A function h is continuous on the closed interval [a,b] and twice differentiable on the open interval (a,b). The function possesses exactly one critical number c with a<c<b, and it is given that h′(c)=0 and h′′(c)>0.
a) Explain, using the geometric interpretations of the first and second derivatives, why h must attain a local minimum at x=c.
b) Under what additional condition(s) on h can you conclude that the value of h at x=c is also the absolute minimum value of h on [a,b]? Justify your response.