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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)=4 for at least one 
c between -2 and 1
(B) 
f(c)=4 for at least one 
c between 3 and 6
(C) 
f(c)=0 for at least one 
c between -2 and 1
(D) 
f(c)=0 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)=4 f(c)=4 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 33 and 66\newline(C) f(c)=0 f(c)=0 for at least one c c between 2-2 and 11\newline(D) f(c)=0 f(c)=0 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)=4 f(c)=4 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 33 and 66\newline(C) f(c)=0 f(c)=0 for at least one c c between 2-2 and 11\newline(D) f(c)=0 f(c)=0 for at least one c c between 33 and 66
  1. The 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 Values: We are given that f(2)=3f(-2) = 3 and f(1)=6f(1) = 6. This means that the function ff takes on the value 33 at x=2x = -2 and the value 66 at x=1x = 1.
  3. Checking Answer Choices: We need to determine if there is a value cc in the interval [2,1][-2, 1] such that f(c)f(c) is equal to a certain value. We will check each answer choice to see which one is guaranteed by the Intermediate Value Theorem.
  4. Choice (A) Analysis: For choice (A), we are looking for a value cc between 2-2 and 11 such that f(c)=4f(c) = 4. Since 44 is between f(2)=3f(-2) = 3 and f(1)=6f(1) = 6, the Intermediate Value Theorem guarantees that there is at least one such cc in the interval [2,1][-2, 1].
  5. Choice (B) Analysis: For choice (B), we are looking for a value cc between 33 and 66 such that f(c)=4f(c) = 4. This choice is not relevant to the Intermediate Value Theorem because it refers to values of f(c)f(c), not values of cc.
  6. Choice (C) Analysis: For choice (C), we are looking for a value cc between 2-2 and 11 such that f(c)=0f(c) = 0. Since 00 is not between f(2)=3f(-2) = 3 and f(1)=6f(1) = 6, the Intermediate Value Theorem does not guarantee that there is a cc such that f(c)=0f(c) = 0 in the interval [2,1][-2, 1].
  7. Choice (D) Analysis: For choice (D), we are looking for a value cc between 33 and 66 such that f(c)=0f(c) = 0. This choice is also not relevant to the Intermediate Value Theorem because it refers to values of f(c)f(c), not values of cc.

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