Question

5 mark for review
the table above shows selected values of a continuous function f. for ( 0 leq x leq 13 ), what is the fewest possible number of times ( f(x)=4 )?
a one
b two
c three
d four

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 \(f(c)=k\).
For the interval \([0,4]\): \(f(0) = 3\) and \(f(4)=4.5\). Since \(3<4<4.5\), by the Intermediate - Value Theorem, there is at least one \(x_1\in(0,4)\) such that \(f(x_1) = 4\).
For the interval \([4,6]\): \(f(4)=4.5\) and \(f(6)=3\). Since \(3<4<4.5\), by the Intermediate - Value Theorem, there is at least one \(x_2\in(4,6)\) such that \(f(x_2)=4\).
For the interval \([8,13]\): \(f(8) = 2.5\) and \(f(13)=4.4\). Since \(2.5<4<4.4\), by the Intermediate - Value Theorem, there is at least one \(x_3\in(8,13)\) such that \(f(x_3)=4\).

Answer:

C. three