Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Let 
h be a continuous function on the closed interval 
[0,4], where 
h(0)=2 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 -2 and 2
(B) 
h(c)=3 for at least one 
c between -2 and 2
(c) 
h(c)=-1 for at least one 
c between 0 and 4
(D) 
h(c)=3 for at least one 
c between 0 and 4

Let h h be a continuous function on the closed interval [0,4] [0,4] , where h(0)=2 h(0)=2 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 2-2 and 22\newline(B) h(c)=3 h(c)=3 for at least one c c between 2-2 and 22\newline(C) h(c)=1 h(c)=-1 for at least one c c between 00 and 44\newline(D) h(c)=3 h(c)=3 for at least one c c between 00 and 44

Full solution

Q. Let h h be a continuous function on the closed interval [0,4] [0,4] , where h(0)=2 h(0)=2 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 2-2 and 22\newline(B) h(c)=3 h(c)=3 for at least one c c between 2-2 and 22\newline(C) h(c)=1 h(c)=-1 for at least one c c between 00 and 44\newline(D) h(c)=3 h(c)=3 for at least one c c between 00 and 44
  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 cc in the interval (a,b)(a, b) such that f(c)=Nf(c) = N.
  2. Given Function Values: We are given that h(0)=2h(0) = 2 and h(4)=2h(4) = -2. This means that the function hh changes from a positive value to a negative value over the interval [0,4][0, 4].
  3. Application of Theorem: Since 1-1 is a value between 22 and 2-2, by the Intermediate Value Theorem, there must be at least one value cc in the interval (0,4)(0, 4) such that h(c)=1h(c) = -1.
  4. Verification of Another Value: The value 33 is not between 22 and 2-2, so the Intermediate Value Theorem does not guarantee that there is a value cc in the interval (0,4)(0, 4) such that h(c)=3h(c) = 3.
  5. Correct Answer: Therefore, the correct answer is that h(c)=1h(c) = -1 for at least one cc between 00 and 44, which corresponds to choice (C).

More problems from Domain and range of quadratic functions: equations