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burhan101

  • one year ago

Solve using first derivative principle

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  1. burhan101
    • one year ago
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    \[\huge y=\sqrt{2x-1}\]

  2. burhan101
    • one year ago
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    |dw:1371162568007:dw|

  3. phi
    • one year ago
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    try multiplying top and bottom by the conjugate ( sq rt + sq rt) so you get a difference of squares up top that can be simplified.

  4. burhan101
    • one year ago
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    conjugate being : \[\huge \frac{ \sqrt{2x+h-1} + \sqrt{2x-1}}{ \sqrt{2x+h-1} + \sqrt{2x-1} }\] ?

  5. burhan101
    • one year ago
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    i forgot the "2h"

  6. phi
    • one year ago
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    yes. the idea is (a-b)*(a+b) = a^2 - b^2 which will get rid of the square roots up top

  7. burhan101
    • one year ago
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    {2x+2h-1}

  8. burhan101
    • one year ago
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    thank you @phi :)

  9. phi
    • one year ago
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    yw

  10. burhan101
    • one year ago
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    Im getting |dw:1371163702503:dw|

  11. phi
    • one year ago
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    yes, but what you want to do is \[\lim_{h \rightarrow 0}\frac{f(x+h) - f(x)}{h} \] the next step is let h->0 so the bottom becomes \[ \sqrt{2x-1} + \sqrt{2x-1} \]

  12. burhan101
    • one year ago
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    |dw:1371164005685:dw|

  13. burhan101
    • one year ago
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    would that be right

  14. phi
    • one year ago
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    yes if we use calculus \[ \frac{d}{dx} (2x -1)^{\frac{1}{2}} = \frac{1}{2} (2x -1)^{-\frac{1}{2}} 2\\= (2x -1)^{-\frac{1}{2}}= \frac{1}{\sqrt{2x-1} }\]

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