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Let 
f be a continuous function on the closed interval 
[1,5], where 
f(1)=1 and 
f(5)=-3.
Which of the following is guaranteed by the Intermediate Value Theorem?
Choose 1 answer:
(A) 
f(c)=2 for at least one 
c between -3 and 1
(B) 
f(c)=-2 for at least one 
c between -3 and 1
(C) 
f(c)=2 for at least one 
c between 1 and 5
(D) 
f(c)=-2 for at least one 
c between 1 and 5

Let f f be a continuous function on the closed interval [1,5] [1,5] , where f(1)=1 f(1)=1 and f(5)=3 f(5)=-3 .\newlineWhich of the following is guaranteed by the Intermediate Value Theorem?\newlineChoose 11 answer:\newline(A) f(c)=2 f(c)=2 for at least one c c between 3-3 and 11\newline(B) f(c)=2 f(c)=-2 for at least one c c between 3-3 and 11\newline(C) f(c)=2 f(c)=2 for at least one c c between 11 and 55\newline(D) f(c)=2 f(c)=-2 for at least one c c between 11 and 55

Full solution

Q. Let f f be a continuous function on the closed interval [1,5] [1,5] , where f(1)=1 f(1)=1 and f(5)=3 f(5)=-3 .\newlineWhich of the following is guaranteed by the Intermediate Value Theorem?\newlineChoose 11 answer:\newline(A) f(c)=2 f(c)=2 for at least one c c between 3-3 and 11\newline(B) f(c)=2 f(c)=-2 for at least one c c between 3-3 and 11\newline(C) f(c)=2 f(c)=2 for at least one c c between 11 and 55\newline(D) f(c)=2 f(c)=-2 for at least one c c between 11 and 55
  1. Theorem Statement: 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(1)=1f(1) = 1 and f(5)=3f(5) = -3. Since ff is continuous on [1,5][1, 5], by the Intermediate Value Theorem, for any value NN between 11 and 3-3, there must be some cc in [1,5][1, 5] such that f(c)=Nf(c) = N.
  3. Option (A) Analysis: We need to determine which of the given options is guaranteed by the Intermediate Value Theorem. Option (A) suggests f(c)=2f(c) = 2 for some cc between 3-3 and 11, but this is outside the interval [1,5][1, 5], so it cannot be guaranteed by the theorem.
  4. Option (B) Analysis: Option (B) suggests f(c)=2f(c) = -2 for some cc between 3-3 and 11, which is also outside the interval [1,5][1, 5], so it cannot be guaranteed by the theorem.
  5. Option (C) Analysis: Option (C) suggests f(c)=2f(c) = 2 for some cc between 11 and 55. Since 22 is not between f(1)=1f(1) = 1 and f(5)=3f(5) = -3, the Intermediate Value Theorem does not guarantee this.
  6. Option (D) Analysis: Option (D) suggests f(c)=2f(c) = -2 for some cc between 11 and 55. Since 2-2 is between f(1)=1f(1) = 1 and f(5)=3f(5) = -3, the Intermediate Value Theorem guarantees that there is at least one cc in [1,5][1, 5] such that f(c)=2f(c) = -2.

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