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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 . Which 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 [2,1] [-2,1] 44 and [2,1] [-2,1] 55\newline(D) f(c)=4 f(c)=4 for at least one c c between [2,1] [-2,1] 44 and [2,1] [-2,1] 55

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 . Which 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 [2,1] [-2,1] 44 and [2,1] [-2,1] 55\newline(D) f(c)=4 f(c)=4 for at least one c c between [2,1] [-2,1] 44 and [2,1] [-2,1] 55
  1. IVT Explanation: 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. We are given that ff is continuous on the interval [2,1][-2, 1], [a,b][a, b]00, and [a,b][a, b]11.
  2. Option Analysis (A): We need to determine which of the given options is guaranteed by the Intermediate Value Theorem. Let's analyze each option:\newline(A) f(c)=0f(c)=0 for at least one cc between 2-2 and 11: Since 00 is not between f(2)=3f(-2) = 3 and f(1)=6f(1) = 6, the theorem does not guarantee a cc such that f(c)=0f(c) = 0 in the interval [2,1][-2, 1].
  3. Option Analysis (B): (B) 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 theorem guarantees that there is at least one cc in the interval [2,1][-2, 1] such that f(c)=4f(c) = 4.
  4. Option Analysis (C): (C)f(c)=0(C) f(c)=0 for at least one cc between 33 and 66: This option is not relevant because the interval between 33 and 66 does not correspond to the xx-values but to the yy-values (function values). The theorem applies to the xx-values in the interval [2,1][-2, 1].
  5. Option Analysis (D): (D) f(c)=4f(c)=4 for at least one cc between 33 and 66: Similar to option (C), this option is not relevant because it refers to the yy-values. The theorem guarantees the existence of a cc for a value between f(2)f(-2) and f(1)f(1), not between 33 and 66.

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