\pm\frac{c}{d} = \frac{ad\pm bc}{bd}
[/tex]
It is the same idea. I don't think it is uncommon to apply the word to the technique of adding two fractions or rational expressions. So by giving the hint of combining the rational expressions in the numerator, you gave the same hint I did.
 

1. What is the difference quotient?

The difference quotient is a mathematical formula used to find the average rate of change of a function over a given interval. It is represented as (f(x+h)-f(x))/h, where h is the change in the input variable x.

2. How do you evaluate the difference quotient?

To evaluate the difference quotient, you need to plug in the values of the function at x and x+h into the formula (f(x+h)-f(x))/h. Simplify the expression and then take the limit as h approaches 0 to get the instantaneous rate of change at x.

3. What is the significance of the difference quotient?

The difference quotient is significant because it allows us to measure the average rate of change of a function at a specific point. It is also an important concept in calculus, as it is used to find the derivative of a function.

4. Can the difference quotient be applied to any type of function?

Yes, the difference quotient can be applied to any type of function, including polynomial, exponential, logarithmic, and trigonometric functions. It is a general formula that can be used to find the average rate of change for any function.

5. How is the difference quotient related to the slope of a line?

The difference quotient is essentially the slope of the secant line between two points on a function. As the interval between the two points becomes smaller (h approaches 0), the secant line becomes closer to the tangent line at that point, which represents the instantaneous rate of change or slope of the function at that point.

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