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Difference Quotient

  1. Jun 5, 2007 #1
    The question is "find and simplify the difference quotient."

    Given function

    f(x)=sq root of x

    So what I did is insert (x+h) under the radical & got

    sq root of (x+h), then I subtracted the sq root of x (original function)

    My answer was sq root [(x+h) - sq root (x)] / h

    The answer in the back of the book is 1 / [sq root (x+h) + sq root (x)]

    I'm not understanding where I'm going wrong, the other problems didn't give me a problem, just this one.
     
  2. jcsd
  3. Jun 5, 2007 #2
    multiply numerator and denominator by [sqrt(x+h) + sqrt(x)] ...

    [sqrt(x+h) - sqrt(x)]/h * [sqrt(x+h) + sqrt(x)]/[sqrt(x+h) + sqrt(x)] =

    [(x+h) - x]/[h[sqrt(x+h) + sqrt(x)]] =

    h/[h[sqrt(x+h) + sqrt(x)]] =

    1/[sqrt(x+h) + sqrt(x)]
     
  4. Jun 5, 2007 #3
    Ok, thanks!

    So is there a reason why its listed in that form instead the other? I think that is what I'm not understanding.
     
  5. Jun 6, 2007 #4
    you'll find out why when you have to take the limit of the difference quotient as h -> 0 ... it's a calculus concept.
     
  6. Jun 11, 2007 #5

    HallsofIvy

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    Not sure if this is a typo of a major misunderstanding. What you said you did is correct but you should have [sq root(x+h)- sq root(x)]/h. Do you see the difference? That is:
    [tex]\frac{\sqrt{x+h}- \sqrt{x}}{h}[/tex]
    where yours is
    [tex]\frac{sqrt{x+h- sqrt{x}}}{h}[/tex]

    As was said before, rationalize the numerator: multiply numerator and denominator by sq root(x+h)+ sqrt(x).
    [tex]\frac{\sqrt{x+h}-sqrt{x}}{h}\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+y}+\sqrt{x}}[/tex]
     
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