Results 1 to 3 of 3

Math Help - difference quotient and absolute value

  1. #1
    Member sinewave85's Avatar
    Joined
    Oct 2006
    From
    Lost in a series of tubes.
    Posts
    218

    difference quotient and absolute value

    "Evaluate the difference quotient for the given functions. Show all of your work."

    The ones I am stuck on are:

    a. f(x) = lxl if x<-1 and 0<h<1

    b. f(x) = lxl if x>1 and 0<h<1

    I know the answers -- a. -1, b. 1 -- but I got them by plugging numbers that fit the parameters into the difference quotient and then evaluating it, which would be fine if I didn't have to show my work. As far as evaluating it directly, thought, I have no idea what to do. (My latex skills are pathetic, so please excuse my not typing all of my work out.)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,889
    Thanks
    326
    Awards
    1
    Quote Originally Posted by sinewave85 View Post
    "Evaluate the difference quotient for the given functions. Show all of your work."

    The ones I am stuck on are:

    a. f(x) = lxl if x<-1 and 0<h<1

    b. f(x) = lxl if x>1 and 0<h<1

    I know the answers -- a. -1, b. 1 -- but I got them by plugging numbers that fit the parameters into the difference quotient and then evaluating it, which would be fine if I didn't have to show my work. As far as evaluating it directly, thought, I have no idea what to do. (My latex skills are pathetic, so please excuse my not typing all of my work out.)
    Let's do a):

    \frac{|x + h| - |x|}{h}

    Now, x is less than -1 and 0 < h < 1 we know that x + h is negative. Thus |x + h| = -(x + h). Again, since x is negative |x| = -x. Thus
    \frac{|x + h| - |x|}{h} = \frac{-(x + h) - (-x)}{h} = \frac{-x - h + x}{h} = \frac{-h}{h} = -1

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member sinewave85's Avatar
    Joined
    Oct 2006
    From
    Lost in a series of tubes.
    Posts
    218
    Quote Originally Posted by topsquark View Post
    Now, x is less than -1 and 0 < h < 1 we know that x + h is negative. Thus |x + h| = -(x + h). Again, since x is negative |x| = -x. Thus

    \frac{|x + h| - |x|}{h} = \frac{-(x + h) - (-x)}{h} = \frac{-x - h + x}{h} = \frac{-h}{h} = -1

    -Dan
    Oh, wow. I never really understood absolute value correctly untill just now. That is both amazing and sad. Oh, well. Glad I decided to do calculus, or I would have gone my whole life not understanding absolute value. Just sad. Thanks for the enlightenment.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: June 29th 2011, 05:57 PM
  2. Difference quotient help!
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: October 1st 2009, 02:34 PM
  3. Difference quotient!
    Posted in the Algebra Forum
    Replies: 2
    Last Post: August 5th 2009, 03:47 AM
  4. How to use the difference quotient..
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 27th 2008, 01:13 PM
  5. Difference Quotient
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: July 19th 2008, 06:58 AM

Search Tags


/mathhelpforum @mathhelpforum