Anzahl der Wendepunkte/Extrempunkte ohne Berechnung bei zusammengesetzten Funktionen?

Wie könnte man ohne Berechnung begründen, dass der Graph von f drei Wendepunkte haben muss?

Die Funktion lautet: 𝑓(𝑥) = −𝑥 ∙ 𝑒^-(x)^2

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mihisu
11 months ago

Not at all. The graph does not have three turning points, but only one single…

mihisu
11 months ago
Reply to  POWER194

It’ll be difficult without invoice. At least without invoice, one could still quite clearly justify that it must be at least three turning points. It is difficult to arrive at exactly three turning points (without checking the invoice).

I’d try that, maybe…

  • Obviously f(x) converged for x against -∞ and for x against +∞ each against 0. [The behaviour of the exponential function prevails here in the infinite.]
  • For x < 0, the functional values are obviously positive, and for x > 0 the functional values are obviously negative. (The exponential function is always positive, which is why the sign is determined by the factor -x.)

So…

  • Coming from x→−∞, the graph extends asymptotically approached from the x axis (due to limit value 0) upwards from the x axis (due to the positive function values). In order to get away from the x-axis, the graph must first be left crumbled be. [In order to obtain a positive slope (because the function extends upwards into the positive functional value range) from gradient close to 0 (because of the asymptotic approach).]
  • Since the function then eventually assumes negative function values, the graph must fall again. In order to get from a positive slope to a negative slope, the function must be right curved so that the gradient can decrease.
  • In order that the graph can approach the function value 0 for x→∞ from negative function values, a positive gradient is required. In order to achieve a positive slope from a negative gradient, the function must left crumbled so that the gradient can increase.
  • In order for the graph to approach the x-axis asymptotically, the gradient must approach 0. The positive slope must thus approach 0, i.e. the slope must decrease, so that Legal curvature is present.

Thus, at least three turning points are required between the curvature changes…

left curvature, Point, right-wing, Point, left curvature, Point, right-wing

It is difficult, however, without further checking (by appropriate calculation) to justify adequately that no change of curvature has been overlooked. Therefore, the consequence of the reasons I have mentioned: There are “at least” 3 turning points.

Halbrecht
11 months ago
Reply to  POWER194

tja, bad

Halbrecht
11 months ago

It is interesting to see how it is possible to delimit the account of arguments.

What is bill and what is not.

ChrisGE1267
11 months ago

This is an uneven, nonlinear function – therefore, a turning point is already in the origin. At the same time, Exp(-x^2) for x -> +/- Infinitely decreases more than -x; therefore, the x-axis is asymptote of the graph for x -> +/- Infinite. For x -> + Infinitely, the graph approaches the x axis from below, for x -> – Infinitely from above. Therefore, two further turning points must be present which are symmetrical to the y-axis…

Halbrecht
11 months ago
Reply to  ChrisGE1267

It is interesting to see how it is possible to delimit the account of arguments.

What is bill and what is not?

Could we also argue in this way that a factor with x3 must arise in the second derivation?

ChrisGE1267
11 months ago
Reply to  Halbrecht

That’s not enough because the polynomial 3. Grades before the factor Exp(-x^2) does not necessarily have to have 3 real zeros…

ChrisGE1267
11 months ago

I think “invoice” is understood here to find the turning points about the zero points of the second derivation…

Halbrecht
11 months ago

You could talk about at least one WP.

But what is Rechnen?

Halbrecht
11 months ago

e^x2 resembles an upwardly opened parabola with a crown at (0/1)

-x : left branch (x < 0 ) is multiplied by PLUS and remains positive. Right branch, on the other hand, becomes negative.