1. can someone please explain the significance of the second derivative in regards to interpreting the shape of the function graphed? For the function f(x)=(20x^2-20)/(x^3+x^2) I got f'(x) = (-20x^4 + 60x^2 + 40x)/(x^3 + x^2)^2, where i found x=0,2,-1...and a local minimum at x=0 and a local maximum at x=2...Lastly i got f''(x)=(-20x^6+80x^5-60x^4-80x^3-160x^2)/(x^3+x^2)^4...Can anyone confirm these to be right and/or tell me the significance of finding the second derivative and when to find it thanks!!![/img]  2.

3. The second derivative gives the concavity information of the function.

f''(x) > 0 implies the function is concave up ( i.e. this shape \/ ) at that point while f''(x) < 0 implies the function is concave down ( i.e. this shape /\ ) at the point ( x, f(x) ). If f''(x) = 0 we have a point of inflection, where the concavity changes from concave up to concave down (like what happens at the origin for the function f(x) = x<sup>3</sup>)  4. Ok i get all that, but what good does it do for me to find it and when should i know to do it...(Someone told me that u could just as easily use the first derivative to find this out)  5. If you're given the length and time on the x and y axis.

Derive it and you'll get the velocity for a given time.

Derive it again (second derivative) it gives you the acceleration  6. I get this but whats the point? When should i know to look for it?  7. Well a table of signs of the first derivative gives the same information, so its up to you.  8. Well, the only reason why you knew 2 was the maximum point and 0 was the minimum is because 2>0. Sometimes there are certain weird graphs where the maximum point has a lower value compared to the minimum point. Normally people who are very good in math can recognize it straight away, but as I am very stupid, I can't roughly tell the shape of a graph when you give me a function. So I would use the second derivative to determine which is the maximum point and which is the minimum point.  9. In physics, one use of the second derivative is when analyzing potential. By taking the second derivative with respect to a parameter (to be determined by the problem), gives the conditionally and unconditionally stable points.

To picture this, imagine a pendulum and let alpha be the angle from the stable points (i.e., hanging down and perfectly pointing up). Take the derivatives WRT alpha of the potential and you should find that the first derivative (where it equals zero) gives the stable points and the second derivative where < 0 and > 0 tells you where the stable points are conditionally and unconditionally stable, respectively. This becomes more clear if you picture the potential as a "terrain." Where you sit at the bottom of a "hill" (or a "valley") you are unconditionally stable and where you sit perfectly on top of a "hill" you are conditionally stable. Everywhere else is unstable.

This really becomes handy with more complicated configurations where the stable points are not so obvious.

Cheers,
william  Bookmarks
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