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

Let h h be a continuous function on the closed interval [3,4] [-3,4] , where h(3)=1 h(-3)=-1 and h(4)=2 h(4)=2 .\newlineWhich of the following is guaranteed by the Intermediate Value Theorem?\newlineChoose 11 answer:\newline(A) h(c)=1 h(c)=1 for at least one c c between 3-3 and 44\newline(B) h(c)=2 h(c)=-2 for at least one c c between 3-3 and 44\newline(C) h(c)=2 h(c)=-2 for at least one c c between 1-1 and 22\newline(D) h(c)=1 h(c)=1 for at least one c c between 1-1 and 22

Full solution

Q. Let h h be a continuous function on the closed interval [3,4] [-3,4] , where h(3)=1 h(-3)=-1 and h(4)=2 h(4)=2 .\newlineWhich of the following is guaranteed by the Intermediate Value Theorem?\newlineChoose 11 answer:\newline(A) h(c)=1 h(c)=1 for at least one c c between 3-3 and 44\newline(B) h(c)=2 h(c)=-2 for at least one c c between 3-3 and 44\newline(C) h(c)=2 h(c)=-2 for at least one c c between 1-1 and 22\newline(D) h(c)=1 h(c)=1 for at least one c c between 1-1 and 22
  1. Theorem Application: The Intermediate Value Theorem states that if a function 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 cc in the interval [a,b][a, b] such that f(c)=Nf(c) = N. We need to apply this theorem to the function hh and the given values.
  2. Endpoint Values: First, we check the value of hh at the endpoints of the interval. We have h(3)=1h(-3) = -1 and h(4)=2h(4) = 2. This means that the function hh takes on the values 1-1 and 22 at the ends of the interval [3,4][-3, 4].
  3. Option (A) Analysis: Now, we look at the options given and apply the Intermediate Value Theorem. Option (A) suggests that h(c)=1h(c) = 1 for some cc between 3-3 and 44. Since 11 is between 1-1 and 22, the Intermediate Value Theorem guarantees that there is at least one cc in the interval [3,4][-3, 4] such that h(c)=1h(c) = 1.
  4. Option (B) Analysis: Option (B) suggests that h(c)=2h(c) = -2 for some cc between 3-3 and 44. However, since 2-2 is not between h(3)=1h(-3) = -1 and h(4)=2h(4) = 2, the Intermediate Value Theorem does not guarantee that there is a cc in the interval [3,4][-3, 4] such that h(c)=2h(c) = -2.
  5. Option (C) Consideration: Option (C) is not applicable because the Intermediate Value Theorem requires the interval to be the same as where the function is continuous and the values are given. Since we are given the interval [3,4][-3, 4] and not [1,2][-1, 2], we cannot apply the theorem to the interval [1,2][-1, 2] directly.
  6. Option (D) Consideration: Option (D) is similar to option (C) in that it refers to an interval [1,2][-1, 2] which is not the interval we are considering for the application of the Intermediate Value Theorem. Therefore, it cannot be guaranteed by the theorem based on the information given.

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