Question: 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.

Answer: 
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](http://www.wolframalpha.com/input/?i=y%20=%20%281/3!%29%28x%20-%205%29%5E3) are [some](http://www.wolframalpha.com/input/?i=y%20=%20%281/4!%29%28x%20-%205%29%5E4) [graphs](http://www.wolframalpha.com/input/?i=y%20=%20-1/%284!%29%28x%20-%205%29%5E4) to help with the visualization.
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 $x = 0$.