1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

General Calculus question

  1. Jan 27, 2008 #1
    1. The problem statement, all variables and given/known data

    Okay, I have here the original equation that we start out with:

    f(x) = -2x + 4


    2. Relevant equations

    I'm supposed to find the slope of that equation. Which is obvious, since it's -2.

    But I was supposed to show how it's consistent with the limit definition of slope, which originally looks like this:

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

    The way they showed it, was by plugging in (x + h) into the x in -2x + 4. My question is, why exactly did they do that? Sorry if it's a stupid question, but it's been bugging me and I would like to know when to plug in the x + h, and when not to when determining slopes.

    Thanks in advance!
     
  2. jcsd
  3. Jan 27, 2008 #2
    Well, the value of the function at the point x+h is -2(x+h) + 4.

    f(x) = -2x+4
    f(x+h) = -2(x+h) + 4
     
  4. Jan 27, 2008 #3
    well look first of all the slope of the line at any point in the curve is basically the derivative of that function at that point. Now to fully understand this you need to look at the concept of the derivative. However, look here, f(x)-f(a) is merely the change of the function in the y-axes, while h=x-a is the change in the x-axes. now the slope of a line drawn at a poin in the curve is the limit of the change of the function on y-axes over the change of argument in the x-axes, as x-->a or, as h-->0,
    this way:
    [tex]\frac{f(x)-f(a)}{x-a}[/tex] the limit of this as x--->a or


    [tex]\frac{f(a+h)-f(a)}{h}[/tex] the limit of this as h-->0
     
  5. Jan 27, 2008 #4
    f'(x) = limit [f(x+h) - f(x)]/h
    (as h->0)
    f(x) = -2x+4

    f'(x) = limit [f(-2(x+h) + 4) - (-2x + 4)]/h
    h -> 0

    f'(x) = limit [-2x - 2h+ 4 +2x - 4]/h
    h -> 0

    f'(x) = limit [ -2 h]/h
    h-> 0

    f'(x) = limit [-2]
    h -> 0

    f'(x) = -2

    Hope this helps.
     
  6. Jan 27, 2008 #5
    As stupid math said [f(x+h) - f(x)] / h is just the average slope of a function.

    However according to the mean value theorem there is some point on a closed interval where the average slope is the the same as the instantaneous rate of change (slope) of the function or:

    f'(c) = [f(x+h) - f(x)] / h

    Since your function is a line this happens to occur everywhere.
     
  7. Jan 27, 2008 #6

    jambaugh

    User Avatar
    Science Advisor
    Gold Member

    Wo there! you're invoking a jackhammer to pull a staple. It is just a trivial instantiation of the definition of the derivative. If he does the algebra correctly and invokes the definition and the limit laws correctly (which I think is the point of the exercise) he gets the correct answer.

    One key step is that the value of the difference quotient is undefined at h=0 but it is "ok" to cancel h's as one is working inside a limit and you invoke the pertinent limit law (f=g except at x=a then limit of f = limit of g at x=a).
     
  8. Jan 27, 2008 #7
    I was just trying to present another way to look at the problem :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: General Calculus question
  1. General question (Replies: 3)

Loading...