Q. 3 - Calculus

Question

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

Step-by-Step Solution

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 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

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