Question

1. classify the following statement as true or false. explain.

if ( f(x) ) is continuous everywhere and ( f(a), f(b)>0 ), then there is no root of ( f(x) ) in the interval
(a, b).

Explanation:

Step1: Recall Intermediate Value Theorem

The Intermediate Value Theorem states that if \(y = f(x)\) is continuous on a closed interval \([a,b]\), and \(k\) is a number between \(f(a)\) and \(f(b)\), then there exists at least one number \(c\in(a,b)\) such that \(f(c)=k\). A root of \(f(x)\) is a value \(x = c\) where \(f(c)=0\).

Step2: Consider a counter - example

Let \(f(x)=(x - \frac{a + b}{2})^2+1\). This function is continuous everywhere (since it is a polynomial). Let \(a = 0\) and \(b = 2\). Then \(f(0)=(0 - 1)^2+1=2>0\) and \(f(2)=(2 - 1)^2+1=2>0\). But if we consider \(f(x)=x(x - 2)+1=x^{2}-2x + 1=(x - 1)^2\), \(f(0)=1>0\), \(f(2)=1>0\), and \(f(1)=0\) (a root in the interval \([0,2]\)).

Answer:

False.