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Evaluate the difference quotient

  1. Jul 23, 2009 #1
    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?
     
  2. jcsd
  3. Jul 23, 2009 #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.
     
  4. Jul 23, 2009 #3
    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. Jul 23, 2009 #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. =]

     
  6. Jul 24, 2009 #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.
     
  7. Jul 25, 2009 #6

    Mark44

    Staff: Mentor

    "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.
     
  8. Jul 25, 2009 #7
    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.
     
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