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)=-2 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)=1 for at least one c between -1 and 2 (D) h(c)=-2 for at least one c between -3 and 4

Apply Intermediate Value Theorem: The Intermediate Value Theorem states that if a function is continuous on a closed interval [a, b] and N is any number between f(a) and f(b), then there exists at least one c in the interval [a, b] such that f(c) = N. We need to apply this theorem to the function h and the given values.Given Function Values: We are given that h(-3) = -1 and h(4) = 2. This means that the function h takes on the value -1 at x = -3 and the value 2 at x = 4.Analyze Given Options: We need to determine which of the given options is guaranteed by the Intermediate Value Theorem. Let's analyze each option: (A) h(c) = -2 for at least one c between -1 and 2. This option is not guaranteed because the interval [-1, 2] does not include both endpoints where the function values are given, which are -3 and 4. (B) h(c) = 1 for at least one c between -3 and 4. This option is possible because 1 is between the function values -1 and 2, and the interval [-3, 4] includes both endpoints where these function values are given. (C) h(c) = 1 for at least one c between -1 and 2. Similar to option (A), this option is not guaranteed because the interval [-1, 2] does not include both endpoints where the function values are given. (D) h(c) = -2 for at least one c between -3 and 4. This option is not guaranteed because -2 is not between the function values -1 and 2.Correct Answer: Based on the Intermediate Value Theorem, the correct answer is option (B) because it is the only option that satisfies the conditions of the theorem: the value 1 is between the given function values -1 and 2, and the interval [-3, 4] includes the endpoints where these function values occur.

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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)=-2 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)=1 for at least one
c between -1 and 2
(D)
h(c)=-2 for at least one
c between -3 and 4

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

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Math Problems

Calculus

Intermediate Value Theorem

Full solution

Q. Let

h

be a continuous function on the closed interval

[-3,4]

, where

h(-3)=-1

and

h(4)=2

\newline

Which of the following is guaranteed by the Intermediate Value Theorem?

\newline

Choose

1

answer:

\newline

(A)

h(c)=-2

for at least one

c

between

-1

and

2

\newline

(B)

h(c)=1

for at least one

c

between

-3

and

4

\newline

(C)

h(c)=1

for at least one

c

between

-1

and

2

\newline

(D)

h(c)=-2

for at least one

c

between

-3

and

4

Apply Intermediate Value Theorem: The Intermediate Value Theorem states that if a function is continuous on a closed interval $[a, b]$ and $N$ is any number between $f(a)$ and $f(b)$ , then there exists at least one $c$ in the interval $[a, b]$ such that $f(c) = N$ . We need to apply this theorem to the function $h$ and the given values.
Given Function Values: We are given that $h(-3) = -1$ and $h(4) = 2$ . This means that the function $h$ takes on the value $-1$ at $x = -3$ and the value $2$ at $x = 4$ .
Analyze Given Options: We need to determine which of the given options is guaranteed by the Intermediate Value Theorem. Let's analyze each option: $\newline$ (A) $h(c) = -2$ for at least one $c$ between $-1$ and $2$ . This option is not guaranteed because the interval $[-1, 2]$ does not include both endpoints where the function values are given, which are $-3$ and $4$ . $\newline$ (B) $h(c) = 1$ for at least one $c$ between $-3$ and $4$ . This option is possible because $c$ $1$ is between the function values $-1$ and $2$ , and the interval $c$ $4$ includes both endpoints where these function values are given. $\newline$ (C) $h(c) = 1$ for at least one $c$ between $-1$ and $2$ . Similar to option (A), this option is not guaranteed because the interval $[-1, 2]$ does not include both endpoints where the function values are given. $\newline$ (D) $h(c) = -2$ for at least one $c$ between $-3$ and $4$ . This option is not guaranteed because $-1$ $4$ is not between the function values $-1$ and $2$ .
Correct Answer: Based on the Intermediate Value Theorem, the correct answer is option (B) because it is the only option that satisfies the conditions of the theorem: the value $1$ is between the given function values $-1$ and $2$ , and the interval $[-3, 4]$ includes the endpoints where these function values occur.

More problems from Intermediate Value Theorem

Question

Let

f

be a continuous function on the closed interval

[1,5]

, where

f(1)=1