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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 -1 and 2
(B) 
h(c)=1 for at least one 
c between -3 and 4
(c) 
h(c)=-2 for at least one 
c between -3 and 4
(D) 
h(c)=-2 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 1-1 and 22\newline(B) h(c)=1 h(c)=1 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 3-3 and 44\newline(D) h(c)=2 h(c)=-2 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 1-1 and 22\newline(B) h(c)=1 h(c)=1 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 3-3 and 44\newline(D) h(c)=2 h(c)=-2 for at least one c c between 1-1 and 22
  1. Understand IVT: Understand the Intermediate Value Theorem (IVT). The Intermediate Value Theorem states that if a function ff is continuous on a closed interval [a,b][a, b] and kk 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)=kf(c) = k.
  2. Apply to function hh: Apply the IVT to the given function hh. We are given that hh is continuous on the closed interval [3,4][-3, 4], h(3)=1h(-3) = -1, and h(4)=2h(4) = 2. We need to find if there is a value cc in the interval [3,4][-3, 4] such that h(c)h(c) equals a certain value.
  3. Evaluate based on IVT: Evaluate the choices based on the IVT.\newline(A) h(c)=1h(c) = 1 for at least one cc between 1-1 and 22. This choice is not relevant because the interval [1,2][-1, 2] is not the interval we are considering for the function hh.
  4. Continue evaluating choices: Continue evaluating the choices.\newline(B) h(c)=1h(c) = 1 for at least one cc between 3-3 and 44. Since 11 is between h(3)=1h(-3) = -1 and h(4)=2h(4) = 2, and the function hh is continuous on [3,4][-3, 4], the IVT guarantees that there is at least one cc in the interval [3,4][-3, 4] where h(c)=1h(c) = 1.
  5. Evaluate remaining choices: Evaluate the remaining choices.\newline(C) h(c)=2h(c) = -2 for at least one cc between 3-3 and 44. Since 2-2 is not between h(3)=1h(-3) = -1 and h(4)=2h(4) = 2, the IVT does not guarantee that there is a cc where h(c)=2h(c) = -2 in the interval [3,4][-3, 4].
  6. Evaluate last choice: Evaluate the last choice.\newline(D) h(c)=2h(c) = -2 for at least one cc between 1-1 and 22. This choice is also not relevant because the interval [1,2][-1, 2] is not the interval we are considering for the function hh, and 2-2 is not between h(3)=1h(-3) = -1 and h(4)=2h(4) = 2.

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