Second derivative criteria for maxima and minima. Why is $\frac{d^2y}{dx^2} > 0$ for a minimum point and $\frac{d^2y}{dx^2} <0$ for a maximum? Also why does the second derivative not provide a reliable nature of the point of inflection?
Sorry I searched around and couldn't find any results.
 A: Intuitively: if you go down and then up, then your gradient must be increasing (it's negative going down, then positive going up again). Since $\dfrac{d^2y}{dx^2}$ is the rate of change of the gradient, this means that it must be positive at a minimum value.
As for points of inflection, if you have one then your second derivative is zero. But it's possible that your second derivative can be zero without having a point of inflection, like what happens with the graph of $y=x^4$ at $x=0$. Why? Because the gradient might itself have a point of inflection!
A: It is not clear to me whether the OP is looking for a formal statement of a result and a proof or not.  But s/he might be, and if not the OP than someone else probably will be.
The result you have in mind is:

Theorem (Second Derivative Test): Let $f$ be a real-valued function defined and twice differentiable on an interval which contains a real number $a$ as an interior point.  Suppose further that $f'(a) = 0$.  Then:
(i) If $f''(a) > 0$, then $f$ has a local minimum at $a$.
(ii) If $f''(a) < 0$, then $f$ has a local maximum at $a$.

The proof can be found, for instance, in $\S$ 5.5.3 of these notes.  But let me say something about it here.  It suffices to treat the case in which $f''(a) > 0$; otherwise replace $f$ with $-f$.  We observe that since $f''(a) = (f')'(a) > 0$, the derivative $f'$ is increasing through $a$.  Since it is zero at $a$, there is some small interval $[a-\delta,a+\delta]$ around $a$ such that in this interval and to the left of $a$, $f'$ is negative, and in this interval and to the right of $a$, $f'$ is positive.  Using the fact that a function which has a derivative of constant sign on an interval is monotone there, we find that $f$ is decreasing on $[a-\delta,a]$ and increasing on $[a,a+\delta]$.  This means it has a local minimum at $a$.
One of the merits of seeing the proof of a result is that it helps you understand the result's limitations and the necessity of the hypotheses (if all the hypotheses are indeed necessary; for these basic calculus facts you can be pretty confident that time would have eroded away any truly extraneous hypotheses, although if you're a sufficiently serious student of analysis you'll find that many if not most of the results of basic calculus could be technically strengthened in one way or another).  Here you ask what goes wrong if $f''(a) = 0$.  Well, everything: we no longer know that the graph of the derivative is either turning upward at $a$ or turning downward at $a$.  In other words the "sign analysis" of $f'$ need not take either of the following simple forms:
$----- 0 +++++ $ when $f''(a) > 0$
or
$+++++ 0 -----$ when $f''(a) < 0$.
Without that information we're helpless to argue the existence of a local minimum or local maximum.  Really helpless, in the sense that simple examples like $f(x) = x^3$ at $a = 0$ show that such things need not exist!
Finally, this is not the point, but your question leads me to suspect that you think "inflection point" means $f''(a) = 0$.  It doesn't: it implies $f''(a)  = 0$ but is stronger [in just the same way that "local extremum" implies $f'(a) = 0$ but is stronger]: we need the sign of $f''$ to change through the point $a$ just as $f'$ does in the two diagrams above.    An inflection point has a geometric meaning: it signals a change in the concavity.  Thus for instance $f(x) = x^4$ has $f''(0) = 0$ but does not have an inflection point at $0$, and indeed it is convex (or as we say in freshman calculus, "concave up") on its entire domain.
A: The criteria is the other way around. If $f'(a)=0$ and $f''(a)>0$ ($f''(a)<0$) then $f$ has a minimum (maximum) at $a$. Geometrically, if $f''(a)<0$ then the function will curve upwards, like $1-x^2$ on the origin, and if $f''(a)>0$ then the function will curve downwards, like $x^2$ on the origin. 
A: In addition to Peter's answer, why don't you sketch a third degree function with a maximum and a minimum? Then again trace the curve (you may want to draw a couple of tangents) as you read Peter's answer. This question becomes much more obvious with a graph, at least that's the experience with my students...
A: The second derivative gives the rate of change of the first derivative. Recall that the first derivative tells you the gradient of the graph.
Draw yourself a picture of the parabola $y=x^2$. This has a (global) minimum at the $x=0$. Draw yourself a few tangent lines, e.g. $x=-2,-1,0,1,2$.
At $x=-2$, the gradient is $-4$, at $x=-1$ the gradient is $-2$, at $x=0$ the gradient is $0$, at $x=1$ the gradient is $2$ and at $x=2$ the gradient is $4$.
The gradient is increasing as we move from left to right. That means that the derivative is increasing. That means that the derivative of the derivative is positive.
In the case of a local/global maximum you will see that the gradient decreases as you pass through the maximum. That means the derivative decreases, which means the derivative of the derivative is negative.
If you have both the derivative (i.e. the gradient) being zero and the second derivative being zero then you have an inflexion or something worse. These get a bit complicated to classify.
A: The most interesting part of the question is why the method fails for some functions.  It's pointed out that it doesn't tell us that x3 has an inflection point at 0 or give us any information about x4, because for those functions f''(0) = 0.
The reason the second derivative test works (where it works), is that we are comparing our given function f to a quadratic function q through the same point and with the same first and second derivatives at that point.  Another way to say this is that we are expanding f into a Taylor's series up to f''.
Now to the extent f is comparable to a quadratic at our point, the second derivative test tells you the story.  But if the f, f', and f'' are all zero at a point,then what we've seen is that the function looks like a cubic or higher order polynomial.  Our test that is based on a close quadratic approximation doesn't work because there isn't any close quadratic approximation.
