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Math Help - Using the definition of derivative...

  1. #1
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    Using the definition of derivative...

    This is going to be an easy one

    but I have to solve the square root of (2x+1) using the definition of limit.

    I set the problem up f(x) lim h-> 0 sq root (2x + h + 1) - sq root (2x + 1) / h then I multiplied by the conjugate.

    That gave me (2x + h + 1) - (2x + 1) / h (sq root 2x + h + 1) + (sq root 2x+1)

    I am not sure where to go from here. How do I simplify it? I know the answer I just don't see how to get there. Thanks!!
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  2. #2
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    Quote Originally Posted by KarlosK View Post
    This is going to be an easy one

    but I have to solve the square root of (2x+1) using the definition of limit.

    I set the problem up f(x) lim h-> 0 sq root (2x + h + 1) - sq root (2x + 1) / h then I multiplied by the conjugate.

    That gave me (2x + h + 1) - (2x + 1) / h (sq root 2x + h + 1) + (sq root 2x+1)

    I am not sure where to go from here. How do I simplify it? I know the answer I just don't see how to get there. Thanks!!
    \lim_{h \to 0} \, \frac{\sqrt{2(x+h)+1} - \sqrt{2x+1}}{h}

    \lim_{h \to 0} \, \frac{2(x+h)+1 - (2x+1)}{h \sqrt{2(x+h)+1} + \sqrt{2x+1}}

    combine like terms in the numerator ... one h in the numerator and denominator will divide out ... then determine the limit.
    Last edited by skeeter; March 17th 2010 at 05:38 PM.
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  3. #3
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    Quote Originally Posted by skeeter View Post
    \lim_{h \to 0} \, \frac{\sqrt{2(x+h)+1} - \sqrt{2x+1}}{h}

    \lim_{h \to 0} \, \frac{2(x+h)+1 - (2x+1)}{h \sqrt{2(x+h)+1} + \sqrt{2x+1}}

    combine like terms in the numerator ... the h's in the numerator and denominator will divide out ... then determine the limit.
    I guess what I am having trouble with is how to cancel out the square roots on the bottom of the fraction.
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  4. #4
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    Quote Originally Posted by KarlosK View Post
    I guess what I am having trouble with is how to cancel out the square roots on the bottom of the fraction.
    do not mess with the radicals in the denominator, they are part of the solution.

    follow my previous directions as given.
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    Quote Originally Posted by skeeter View Post
    do not mess with the radicals in the denominator, they are part of the solution.

    follow my previous directions as given.
    Ok well taking your advice... I wind up with

    2h / (bottom part)

    I guess you could cancel the H's and have

    h/ (bottom part)

    but the answer to the problem is 1/ sq root (2x + 1)
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  6. #6
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    Quote Originally Posted by KarlosK View Post
    Ok well taking your advice... I wind up with

    2h / (bottom part)

    I guess you could cancel the H's and have

    2/ (bottom part) ... correction

    but the answer to the problem is 1/ sq root (2x + 1)

    ... and that is the result
    ...
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