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Math Help - limit definition of a derivative?

  1. #1
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    limit definition of a derivative?

    consider the following piecewise defined function:

    f(x) = {x + 5 if x < 3} and {4*sqrt(x+1) if x >= 3}

    Using the limit definition of the derivative, and computing the limits from both sides, show that f(x) is differentiable at x=3.


    I have no clue what its asking, can someone help?
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by break View Post
    consider the following piecewise defined function:

    f(x) = {x = 5 if x < 3} and {4*sqrt(x+1) if x >= 3}

    Using the limit definition of the derivative, and computing the limits from both sides, show that f(x) is differentiable at x=3.


    I have no clue what its asking, can someone help?
    I'm not sure I do either. How can x=5\text{ if }x<3?
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  3. #3
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    Quote Originally Posted by VonNemo19 View Post
    I'm not sure I do either. How can x=5\text{ if }x<3?
    sorry, typo, was supposed to be a + not =
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  4. #4
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    Quote Originally Posted by break View Post
    sorry, typo, was supposed to be a + not =
    Note: this is the same as saying

    \lim_{x\to 3^{-1}} x + 5
    Saying the limit of this function as its approaches values less than 3 to 3, or as the limit as x approaches 3 from the left.

    \lim_{x\to 3^{+1}} 4\cdot \sqrt{x+1}
    Saying the limit of this function as its approaches values more than or equal to 3, or as the limit as x approaches 3 from the right.

    Now compute the limit definition of a derivative

    \lim_{h\to 0}\frac{f(x+h)-f(x)}{h^2}
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  5. #5
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by break View Post
    sorry, typo, was supposed to be a + not =
    You need to show that

    \lim_{x\to3^-}\frac{(x+5)-(3+5)}{x-3}=\lim_{x\to3^+}\frac{4\sqrt{x+1}-4\sqrt{3+1}}{x-3}
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  6. #6
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    Quote Originally Posted by VonNemo19 View Post
    You need to show that

    \lim_{x\to3^-}\frac{(x+5)-(3+5)}{x-3}=\lim_{x\to3^+}\frac{4\sqrt{x+1}-4\sqrt{3+1}}{x-3}

    uhhh once i simplify it down i get

    \lim_{x\to3^-}\frac{x-3}{x-3}=\lim_{x\to3^+}\frac{4\sqrt{x+1}-8}{x-3}

    do i plug in 3 for the x?
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  7. #7
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by break View Post
    uhhh once i simplify it down i get

    \lim_{x\to3^-}\frac{x-3}{x-3}=\lim_{x\to3^+}\frac{4\sqrt{x+1}-8}{x-3}

    do i plug in 3 for the x?
    Note that the left hand side simplifies to, \lim_{x\to3^-}1=1


    For the right hand side, multiply both numerator and denominator by the conjugate of the numerator ( \sqrt{x+1}+2) and see where that gets you.
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  8. #8
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    Quote Originally Posted by VonNemo19 View Post
    Note that the left hand side simplifies to, \lim_{x\to3^-}1=1


    For the right hand side, multiply both numerator and denominator by the conjugate of the numerator ( \sqrt{x+1}+2) and see where that gets you.
    where did you get ( \sqrt{x+1}+2) from?
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  9. #9
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by break View Post
    where did you get ( \sqrt{x+1}+2) from?
    I factored a four out of the numerator.
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