How to find the Global maxima and minima of the function on an open interval? How can we confirm that the local extrema we have found by putting the derivative of the function equal to zero are absolute extrema or not ? if they are not absolute extrema then how can we find the absolute extrema ?
 A: If your problem is unconstrained, and assuming the objective function is twice differentiable, if the Hessian of the objective function at the stationary point (i.e., a point at which gradient of objective function = vector of zeros) is positive semi-definite, the stationary point is a local minimum. If the Hessian is positive definite, the stationary point is a strict local minimum.  If the objective function is convex, i.e., the Hessian is positive semi-definite everywhere, the stationary point is a global minimum, and if the Hessian is positive definite everywhere, the stationary point is the unique global minimum.
Local and global maximum are determined the same as above, but substitute concave, negative semi-definite, and negative definite for convex, positive semi-definite, and positive definite.
If the Hessian at the stationary point is indefinite, i.e., has at least one positive eigenvalue and at least one negative eigenvalue, the stationary point is a saddlepoint, which is neither a local minimum nor a local maximum, but means that in some direction(s) the point is a local minimum with other variables fixed, and in other direction(s) is a local maximum with other variables fixed; the directions are the eigenvectors corresponding respectively to the positive and negative eigenvalues of the Hessian. 
I have provided sufficient conditions, but not necessary conditions. In general, the problem of finding the global minimum of a non-convex function or a global maximum of a non-concave function is difficult.  However, if you can verify the convexity or concavity of the objective function, then matters simplify as I have described them.
In the event that your optimization problem is constrained, matters get more complicated - see my answer in How can I determine the type of the critical point?
You really need to clarify what you mean by open interval.  If you mean -infinity to +infinity (i.e., open with respect to extended reals), then the information provided for the problem being unconstrained applies.  If you mean that one or more variables have a finite lower or upper bound, then the optimization problem is a (very simple case of) constrained optimization. The global minimum or maximum may occur at a point in which one or more of the variables is at a lower or upper bound; in such case, the gradient will not generally be zero at the optimal solution, not even for a local minimum or maximum. Indeed, my answer linked to above addresses that. If there are only bound constraints (what I just described) and no general linear or nonlinear constraints, then projection of the gradient and Hessian into the nullspace of the Jacobian of active constraints amounts to removing those variables on a bound from the gradient and Hessian (for the Hessian, that means deleting the row and column corresponding to that variable). So in general, only the gradient components for variables not on a bound will be zero at a local or global minimum or maximum.  Then there is still the matter of the Lagrange multiplier for each active bound constraint having the correct sign.  For minimization, this boils down to the gradient component for a variable having to be positive if on its lower bound, and negative if on its upper bound, and the opposite for maximization. 
A: Since the domain of your function is an open interval $J$ global extrema need not exist. The function might be unbounded, or might converge monotonically to finite limits at the ends of the interval. The example
$$f(x):=\cos x\>\tanh x\qquad(-\infty<x<\infty)$$
shows that we may have an infinity of local extrema, while the "real" $\inf$ and $\sup$ are eluding. Etcetera.
In the following case a final diagnosis is possible: If $f$ has finite limits at the endpoints $a$, $b$ of $J$, whereby $a=-\infty$ or $b=\infty$ is allowed, then we essentially have a continuous function on the compact interval $[a,b]$ which is differentiable in the interior. If there is a finite set $\{c_1,\ldots,c_m\}$ of critical points in $\>]a,b[\>$ then
$$\max_{a<x<b}f(x)=\max_{1\leq i\leq m}f(c_i)\ ,\tag{1}$$
under the condition that none of the end limits is greater than the $\max$ on the RHS of $(1)$.
