Justifying existence with the IVT | Code Scholars Tutoring

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\in C[a,b],\quad f(a)<N<f(b)\ \Longrightarrow\ \exists c\in(a,b):f(c)=N

Worked example

Prove $\displaystyle x^3+x-1=0$ has a solution in $\displaystyle (0,1)$ .

The polynomial

\displaystyle f(x)=x^3+x-1

is continuous on

\displaystyle [0,1]

. Since

\displaystyle f(0)=-1

and

\displaystyle f(1)=1

, zero lies between the endpoint values. The IVT guarantees a point

\displaystyle c\in(0,1)

with

\displaystyle f(c)=0

.

Try it yourself

Does a sign change for $\displaystyle f(x)=1/x$ on $\displaystyle [-1,1]$ prove a root?

Show solution and reasoning

No. The function is undefined at $\displaystyle 0$ , so it is not continuous on the interval. Indeed $\displaystyle 1/x$ is never zero.