when bisection method doesnt work for finding roots is there an example of a function f(x) which is continuous on the interval [-b,b] and has a root in the interval (-b,b) but for which the Bisection method with
starting values of -b and b would not work for finding a root
 A: If you assume that the calculations for $f(x)$ are accurate enough (which is not always the case on a computer: I have come up against this exception multiple times), and if you assume that $f(-b)$ and $f(b)$ have opposite sign, then the Bisection method always works. That is why it is popular. In fact, the common proof of the Intermediate Value Theorem uses the Bisection method. Other root-finding methods such as Newton-Raphson, Secant method, or False Position are usually faster but are less certain.
The main way Bisection fails is if the root is a double root; i.e. the function keeps the same sign except for reaching zero at one point. In other words, $f(a)$ and $f(b)$ have the same sign at each step. Then it is not clear which half of the interval to take at each step. In this case, a method for finding the minimum or maximum is better. But of course, that is often unclear when beginning the problem.
Perhaps I should explain how the inaccurate calculation of $f(x)$ can ruin the Bisection method. There is the obvious problem of the computer not correctly comparing the values of $f$ at two points. But the computer calculation can also make $f(x)$ no longer continuous. The computer may see many sub-intervals where $f(x)$ changes sign where $f(x)$ does not actually do that. The computer routine then may zoom in on the wrong spot for $x$.
ADDED:
I should answer a point raised by John Hughes. The OP asked if the bisection method "would not work for finding a root". We should clarify that the purpose of the bisection method, as with any other iteration method for finding real roots, is not to get the exact root. Rather, it is to find a "sufficiently small" interval that definitely contains the root. Given the two assumptions I give at the top of this answer, the bisection method is guaranteed to work in that sense.
A: Consider the sequence of points
$$
x_1 = -1, x_2 =  1, x_3 = 0, x_4 = \frac{1}{2}, x_5 = \frac{1}{4}, x_6 = \frac{3}{8}, \ldots
$$
in which each number is the average of the two previous numbers. After the first two numbers, the denominator of the $i$th number is $d_i = 2^{i-3}$ (with indices starting at $i = 1$).  
Now define a function $f$ with 
$$
f(-1) = -1\\
f(1) = 1 \\
f(0) = -\frac{1}{2} \\
f(\frac{1}{2}) = \frac{1}{4} 
$$
and so on, with $f(x_i) = (-1)^i \frac{1}{2^{i-2}}$. 
Extend this function, piecewise linearly, to a continuous function $g$ on $[-1, 1]$ (i.e., connect the dots). At the limit $s$ of the $x_i$s, define $g(s) = 0$. 
Evidently $g(x_i)$ and $g(x_{i+1}$ have opposite signs, so bisection will proceed for another step. But it will never terminal, so in this sense, "bisection" will not find at zero of the function $g$, although it will get arbitrarily close to one (namely $s$). 
You cannot hope that bisection will have a limit point $u$ at which, in an arbitrarily small neighborhood of $u$, your function $h$ takes on "large" values, for such an $h$ would fail to be continuous at $u$. 
