Question

suppose that $f(x)$ is a continuous function with the following values:

$x$	0	1	2	3	4	5
$f(x)$	2	5	10	8	9	-4

what is the minimum number of solutions to $f(x)=0$ in the interval $0leq xleq5$
there are at least solutions.

Explanation:

Step1: Apply 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 the interval \((a,b)\) such that \(f(c)=k\). We want to find where \(f(x) = 0\).

Step2: Check sign - changes

We have \(f(0)=2\) and \(f(5)= - 4\). Since \(f(x)\) is continuous on \([0,5]\) and \(0\) is between \(f(0) = 2\) and \(f(5)=-4\), by the Intermediate - Value Theorem, there must be at least one value \(c\in(0,5)\) such that \(f(c) = 0\).

Answer:

1