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anonymous
 3 years ago
Solve using first derivative principle
anonymous
 3 years ago
Solve using first derivative principle

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\huge y=\sqrt{2x1}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1371162568007:dw

phi
 3 years ago
Best ResponseYou've already chosen the best response.1try multiplying top and bottom by the conjugate ( sq rt + sq rt) so you get a difference of squares up top that can be simplified.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0conjugate being : \[\huge \frac{ \sqrt{2x+h1} + \sqrt{2x1}}{ \sqrt{2x+h1} + \sqrt{2x1} }\] ?

phi
 3 years ago
Best ResponseYou've already chosen the best response.1yes. the idea is (ab)*(a+b) = a^2  b^2 which will get rid of the square roots up top

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Im getting dw:1371163702503:dw

phi
 3 years ago
Best ResponseYou've already chosen the best response.1yes, 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 \[ \sqrt{2x1} + \sqrt{2x1} \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1371164005685:dw

phi
 3 years ago
Best ResponseYou've already chosen the best response.1yes if we use calculus \[ \frac{d}{dx} (2x 1)^{\frac{1}{2}} = \frac{1}{2} (2x 1)^{\frac{1}{2}} 2\\= (2x 1)^{\frac{1}{2}}= \frac{1}{\sqrt{2x1} }\]
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