Derivatives on HP 42S

08212018, 01:35 AM
Post: #8




RE: Derivatives on HP 42S
(08202018 11:54 PM)Albert Chan Wrote: Why would search for slope of 0 get the maximum ? Cf. Maxima and minima We search for a critical point in the interval [14.13,14.14], that is x where f'(x) = 0. Quote:Should it get the minimum, at x = 14.13 ? The function is somewhat pathological in that it's close to 0 most of the time and only 1 at the maximum which is where \(\sin(x)=1\). That means \(x=\frac{\pi}{2}+k\cdot2\pi\) for \(k\in\mathbb{Z}\). It's not defined (or then takes complex values) where \(\sin(x)<0\). Thus the minimal value is 0 when \(x=k\cdot\pi\) for \(k\in\mathbb{Z}\). So the minimal values in that section are \((4\pi, 0)\) and \((5\pi, 0)\) and the maximal value is \((\frac{9}{2}\pi, 1)\). The flatness of the function makes it hard to numerically find the stationary points. Cheers Thomas Since I used Free42 instead of the HP42S it could very well be that the results don't agree. 

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