Inflection points are where f"(x) = 0 (derivitive of the derivitive of f(x)). Local minima or maxima are where f'(x) = 0.

EDIT: Actually, it depends what f(x) is. If it's a position-time function, then it's the second derivitive; a velocity-time function, then the first derivitive; an acceleration-time graph, and it's just f(x).

[edited by - Zipster on May 8, 2002 6:51:00 PM]

i know ive had a problem before that was based on noting that the you had an angle of 45 degrees when the first derivative was 1. but i cant fricken remember it off the top of my head.

i dont understand the car context but he used the buzz word "efficiency" and that screams second derivative = 0 type problem. comes up in econ probs alot.

Dean Harding has the essense of my problem.
Signifigance ? I dunno...I dreamed up the prob in calc class...
echeslacke also seems to grasp it.

Heres some functions:

y = x^2

y'' = 2x

y'''' = 2

There are no inflection points.

Minimum is at x = 0
''Bends'' would be at x = 1/2, x = -1/2
(there are 2 places where |f''(x)| = 1- you eval it at
-f''(x) and f''(x))

Now, ''bends'' may be inflection points.

Inflection pts are where f''''(x) = 0 or undef at x, and checking for concavity shows that concavity changes at x.

So in a sin/cos function type, you would run into inflections and bends at the same points, I''m pretty sure.

I''m amused by seeing I now have two pages on this topic.:D

~V''lion

Bugle4d

quote:

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Original post by Zipster
EDIT: Actually, it depends what f(x) is. If it''s a position-time function, then it''s the second derivitive; a velocity-time function, then the first derivitive; an acceleration-time graph, and it''s just f(x).

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A point of inflection on the graph of f(t) has nothing to do with what f(t) measures... only the shape of the graph. In each of your examples, your thinking of a point of inflection on the function f(t) and relating it to derivative graphs of df/dt and d²/dt².

I suspect what you meant was that taking the first derivative of a velocity-time graph and looking for its roots indicates the location of a point of inflection on the displacement time graph... and similarly, the roots of the acceleration-time graph indicate points of inflection on the displacement-time graph.

Cheers,

Timkin

Find local maxima and minima by setting f''(x)=0 and solving for x. Find points of inflection by setting f''''(x)=0 and solving for x. It''s that simple.

quote:

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I suspect what you meant was that taking the first derivative of a velocity-time graph and looking for its roots indicates the location of a point of inflection on the displacement time graph...

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Correct, I''m using displacement-time as a base f(x).