Question

let f be a continuous function on the closed interval -3, 2. a few values of f are given in this table:

x -3 -1 0 2
f(x) 6 1 5 3

which intervals must contain a solution to f(x)=2?

choose all answers that apply:
a -3, -1
b -1, 0
c 0, 2

Explanation:

Step1: Recall Intermediate - Value Theorem

If \(y = f(x)\) is continuous on \([a,b]\) and \(k\) is a number between \(f(a)\) and \(f(b)\), then there exists at least one number \(c\) in \((a,b)\) such that \(f(c)=k\).

Step2: Check interval \([-3,-1]\)

\(f(-3) = 6\) and \(f(-1)=1\). Since \(2\) is between \(1\) and \(6\) (i.e., \(1<2<6\)), by the Intermediate - Value Theorem, there must be a \(c\in[-3,-1]\) such that \(f(c) = 2\).

Step3: Check interval \([-1,0]\)

\(f(-1)=1\) and \(f(0)=5\). Since \(2\) is between \(1\) and \(5\) (i.e., \(1<2<5\)), by the Intermediate - Value Theorem, there must be a \(c\in[-1,0]\) such that \(f(c)=2\).

Step4: Check interval \([0,2]\)

\(f(0)=5\) and \(f(2)=3\). Since \(2\) is not between \(3\) and \(5\) (i.e., \(2<3<5\) or \(3 < 5\) and \(2\) is less than both), the Intermediate - Value Theorem does not guarantee a solution in \([0,2]\).

Answer:

A. \([-3,-1]\), B. \([-1,0]\)