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Homework Help: Differentiating absolute quadratic

  1. Sep 18, 2010 #1
    1. The problem statement, all variables and given/known data
    For the function y= |x2-4x+3|
    a. Sketch the function
    b. Is the function continuous? Explain.
    c. Is the function differentiable everywhere? explain
    d. find gradient of curve at the point where x = 3/2

    2. Relevant equations
    chain rule, differentiating...

    3. The attempt at a solution
    a. easy enough.
    b. I said the function was continuous as i dont think it would be undefined at any point..?
    c. I know the derivative is undefined at x=3 and x=1 (the x intercepts) but I am not sure why...
    d. Not sure where to go here, all i know is that i can put it in the form f(x)=|(x-3)(x-1)|, let u=x-3, let v=x-1, and that |u|=sqrt(u2), and |v|=sqrt(v2)
    But im not sure where to go from there..
  2. jcsd
  3. Sep 18, 2010 #2


    Staff: Mentor

    A derivative formula would be helpful here.
    [tex]\frac{d|x|}{dx} = \frac{x}{|x|}[/tex]
    Notice that this reduces to 1 if x > 0, and -1 if x < 0.

    Can you come up with a chain rule form of this formula? I.e.,
    [tex]\frac{d|u|}{dx} [/tex]

    If so, that would give you a justification for saying that the derivative was undefined at x = 3 and x = 1, and you could use it to find the slope of the tangent line at x = 3/2.
  4. Sep 18, 2010 #3


    User Avatar
    Science Advisor


    Yes, the function is continuous but that's not a very good explanation! A function does not have to be "undefined" any where in order not to be continuous. The definition of "continuous" is that [itex]\lim_{x\to a} f(x)= f(a)[/itex]. The function f(x)= 0 if x< 0 , 1 if x>= 0 is defined for all x but not continous at x= 0. Since you have already graphed the function what does the graph tell you about continuity? Continuity should be trivial except possibly at two points. What are those points?

    Again, look at the graph. The derivative is the "slope" of the tangent line" isn't it? What would a tangent line look like at x= 1 and x= 3? What do tangent lines look like at points just on either side of those?

    At x= 3/2, you are well away from either x= 1 or x= 3. For all x close to 3/2 (x= 3/2+ h for some small h), f(x) is either [itex]x^2- 4x+ 3[/itex] or [itex]-(x^2- 4x+ 3)[/itex] Which is it? Can you differentiate that?
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