MATHEMATICS · Single correct · AP · Calculus AB · Limits and Continuity · Limits and Continuity
Problem
A function p is continuous on [2,7] with p(2)=5 and p(7)=−3. Which statement best justifies that the equation p(x)=0 has at least one solution in (2,7)?
(a) Because p is differentiable on (2,7), its derivative must equal zero for some x in (2,7), implying a root.
(b) Because p is continuous on [2,7] and changes sign between x=2 and x=7, a zero must occur in the interval.
(c) Because p attains an absolute minimum and maximum on [2,7], it must cross the x-axis within that interval.
(d) Because p(2)+p(7)=2, the average value theorem ensures p equals its average value at some point, guaranteeing a root.