Unit 1 · Topics 1.16. Original AP Calculus AB notes, worked example, and self-check.
Understand the idea
A function continuous on a closed interval takes every value between its endpoint outputs. To prove a root exists, establish continuity and opposite endpoint signs. The theorem guarantees at least one solution, not uniqueness or its location. Graphs and tables help choose an interval, but the written argument must state why continuity holds.
f∈C[a,b],f(a)<N<f(b)⟹∃c∈(a,b):f(c)=N
Worked example
Prove x3+x−1=0 has a solution in (0,1).
The polynomial f(x)=x3+x−1 is continuous on [0,1]. Since f(0)=−1 and f(1)=1, zero lies between the endpoint values. The IVT guarantees a point c∈(0,1) with f(c)=0.
Try it yourself
Does a sign change for f(x)=1/x on [−1,1] prove a root?
Show solution and reasoning
No. The function is undefined at 0, so it is not continuous on the interval. Indeed 1/x is never zero.
