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

let f be a continuous function on the closed interval -3, 2. a few values of f are given in this table:\n\nx -3 -1 0 2\nf(x) 6 1 5 3\n\nwhich intervals must contain a solution to f(x)=2?\n\nchoose all answers that apply:\na -3, -1\nb -1, 0\nc 0, 2

let f be a continuous function on the closed interval -3, 2. a few values of f are given in this table:\n\nx -3 -1 0 2\nf(x) 6 1 5 3\n\nwhich intervals must contain a solution to f(x)=2?\n\nchoose all answers that apply:\na -3, -1\nb -1, 0\nc 0, 2

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).

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])