Question

suppose that f is continuous on the interval 0,5 and some of its values are given as follows:

x	0	1	2	3	4	5
f(x)	3	-6	6	2	-2	4

for each of the following statements, determine whether it is necessarily true (enter t), necessarily false (enter f) or possibly but not necessarily true (enter u).

1. f(x)=0 has a root in the interval 0,1:
2. f(x)=0 has a root in the interval 2,5:
3. f(x)=0 has no roots in the interval 2,3:

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 \(f(c)=k\). When \(k = 0\), if \(f(a)\) and \(f(b)\) have opposite signs, \(f(x)=0\) has a root in \((a,b)\).

Step2: Analyze the interval \([0,1]\)

We have \(f(0)=3\) and \(f(1)= - 6\). Since \(f(0)>0\) and \(f(1)<0\) and \(f(x)\) is continuous on \([0,1]\), by the Intermediate - Value Theorem, \(f(x) = 0\) has a root in the interval \([0,1]\). So the answer for statement 1 is \(T\).

Step3: Analyze the interval \([2,5]\)

We have \(f(2)=6\), \(f(3)=2\), \(f(4)=-2\), \(f(5)=4\). Since \(f(3)>0\) and \(f(4)<0\) (and \(f(x)\) is continuous on \([2,5]\)), by the Intermediate - Value Theorem, \(f(x)=0\) has a root in the interval \([2,5]\). So the answer for statement 2 is \(T\).

Step4: Analyze the interval \([2,3]\)

We have \(f(2)=6\) and \(f(3)=2\). Both \(f(2)\) and \(f(3)\) are positive. But we cannot be sure that \(f(x)\) does not cross the \(x\) - axis in the open - interval \((2,3)\) (for example, the function could dip below the \(x\) - axis and then come back up). So \(f(x) = 0\) possibly has roots in \([2,3]\), and the answer for statement 3 is \(U\).

Answer:

1. \(T\)
2. \(T\)
3. \(U\)