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.
question help: video written example message instructor

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 \(f(c)=k\).

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 \(f(5)=-4\), there must be at least one value \(x_1\in(0,5)\) such that \(f(x_1)=0\). Also, \(f(3) = 8\) and \(f(5)=-4\), and since \(0\) is between \(f(3)\) and \(f(5)\), there must be at least one value \(x_2\in(3,5)\) such that \(f(x_2)=0\).

Answer:

2