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Homework Help: Derivative absolute value and also curve help !

  1. Jun 26, 2007 #1
    Hey i have 2 questions here that i have finished and i have no answers to my solutions so i would like someone to check it over and see where i went wrong , im not 100% sure on this stuff and need some help , thanks alot !

    Question 1: Using the definition of a derivative, show that ƒ (x) = |x| is not differentiable at x = 0.

    My soluition: ( aside; √ = square root)

    I know | u | = √u²

    Let x = u

    ƒ(x) = y = √u²
    ƒ¹(x) = ½(2u) / √u² (du/dx)
    = u / |u|
    = x / |x|

    Since subbing in x would make this undefined , x ≠ 0


    Question #2: Find the points on the curve y=(x+1) ³(1-3x) ²

    For horizontal dy/dx = 0

    Dy/dx = 3(x+1) ²(1-3x) ² + 2(1-3x)(-3)(x+1) ³

    Dy/dx = 3(x+1) ²(1-3x) ² + (-6)(1-3x)(x+1) ³
    Dy/dx = 0
    0 = 3(x+1) ²(1-3x) [ (1-3x) – 2(x+1) ]

    Therefore
    0 = (x+1) ²
    X = -1

    0 = (1-3x)
    X = 1/3

    0 = 1-3x-2x-2
    0 = -5x -1
    X = -1/5

    Therefor the x-values of the points on the curve are -1 , 1/3 , and -1/5 (sub them into y=(x+1) ³(1-3x) ² to get corresponding y values)
     
  2. jcsd
  3. Jun 26, 2007 #2

    nrqed

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    what you have to show is that the limit from the left (as you approach [itex] x \rightarrow x^- [/itex] ) gives a different value than the limit from the right. Just use your answer for the derivative an dit will be obvious that one limit gives +1 and the other gives -1.

    :confused: Are there some words missing in the question?? The points of inflexion? the extrema?
    Looks like you are looking for the extrema.
    At a quick glance it seems right to me.
     
  4. Jun 26, 2007 #3
    hey sry i forgot the second part

    Question #2: Find the point on the curve y=(x+1) ³(1-3x) ² where the slope of the tagent line is horizontal

    sorry about that
     
  5. Jun 26, 2007 #4
    okay and for the first one you are saying that i should break it up into 2 parts

    the first being:
    lim [itex] x \rightarrow x^- [/itex]

    so i would get x/-x
    =-1

    and then:
    lim x -> x
    gettinng x/x
    = 1

    ? is this what you are referring too?
     
  6. Jun 26, 2007 #5
    Your original proof for #1 is good enough. However, the question asked you to do it by the definitions.

    A function is differentiable at x=a IF AND ONLY IF

    [tex]\lim_{x \to a}\frac{f(x)-f(a)}{x-a}[/tex] exists.

    In general, limits of a function g at x=a exist IF AND ONLY IF

    [tex]\lim_{x \to a^-} g(x) = \lim_{x \to a^+}g(x)[/tex].

    These are your definitions. Be sure to use them in your proof. Explicitly. You seem to have the right idea, but you should elaborate on your thoughts (and notation).

    For #2, what is the slope of a horizontal line? So, when could the slope of the tangent be the same as this number?
     
    Last edited: Jun 26, 2007
  7. Jun 26, 2007 #6

    nrqed

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    that's right.
     
  8. Jun 26, 2007 #7
    What the hell is [tex]\lim x \to x^-[/tex]?

    Don't encourage such bad notation. He has the right idea, but hasn't put forth a good argument.

    If you merely meant that his idea was right, disregard previous comment.

    :)
     
    Last edited: Jun 26, 2007
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