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

suppose that f is continuous on the interval 0,5 and some of its values are given as follows: \n|x|0|1|2|3|4|5|\n|f(x)|3|-6|6|2|-2|4|\nfor 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).\n1. f(x)=0 has a root in the interval 0,1:\n2. f(x)=0 has a root in the interval 2,5:\n3. f(x)=0 has no roots in the interval 2,3:

Answer

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).