Can a point of local minimum be a point of inflection? 
I am having little confusion regarding this graph , as we see that just before close to zero and to the right of zero the function changes its sign from negative to positive so then how come $x=0$ is a point of inflection. According to me $x=0$ should be a point of local minimum here .
 A: The first derivative of the function must take different signed values at two sides of the local minimum point. 
This is not satisfied in the example above. Here, $(0,0)$ is an inflection point because $f''$ changes its sign at $x=0$.
A: Let's try to clear things up.

At the graph you are given the graph of the function $f(x)=x^3$. It is apparently differentiable on $\mathbb{R}$ as polynomial with a derivative of $f'(x)=3x^2 \geq 0$. Note that $f'(x)=0 \Leftrightarrow x=0$. One could speculate here that $x=0$ is an extrema value for $f$. No, this is not the case because the derivative does not change sign around $0$.
Hence $f$ is stricly increasing on $\mathbb{R}$. Now $f'$ is again differentiable as a polynomial. Differentiating once more we get that $f''(x)=6x$. Now, we note that $f''(x)=0 \Leftrightarrow 6x =0 \Leftrightarrow x =0$.
We also see that around zero the second derivative changes sign , meaning that $f$ has an inflection point at $x=0$ according to a known theorem
Regarding your question if a function has an extrema value at $x_0$ then at that value cannot have an inflection point and vice versa.
Proof
I leave the details up to you, but you may begin from $f''$ and study the monotony of $f$. You'll also use the process of elimination. Also you will use Fermat's theorem since $x_0$ is an internal point. 
P.S 1: If for a function $f$ holds that $f'(x_0)=0$ then $x_0$ is not necessarily an extrema value of $f$. See $f(x)=x^3$.
P.S 2: Same holds if $f''(x_0)=0$ then it is not necessary an inflection point.
Hope this clears things up a bit.
