Proof of $n^{th}$ derivative Test Proof needn't be a rigourous , but should give an insight of how $n^{th}$ derivative test (higher order derivative test) works as i know how to use it in application but i don't much understand it ,especially the inflexion point thing.
 A: Let $f$ be $C^k$ (i.e., differentiable $k$ times with continuous $k$th derivative; we can relax this condition a bit, but it's not really useful here) on a neighborhood of a point $x$ , which we'll assume without loss of generality is $0$. Then by Taylor's theorem, we have
$$f(x) = f(0) + f'(0)x + \frac{1}{2}f''(0)x + \cdots + \frac{1}{k!} f^{(k)}(0) x^k + o(x^k)$$
near $x = 0$. If the first $(k-1)$st derivatives vanish then we're left with
$$f(x) = \frac{1}{k!}f^{(k)}(0) x^k + o(x^k).$$
Near $0$, the term $\frac{1}{k!} f^{(k)}(0) x^k$ dominates the sum above. If $k$ is even, then $x^k$ has a minimum at $x = 0$, which turns into a minimum or maximum of $f$ depending on the sign of $f^{(k)}(0)$. If $k$ is odd, then $f$ behaves like $x^k$ near $0$, which has neither a minimum nor a maximum at $x = 0$.
A: For intuition, you can think about the Taylor series of a function $f$ near a point $x_0 \in \mathbb R$:
\begin{equation}
f(x) \approx f(x_0) + f'(x_0)(x - x_0) + \frac12 f''(x_0)(x - x_0)^2 +
\cdots + \frac{1}{n!} f^{(n)}(x_0)(x - x_0)^n.
\end{equation}
What happens if the first $n - 1$ derivatives of $f$ at $x_0$ are equal to $0$ (and $f^{(n)}(x_0) \neq 0$)?  Then those terms disappear, leaving us with
\begin{equation}
f(x) \approx f(x_0) + \frac{1}{n!} f^{(n)}(x_0)(x - x_0)^n. \tag{$\spadesuit$}
\end{equation}
And this approximation is good when $x$ is close to $x_0$.
We can easily visualize the graph of the right hand side of $(\spadesuit)$. Doing so suggests that if $n$ is even, then $f$ has a local extremum at $x_0$, and if $n$ is odd then $f$ has an inflection point at $x_0$.
Here are some graphs to help with the visualization.
A: The following is based off patterns I have noticed and seems to make some sense (to me at least).
The derivative can be seen as having a special space; related to the original function, such that it's correlation to the original function corresponds to its 'n' . Think of it like using a ruler with a certain degree of accuracy. Each derivative can be seen as using more and more precise units of measurement. Let me try to explain.
The initial measure of change ${\frac {\rm d^0}{{\rm d^0}x}}f \left( x \right)$can be said to measure change at a scale of 
                       "identity space."
Because the projected 
                         identity space
 , space is equal to that of the initial space which the function lies in (ie. Euclidean Orthogonal...other coordinate systems also), its graph will be equal to the original function. 
The next derivative ${\frac {\rm d^1}{{\rm d^1}x}}f \left( x \right)$will set up a space, which resembles the function 
$a+x^0 = a+1$.This means that the "scale" of the derivative becomes sensitive to changes where the original function becomes a constant...(at 0). It takes this point of constant change and maps it to a zero (when 
$f(x)$changes from 
$a+1\mapsto a+x$.[form, imagine creating the function as you ride along it, and the type of degrees you would have to use to mimic the type of changes]). 
The derivative after that ${\frac {\rm d^2}{{\rm d^2}x}}f \left( x \right)$, sets up a space resembling 
$a+x^1=a+x$.Causing the derivative to become 'sensitive' to when the original function changes relative to the degree, meaning it maps points when 
$f(x)$ changes from a $
a+x\mapsto a+x^2
$[form]
This change is called the inflection point.  Note(Imagine plotting and function=
                           $ a+x$
,  onto an inflection point of another function. Notice that the function being analyzed crosses the function 
=$a+x$.)
Each time a derivative is taken it makes this "scale" more exact (while leaving out less exact "scales"), leading to ${\frac {\rm d^n}{{\rm d^n}x}}f \left( x \right)$ measuring at a scale of 
$a+x^{(n-1)}$;
Thus:
$ \frac{d ^{n }}{d ~x ^{
n }}f \left(x \right)=0~,\forall \Delta f \left(x \right)\mid
  \exists \Delta f \left(x \right):~a + x _{
{\it form} }^{n -1}~~~\mapsto ~a + x _{{\it form} }^{n }$
$\therefore\frac{d ^{n }}{d x ^{n }}f \left(x \right):\left\{\Delta 
f \left(x \right): \left( a+  x _{{\it form} }^{n 
-1}\mapsto a+ x _{{\it form} }^{n }, \right )\right\}\mapsto 0~
\mathop{\rm }
$ in initial coordinate system.
(Says the derivative maps points which it is "sensitive to to zero")
