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Math Help - Intro to Calculus - Why are quadratic function differentiable at the vertex?

  1. #1
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    Question Intro to Calculus - Why are quadratic function differentiable at the vertex?

    Good day everybody!!

    I am new to the forum and I just started calculus for a month or two. We are currently learning limits and derivatives and I came across this problem (or perhaps a misunderstanding on my part )

    Why are quadratic functions (or any other functions with a local max/min) differentiable at the vertex (or at their local max/min)?

    __________________________________________________ __________________________________________________ __________________________________________________ ____________________
    Before I start, there are a couple things that I would like to note before going further, I was taught that:
    A limit only exists when:
    1) The left limit exists
    2) The right limit exists
    3) The left and right limits are equal

    If any of these conditions are not met, a limit does not exist (DNE)
    In addition, at least from my instructor, I was told that infinity is regarded as a "DNE". A limit is a real number.


    A function is differentiable at a if f'(a) exist (The tangent exist at x=a)
    All differentiable function must be continuous.
    NOT all continuous functions are differentiable.

    A function is not differentiable at points where:
    1) A tangent does not exist
    2) The function is discontinuous
    3) The left and right limits of the slope of the secant are not equal (<-- The wording of this one is what bothers me the most, does it simply mean the left limit and right limit are not equal as x -> point of tangency?)
    __________________________________________________ __________________________________________________ __________________________________________________ ____________________

    Okay, with that being said, here is my problem.
    For the absolute value function, ex. f(x) = abs(x), it is known that at x = 0, the function is not differentiable as "the tangent does not exist because the left and right limits of the slope of the secant does not equal" We did a proof in class:

    (Note: The bold is to emphasize the numerator for easier reading.
    Left Limit
    lim(h->0-) f(x+h)-f(x) / h
    lim(h->0-) -(x+h)-(-x) / h <- by multiplying -1, the absolute value sign (| |) is replaced
    lim(h->0-) -x-h+x / h
    lim(h->0-) -h / h
    lim(h->0-) -1
    -1

    Right Limit
    lim(h->0+) f(x+h)-f(x) / h
    lim(h->0+) (x+h)-(x) / h <- given x = 0 and h > 0 (Positive value), absolute value is redundant here
    lim(h->0+) x+h-x / h
    lim(h->0+) h / h
    lim(h->0+) 1
    1

    Since LL /= RL as x->0
    The function is not differentiable at x = 0

    From this proof, I am unsure if I can conclude that at x = 0, the function is not differentiable because for the interval x < 0, f(x) has a negative slope and for the interval x > 0, f(x) has a positive slope.

    If the above statement is valid, then I do not see why any function is differentiable at the local max/min (ex. Quadratic Function at the vertex)?
    If the above statement is invalid, then why is the absolute value function f(x) = abs(x) indifferentiable at x = 0?


    Thanks in advance and your time and effort is much appreciated!!
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  2. #2
    Senior Member jakncoke's Avatar
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    Re: Intro to Calculus - Why are quadratic function differentiable at the vertex?

    Its not so much that for x < 0, f(x) has a negative slope and for x > 0 f(x) has positive slope, its more the fact that as you approach 0, the negative slope and positive slopes dont go to the same limit

    if you take f(x) =  x^2 then  lim_{h \to 0} \frac{f(x+h)-f(x)}{h} = lim_{h \to 0} \frac{x^2 + h^2 + 2xh - x^2 }{h} =  lim_{h \to 0} \frac{h(h + 2x)}{h} =  lim_{h \to 0} h + 2x = 2x .

    No matter how you approach anypoint in the domain, this limit will always be 2x. So the slope is defined for any point in the domain even (x=0).
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    Re: Intro to Calculus - Why are quadratic function differentiable at the vertex?

    I'm not sure I understand your question, or, perhaps, why you are asking. What you have done is correct:
    The derivative of f(x), at x= x_0 is, by definition, \lim_{h\to 0}\frac{f(x_0+h)- f(x_0)}{h}.

    Yes, for f(x)= |x|, x_0= 0, we have to consider h< 0 and h> 0 separately because f(h)= h for h> 0, f(h)= -hf for h< 0.

    So, for h> 0, we have \lim_{h\to 0^+}\frac{|h|- 0}{h}= \lim_{h\to 0^+}\frac{h}{h}= 1 while, for h< 0,
    \lim_{h\to 0^-}\frac{|h|- 0}{h}= \lim_{h\to 0^-}\frac{-h}{h}= -1. Those two "one-sided" limits are not the same so the "limit" itself does not exist. |x| is not differentiable at x= 0.

    Now, for f(x)= x^2, x_0= 0, that becomes \lim_{h\to 0}\frac{(0+ h)^2- 0^2}{h}= \lim_{h\to 0}\frac{h^2}{h}= \lim_{h\to 0} h= 0
    Notice that the last "cancellation", \frac{h^2}{h}= h does NOT depend upon whether h is positive or negative, unlike the case with \frac{|h|}{h} which 1 or negative 1 depending upon whether h is positive or negative. Here, the "one sided limits", both "from above" and "from below", are 0. So the limit itself exists and is 0.
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    Re: Intro to Calculus - Why are quadratic function differentiable at the vertex?

    Quote Originally Posted by HallsofIvy View Post
    I'm not sure I understand your question, or, perhaps, why you are asking. What you have done is correct:
    The derivative of f(x), at x= x_0 is, by definition, \lim_{h\to 0}\frac{f(x_0+h)- f(x_0)}{h}.

    Yes, for f(x)= |x|, x_0= 0, we have to consider h< 0 and h> 0 separately because f(h)= h for h> 0, f(h)= -hf for h< 0.

    So, for h> 0, we have \lim_{h\to 0^+}\frac{|h|- 0}{h}= \lim_{h\to 0^+}\frac{h}{h}= 1 while, for h< 0,
    \lim_{h\to 0^-}\frac{|h|- 0}{h}= \lim_{h\to 0^-}\frac{-h}{h}= -1. Those two "one-sided" limits are not the same so the "limit" itself does not exist. |x| is not differentiable at x= 0.

    Now, for f(x)= x^2, x_0= 0, that becomes \lim_{h\to 0}\frac{(0+ h)^2- 0^2}{h}= \lim_{h\to 0}\frac{h^2}{h}= \lim_{h\to 0} h= 0
    Notice that the last "cancellation", \frac{h^2}{h}= h does NOT depend upon whether h is positive or negative, unlike the case with \frac{|h|}{h} which 1 or negative 1 depending upon whether h is positive or negative. Here, the "one sided limits", both "from above" and "from below", are 0. So the limit itself exists and is 0.
    Would you mind showing me that left and right limit are equal as x->0 for f(x) = x^2?
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    Senior Member jakncoke's Avatar
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    Re: Intro to Calculus - Why are quadratic function differentiable at the vertex?

    Quote Originally Posted by reventon View Post
    Would you mind showing me that left and right limit are equal as x->0 for f(x) = x^2?
    as h-->0, not x-->0

    We evaluate the lim as h-->0 for f(x)= x^2 at point x =0.

    if you take f(x) =  x^2 then  lim_{h \to 0} \frac{f(0 + h)-f(0)}{h} = lim_{h \to 0} \frac{h^2  }{h} = lim_{h \to 0} h= 0.
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    Re: Intro to Calculus - Why are quadratic function differentiable at the vertex?

    Quote Originally Posted by jakncoke View Post
    as h-->0, not x-->0

    We evaluate the lim as h-->0 for f(x)= x^2 at point x =0.

    if you take f(x) =  x^2 then  lim_{h \to 0} \frac{f(0 + h)-f(0)}{h} = lim_{h \to 0} \frac{h^2  }{h} = lim_{h \to 0} h= 0.
    My apologies, typo, would you mind showing me the left limit h-> 0- and right limit h-> 0+ and that they are equal for the function f(x) = x^2
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    Re: Intro to Calculus - Why are quadratic function differentiable at the vertex?

    I'm gonna denote h^{-} denote  h \to 0 approaching from the left of 0.
    and h^{+} denote  h \to 0 approaching from the right of 0.

    Now  lim_{h^{-} \to 0} \frac{f(0 + h^{-}) - f(0)}{h^{-}} = \frac{h^{-}*h^{-}}{h^{-}} = h^{-} which goes to zero

     lim_{h^{+} \to 0} \frac{f(0 + h^{+}) - f(0)}{h^{+}} = \frac{h^{+}*h^{+}}{h^{+}} = h^{+} which goes to zero
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    Re: Intro to Calculus - Why are quadratic function differentiable at the vertex?

    Quote Originally Posted by jakncoke View Post
    I'm gonna denote h^{-} denote  h \to 0 approaching from the left of 0.
    and h^{+} denote  h \to 0 approaching from the right of 0.

    Now  lim_{h^{-} \to 0} \frac{f(0 + h^{-}) - f(0)}{h^{-}} = \frac{h^{-}*h^{-}}{h^{-}} = h^{-} which goes to zero

     lim_{h^{+} \to 0} \frac{f(0 + h^{+}) - f(0)}{h^{+}} = \frac{h^{+}*h^{+}}{h^{+}} = h^{+} which goes to zero
    So as long as the left limit and the right limit are approaching the same value, they are considered equal and a limit at 0 (in this case) would exist?
    Because when I was doing this proof, I did not understand how 0+ would actually EQUAL to 0-, since they are only APPROACHING 0
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    Re: Intro to Calculus - Why are quadratic function differentiable at the vertex?

    Quote Originally Posted by reventon View Post
    So as long as the left limit and the right limit are approaching the same value, they are considered equal and a limit at 0 (in this case) would exist?
    Yes if the left and right limits approach the same value value, then f'(0) exists.

    Because when I was doing this proof, I did not understand how 0+ would actually EQUAL to 0-, since they are only APPROACHING 0
    I'm assuming 0^{+} means a sequence approaching 0 from the right and 0^{-} a sequence approaching 0 from the left.
    Absolutely 0^{+} need not equal 0^{-}

    but their limits have to be equal so,
    Lim_{h^{-} \to 0} h^{-} = Lim_{h^{+} \to 0} h^{+}
    Last edited by jakncoke; February 18th 2013 at 02:44 PM.
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    Re: Intro to Calculus - Why are quadratic function differentiable at the vertex?

    Quote Originally Posted by reventon View Post
    So as long as the left limit and the right limit are approaching the same value, they are considered equal and a limit at 0 (in this case) would exist?
    Because when I was doing this proof, I did not understand how 0+ would actually EQUAL to 0-, since they are only APPROACHING 0
    Limits do no "approach" anything- limits are numbers. In \lim_{h\to 0^-} \frac{h^2- 0}{h}, h is "approaching 0", and, because, for h< 0, h^2/h= h, the limit is equal to (not "approaching") 0. Similarly for lim_{h\to 0^+}\frac{h^2- 0}{h}- h "approaches" 0 but the limit is equal to 0.
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    Re: Intro to Calculus - Why are quadratic function differentiable at the vertex?

    Understood. Thank You very much jakncoke and HallsofIvy! I appreciated it.
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