When using the trapezium rule to approximate a definite integral between a and b, the error is bounded by:

(1/12)(b-a)(h^2) * max(a<= x <= b)|f''(x)|

According to my notes, the way to find the max value of f''(x) is to differentiate again to find f'''(x) and find where this is zero (maximum), makes sense but what if the maximum isn't the actual maximum value f''(x) can take ? As in you could have a maximum turning point but then further along the x-axis the graph could go back up, higher than the value of the maximum (still within [a,b]) then f(b) could be the max(a<=x<=b)|f''(x)|. E.g. some graph of x^3

Thanks