# Whether the statement, “If f ( 1 ) &gt; 0 and f ( 3 ) &lt; 0 , then there exists a number c between 1 and 3 such that f ( c ) = 0 ” is true or false. ### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805 ### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

#### Solutions

Chapter 2, Problem 12RQ
To determine

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

Expert Solution

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(a) and f(b), where f(a)f(b). Then there exists a number c in (a, b) such that f(c)=N.

Reason:

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

Suppose f(x)={x2if1x<22ifx=2x2if 2<x3 and take a=1, b=3 and N=0.

Here, f(1)=1<0 and f(3)=1>0.

This implies that, f(1)<N<f(3) whenever f(x) is continuous. But f(x) 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(x) is an oscillating function, then it can be possible to occurs the root in the subinterval of (1,3).

At the interval (1, 2):

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

At the interval (2, 3):

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

At the point 2:

Suppose the function f(x) has a one root at x=2, then f(x) must be zero. But here, f(2)=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(c)=0.

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

Therefore, the statement is false.

### Have a homework question?

Subscribe to bartleby learn! Ask subject matter experts 30 homework questions each month. Plus, you’ll have access to millions of step-by-step textbook answers!