Turning point / saddle point?
Can anyone confirm if this is true?
That, for example, both at the inflection point and at the saddle point, if the third derivative is less than zero then there is a left curvature to a spelling?
Can anyone confirm if this is true?
That, for example, both at the inflection point and at the saddle point, if the third derivative is less than zero then there is a left curvature to a spelling?
Hello If you know what the graph looks like and also certain information about it, such as where the slope is 0 or where the function value is 0 at the first derivative and so on, how can you set up the function? Can someone explain this to me? Or is there a good video…
Is this correct for the a) x1x2 plane? and at b) parallel to the x1x3 plane?
Is my solution correct? If not, please add something. Thank you.
Hello folks, There are certain methods in numerics and the finite element method that allow numerical differentiation (derivation). As far as I know, these are called finite difference methods, or FDM. The actual question is much more complicated. So complicated that I don't even want to ask it. Instead, I'd like to discuss it privately…
Hello please help me quickly! Thank you, please formulate in a medium-length text thank you thank you
I need help with task 3 because I never know when and how to use which formula.
Yeah, that’s right.
That’s right. If the third derivation is less than zero, a high point is present in the first derivation (second derivation is zero), i.e. the slope increases first, then decreases—that is, the output function is first left, then curved right. Analogously in the other case, only then there is a right-left course (first derivation has low point, i.e. only falling slope, then rising). All of course with the condition that the curvature, i.e. the second derivative, is zero at the point
A saddle point is “bloß” a WP with horizontal tangent
.
Turning point at 2/3
f””(2/3) > 0
from right to left curve
.
.
and here the same Fkt with MAL -1
f””(2/3) is < 0