Page 1 of 2 12 LastLast
Results 1 to 15 of 17

Math Help - Delta?

  1. #1
    Member CalcGeek31's Avatar
    Joined
    Aug 2008
    Posts
    83

    Delta?

    what does the symbol delta(the triangle symbol)x mean?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    It is the 4th Greek letter and usually represents 'change in' x
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member CalcGeek31's Avatar
    Joined
    Aug 2008
    Posts
    83
    so what would I do if the problem is:

    (f(2+delta(x))+ f(2)) / delta(x)

    the equation:x^2 - 2x + 1
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member Spec's Avatar
    Joined
    Aug 2007
    Posts
    318
    The problem is most likely to find the derivative of f in the point x=2.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Jun 2008
    Posts
    792
    Quote Originally Posted by CalcGeek31 View Post
    so what would I do if the problem is:

    (f(2+delta(x))+ f(2)) / delta(x)

    the equation:x^2 - 2x + 1
    Delta, as Galactus said, represents change in x. In your case, it represents the change of x as it approaches zero. This is the derivative at the point x = 2. Also, you made a small typo:

    \lim_{\Delta x \to 0} \frac{f(2+\Delta x ) \color{red}- f(2)}{\Delta x}
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member CalcGeek31's Avatar
    Joined
    Aug 2008
    Posts
    83
    Im not sure I understand...

    my problem is f(x) is equal to the second equation... now evaluate the first equation...
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    Jun 2008
    Posts
    792
    Let h = \Delta x

    \lim_{h \to 0} \frac{f(2+h) - f(2)}{h}

    Our objective here is to get rid of the h in the denominator so we can find the limit. Now, let's evaluate f(2+h) and f(2)

    f(2+h) = (2+h)^2 - 2(2+h) - 1

    f(2) = (2)^2 - 2(2) - 1

    Now simply replace and simplify. You'll find that you can cancel the h in the denominator after all.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member CalcGeek31's Avatar
    Joined
    Aug 2008
    Posts
    83
    ohhh... now that makes perfect sense... it takes the form of f(x+h) thank you

    I also have a similar problem in which there is no change in x. does this mean I work it all the way up until I do the limit?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member CalcGeek31's Avatar
    Joined
    Aug 2008
    Posts
    83
    wait... I get down to lim(h-> 0) (2h + h^2 - 2) / h
    then how do I cancel the denominator?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member
    Joined
    Jun 2008
    Posts
    792
    Quote Originally Posted by CalcGeek31 View Post
    wait... I get down to lim(h-> 0) (2h + h^2 - 2) / h
    then how do I cancel the denominator?
    You made a mistake while simplifying:
    \lim_{h \to 0} \frac{f(2+h) - f(2)}{h}

    \lim_{h \to 0} \frac{(2+h)^2 - 2(2+h) - 1 - (-1)}{h}

    \lim_{h \to 0} \frac{4 + 4h + h^2 - 4 +2h - 1 +1}{h}

    ...
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Member CalcGeek31's Avatar
    Joined
    Aug 2008
    Posts
    83
    ahhh... k so my final answer is 2?
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Super Member
    Joined
    Jun 2008
    Posts
    792
    Quote Originally Posted by CalcGeek31 View Post
    ahhh... k so my final answer is 2?
    EDIT: Yes, I made trivial mistake. You're correct.
    Last edited by Chop Suey; August 22nd 2008 at 07:14 PM.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Member CalcGeek31's Avatar
    Joined
    Aug 2008
    Posts
    83
    <br />
f(2+h) = (2+h)^2 - 2(2+h) - 1<br />
    from that... -2(h) = -2 h + 4h
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Super Member
    Joined
    Jun 2008
    Posts
    792
    \lim_{h \to 0} \frac{f(2+h) - f(2)}{h}

    \lim_{h \to 0} \frac{(2+h)^2 - 2(2+h) - 1 - (-1)}{h}

    \lim_{h \to 0} \frac{4 + 4h + h^2 - 4 -2h - 1 +1}{h}

    \lim_{h \to 0} \frac{4h + h^2 -2h}{h}

    \lim_{h \to 0} \frac{h^2 +2h}{h}

    \lim_{h \to 0} \frac{h(h+2)}{h}

    \lim_{h \to 0} (h+2) = 2
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Member CalcGeek31's Avatar
    Joined
    Aug 2008
    Posts
    83
    <br />
\lim_{h \to 0} \frac{4 + 4h + h^2 - 4 +2h - 1 +1}{h}<br />
[
    your +2h in here should be -2h...
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Help. Can I do this? (Delta Epsilon)
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 21st 2010, 07:35 PM
  2. Replies: 0
    Last Post: October 27th 2009, 08:06 AM
  3. Replies: 2
    Last Post: September 29th 2009, 09:59 AM
  4. delta x
    Posted in the Math Topics Forum
    Replies: 3
    Last Post: December 9th 2008, 05:55 PM
  5. delta!!
    Posted in the Math Topics Forum
    Replies: 8
    Last Post: October 2nd 2006, 12:02 AM

Search Tags


/mathhelpforum @mathhelpforum