Question

suppose (f) is continuous on ((-infty,infty)) and the following data is known: (f(-1)=2,f(0)=4,f(1)= - 1,f(2)=2,f(3)=-6). which of the following statements is true? (f(x) = 0) has exactly three solutions. (f(x)=0) has at most two solutions. (f(x)=0) has at least three solutions. (f(x)=0) has at most three solutions. (f(x)=0) has at exactly two solutions.

Explanation:

Step1: Apply Intermediate - Value Theorem

The Intermediate - Value Theorem states that if a function \(y = f(x)\) 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 know \(f(-1) = 2\) and \(f(0)=4\), \(f(1)= - 1\), \(f(2)=2\), \(f(3)=-6\).
Since \(f(0) = 4\) and \(f(1)=-1\), and \(0\) is between \(4\) and \(- 1\), by the Intermediate - Value Theorem, there exists a number \(c_1\in(0,1)\) such that \(f(c_1)=0\).
Since \(f(1)=-1\) and \(f(2) = 2\), and \(0\) is between \(-1\) and \(2\), by the Intermediate - Value Theorem, there exists a number \(c_2\in(1,2)\) such that \(f(c_2)=0\).
Since \(f(2)=2\) and \(f(3)=-6\), and \(0\) is between \(2\) and \(-6\), by the Intermediate - Value Theorem, there exists a number \(c_3\in(2,3)\) such that \(f(c_3)=0\).

Answer:

\(f(x) = 0\) has at least three solutions.