Chapter 2, Problem 12RQ

To determine

Whether the statement, “If $f\left(1\right)>0$ and $f\left(3\right)<0$, then there exists a number c between 1 and 3 such that $f\left(c\right)=0$” is true or false.

Answer to Problem 12RQ

The statement is false.

Explanation of Solution

Theorem used: The Intermediate Value Theorem

Suppose that if f is continuous on the closed interval [a, b] and let N be any number between $f\left(a\right)$ and $f\left(b\right)$, where $f\left(a\right)\ne f\left(b\right)$. Then there exists a number c in (a, b) such that $f\left(c\right)=N$.

Reason:

The given statement is false since the below example disproves the given statement.

Suppose $f\left(x\right)=\{\begin{array}{l}x-2\text{if}1\le x<2\\ -2\text{if}x=2\\ x-2\text{if }2<x\le 3\end{array}$ and take $a=1$, $b=3$ and $N=0$.

Here, $f\left(1\right)=-1<0$ and $f\left(3\right)=1>0$.

This implies that, $f\left(1\right)<N<f\left(3\right)$ whenever $f\left(x\right)$ is continuous. But $f\left(x\right)$ has a hole at $x=2$, then the function is discontinuous at $x=2$. So the Intermediate theorem cannot be used here.

Suppose the function $f\left(x\right)$ is an oscillating function, then it can be possible to occurs the root in the subinterval of $\left(1,3\right)$.

At the interval (1, 2):

Suppose the function $f\left(x\right)$ has at least one root in the interval (1, 2), then $f\left(x\right)$ must be zero. But here, taking any c in the interval (1, 2), then $f\left(x\right)<0$. So there is no root in the interval (1, 2).

At the interval (2, 3):

Suppose the function $f\left(x\right)$ has at least one root in the interval (2, 3), then $f\left(x\right)$ must be zero. But here, taking any c in the interval (2, 3), then $f\left(x\right)>0$. So there is no root in the interval (2, 3).

At the point $2$:

Suppose the function $f\left(x\right)$ has a one root at $x=2$, then $f\left(x\right)$ must be zero. But here, $f\left(2\right)=-2$. So there is no root in at the point 2.

It is clear that from the above subintervals, there is no c between 1 and 3 for which $f\left(c\right)=0$.

Even though $f\left(1\right)>0$ and $f\left(3\right)<0$, there is no c between 1 and 3 for which $f\left(c\right)=0$.

Therefore, the statement is false.