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!

Difference Quotient - average rate of change

  1. Oct 12, 2012 #1
    1. The problem statement, all variables and given/known data

    Original function f(x) = -x^2+5x-2

    I have calculated the difference quotient as -2x-h+5

    Then use difference quotient to calculate the average rate of change:

    for the following x=2 and x=2+h

    I need this to go on to the next part of the question and i wanted to ensure that I calculated this correctly

    2. Relevant equations

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

    3. The attempt at a solution

    Change in x = 2+h-2 = h

    Change in y = plugged f(2+h)-f(2) into the original function

    = -h^2-4h-4+10+5h-2-4 = -h^2+h

    Δy/Δx = (-h^2+h)/(h) = -h+1 or 1-h (not sure which format is correct)?

    Plug answer back into to the difference quotient - answer is consistent with DQ formula

    If this is correct I need to then answer the following question which is:

    as h approaches 0 what numerical value does

    f(2+h)-(2) approach
    h

    The question is where do I start?

    Cheers
     
  2. jcsd
  3. Oct 12, 2012 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    hi zebra1707! :smile:
    yes, and now the limit of that as h -> 0 is … ? :wink:
     
  4. Oct 12, 2012 #3

    Hi TT, thats where I am stuck, Im not sure as to the next step. Can I assume that the answer to the previous question is correct?
     
  5. Oct 12, 2012 #4

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    to find the limit of a continuous function as h -> 0, you can always just put h = 0 :smile:

    (the only exception being that if that gives 0/0)
     
  6. Oct 12, 2012 #5

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Why did you bother to repeat the calculation of [f(2+h)-f(2)]/h? You already stated that you had calculated [f(x+h) - f(x)/h = -2x - h + 5. Why wouldn't you just put x = 2 in that formula?

    As for whether to use 1-h or -h+1: well, what's the problem? They are exactly the same thing, written in different order.

    Finally: surely you can see what happens to the value of 1-h as h gets smaller and smaller and smaller, eventually going to zero.

    RGV
     
  7. Oct 12, 2012 #6
    Ray

    I appreciate your comments. However your reply is bordering on "well aren't you an idiot" I have used this forum for a few years now, and never once have I received a reply like this.

    Not everyone fully understands calculus as you do - Just a simple " did you realise that you have repeated the calculation" and maybe an explanation as to where I was going wrong...

    Sorry all I needed was a little guidance..
     
  8. Oct 12, 2012 #7

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Well, the first question was asked in order to try to get you to "unblock" something. When I assisted students one-on-one in the office, they would often do just what you did, and I would ask them why they did that; usually that would lead eventually to a realization on their part that there is a relation between something like [f(x+h) - f(x)]/h and [f(1+h) - f(1)]/h, and that one really is just a special case of the other. Sometimes just asking the question leads to greater understanding. But, as I said, that was in a one-on-one situation in an office or a classroom.

    The other statement about what happens to -1 + h in the limit h --> 0 could have been put another way, I admit. But the question still stands unchanged, and you really should try to answer it: what DOES happen to -1 + h as h gets smaller, and smaller, and smaller? Use some numbers if you have to. Draw a picture if you have to. Do whatever you feel most comfortable with, but try, try try to answer the question. That is how you learn---by overcoming difficulties.

    RGV
     
  9. Oct 12, 2012 #8

    Mark44

    Staff: Mentor

    IMO, Ray's response was NOT bordering on "aren't you an idiot", so it seems to me that you are reading something into what he said that probably wasn't there.
     
  10. Oct 12, 2012 #9
    Thank you Ray, I am a mature age student trying to nut things out for myself, I don't always have access to teachers, tutors etc - so its a challenge most of the time.

    I appreciate the response.

    Regards
     
  11. Oct 14, 2012 #10
    This is still confusing me..

    The equation represents the difference quotient
    f(2+h)− f(2) /h

    it represents the slope of the secant line through the points (2, f (2)) and
    (2+h, f(2+h)) on the curve.

    But i used this equation to find the DQ = -h+1 in the first place - feels like I am going around in circles

    If you could point me in the direction of an example - that would be helpful

    Regards
     
    Last edited: Oct 14, 2012
  12. Oct 14, 2012 #11

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Well, h can be positive or negative, but let's start with the positive case. For h = 0.1, DQ = -1 + .1 = -0.9. For h = 0.01, DQ = -0.99, and for h = 0.001, DQ = -0.999, etc. Now look at the case of negative h. For h = -0.1, DQ = -1.1. For h = -0.01, DQ = -1.01, etc. So, you can see what is happening as |h|---the absolute value of h----becomes smaller and smaller, going to zero.

    Before, when I said h gets smaller and smaller, I really meant |h|, but if h > 0 it is the same thing anyway.

    RGV
     
  13. Oct 14, 2012 #12

    Mentallic

    User Avatar
    Homework Helper

    Actually,

    [tex]\frac{f(2+h)-f(2)}{h}[/tex]

    and

    [tex]1-h[/tex]

    are equivalent in this example. If you use a different function, you'll probably get a different answer.

    So you have that the secant between the point (2,f(2)) and (2+h, f(2+h)) have a gradient of 1-h, and now we want to do what calculus is all about and that is to find the tangent, which is a secant that has two points which are infinitesimally close to each other (basically in the same spot), so we want to let h=0.

    So the moral of the story is, while both of those expressions above are equivalent, when we let h=0, one of them gives us the result 0/0 which tells us nothing, and the other gives us the answer 1-0=1.
     
  14. Oct 15, 2012 #13
    Thank you Mentallic, greatly appreciated.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Difference Quotient - average rate of change
  1. Difference Quotient (Replies: 4)

  2. Difference Quotient (Replies: 3)

Loading...