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Let 
f be a continuous function on the closed interval 
[-2,1], where 
f(-2)=3 and 
f(1)=6.
Which of the following is guaranteed by the Intermediate Value Theorem?
Choose 1 answer:
(A) 
f(c)=0 for at least one 
c between -2 and 1
(B) 
f(c)=4 for at least one 
c between -2 and 1
(c) 
f(c)=0 for at least one 
c between 3 and 6
(D) 
f(c)=4 for at least one 
c between 3 and 6

Let f f be a continuous function on the closed interval [2,1] [-2,1] , where f(2)=3 f(-2)=3 and f(1)=6 f(1)=6 .\newlineWhich of the following is guaranteed by the Intermediate Value Theorem?\newlineChoose 11 answer:\newline(A) f(c)=0 f(c)=0 for at least one c c between 2-2 and 11\newline(B) f(c)=4 f(c)=4 for at least one c c between 2-2 and 11\newline(C) f(c)=0 f(c)=0 for at least one c c between 33 and 66\newline(D) f(c)=4 f(c)=4 for at least one c c between 33 and 66

Full solution

Q. Let f f be a continuous function on the closed interval [2,1] [-2,1] , where f(2)=3 f(-2)=3 and f(1)=6 f(1)=6 .\newlineWhich of the following is guaranteed by the Intermediate Value Theorem?\newlineChoose 11 answer:\newline(A) f(c)=0 f(c)=0 for at least one c c between 2-2 and 11\newline(B) f(c)=4 f(c)=4 for at least one c c between 2-2 and 11\newline(C) f(c)=0 f(c)=0 for at least one c c between 33 and 66\newline(D) f(c)=4 f(c)=4 for at least one c c between 33 and 66
  1. Intermediate Value Theorem: The Intermediate Value Theorem states that if a function ff is continuous on a closed interval [a,b][a, b] and NN is any number between f(a)f(a) and f(b)f(b), then there exists at least one number cc in the interval (a,b)(a, b) such that f(c)=Nf(c) = N.
  2. Given Function and Interval: We are given that ff is continuous on the closed interval [2,1][-2,1], f(2)=3f(-2)=3, and f(1)=6f(1)=6. We need to determine which statement is guaranteed by the Intermediate Value Theorem.
  3. Option (A) Analysis: Option (A) suggests that f(c)=0f(c)=0 for at least one cc between 2-2 and 11. However, since f(2)=3f(-2)=3 and f(1)=6f(1)=6, and 00 is not between 33 and 66, the Intermediate Value Theorem does not guarantee that f(c)f(c) will be 00 for any cc in the interval cc22.
  4. Option (B) Analysis: Option (B) suggests that f(c)=4f(c)=4 for at least one cc between 2-2 and 11. Since 44 is between f(2)=3f(-2)=3 and f(1)=6f(1)=6, the Intermediate Value Theorem guarantees that there will be at least one cc in the interval [2,1][-2,1] such that f(c)=4f(c) = 4.
  5. Option (C) Analysis: Option (C) suggests that f(c)=0f(c)=0 for at least one cc between 33 and 66. This option is not relevant because the values 33 and 66 are the function values, not the values in the domain of ff.
  6. Option (D) Analysis: Option (D) suggests that f(c)=4f(c)=4 for at least one cc between 33 and 66. This option is also not relevant because, again, 33 and 66 are the function values, not the values in the domain of ff.

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