Derivatives on HP 42S

[Thomas Klemm]

(08-20-2018 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

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Since I used Free42 instead of the HP-42S it could very well be that the results don't agree.