Question

State the Intermediate Value Theorem, and illustrate the theorem with a graph.

Short answer

$$\begin{gathered} \mathrm{The}\mathrm{meaning}\mathrm{behind}\mathrm{the}\mathrm{intermediate}\mathrm{value}\mathrm{theorem}\mathrm{is},\mathrm{when}\mathrm{graph}\mathrm{is}\mathrm{change}\mathrm{from} \\ \mathrm{positive}\mathrm{to}\mathrm{negative},\mathrm{it}\mathrm{must}\mathrm{cross}\mathrm{x}-\mathrm{axis}. \end{gathered}$$

Step-by-step solution

Step 1. Given information is:

If f is continuous on [a,b] and f(a) and f(b) are having opposite signs, then there exist at least one point c in [a,b] such that f(c)=0

Step 2. Graph and Theorem

$$\begin{gathered} \mathrm{Consider}\mathrm{the}\mathrm{above}\mathrm{graph},\mathrm{f}\left(\mathrm{a}\right)\mathrm{is}\mathrm{positive}\mathrm{and}\mathrm{f}\left(\mathrm{b}\right)\mathrm{is}\mathrm{negative},\mathrm{there}\mathrm{exist}\mathrm{a}\mathrm{point}\mathrm{c} \\ \mathrm{such}\mathrm{that}\mathrm{f}\left(\mathrm{c}\right)=0. \\ \mathrm{The}\mathrm{meaning}\mathrm{behind}\mathrm{the}\mathrm{intermediate}\mathrm{value}\mathrm{theorem}\mathrm{is},\mathrm{when}\mathrm{graph}\mathrm{is}\mathrm{change}\mathrm{from} \\ \mathrm{positive}\mathrm{to}\mathrm{negative},\mathrm{it}\mathrm{must}\mathrm{cross}\mathrm{x}-\mathrm{axis}. \end{gathered}$$