Let f be a continuous function on the closed interval [-2,1], where f(-2)=3 and f(1)=6. Which of the following is guaranteed by the Intermediate Value Theorem? Choose 1 answer: (A) f(c)=4 for at least one c between -2 and 1 (B) f(c)=4 for at least one c between 3 and 6 (C) f(c)=0 for at least one c between -2 and 1 (D) f(c)=0 for at least one c between 3 and 6

The Intermediate Value Theorem: The Intermediate Value Theorem states that if a function f 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 number c in the interval (a, b) such that f(c) = N.Given Function Values: We are given that f(-2) = 3 and f(1) = 6. This means that the function f takes on the value 3 at x = -2 and the value 6 at x = 1.Checking Answer Choices: We need to determine if there is a value c in the interval [-2, 1] such that f(c) is equal to a certain value. We will check each answer choice to see which one is guaranteed by the Intermediate Value Theorem.Choice (A) Analysis: For choice (A), we are looking for a value c between -2 and 1 such that f(c) = 4. Since 4 is between f(-2) = 3 and f(1) = 6, the Intermediate Value Theorem guarantees that there is at least one such c in the interval [-2, 1].Choice (B) Analysis: For choice (B), we are looking for a value c between 3 and 6 such that f(c) = 4. This choice is not relevant to the Intermediate Value Theorem because it refers to values of f(c), not values of c.Choice (C) Analysis: For choice (C), we are looking for a value c between -2 and 1 such that f(c) = 0. Since 0 is not between f(-2) = 3 and f(1) = 6, the Intermediate Value Theorem does not guarantee that there is a c such that f(c) = 0 in the interval [-2, 1].Choice (D) Analysis: For choice (D), we are looking for a value c between 3 and 6 such that f(c) = 0. This choice is also not relevant to the Intermediate Value Theorem because it refers to values of f(c), not values of c.

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Let
f be a continuous function on the closed interval
[-2,1], where
f(-2)=3 and
f(1)=6.
Which of the following is guaranteed by the Intermediate Value Theorem?
Choose 1 answer:
(A)
f(c)=4 for at least one
c between -2 and 1
(B)
f(c)=4 for at least one
c between 3 and 6
(C)
f(c)=0 for at least one
c between -2 and 1
(D)
f(c)=0 for at least one
c between 3 and 6

Let $f$ be a continuous function on the closed interval $[-2,1]$ , where $f(-2)=3$ and $f(1)=6$ . $\newline$ Which of the following is guaranteed by the Intermediate Value Theorem? $\newline$ Choose $1$ answer: $\newline$ (A) $f(c)=4$ for at least one $c$ between $-2$ and $1$ $\newline$ (B) $f(c)=4$ for at least one $c$ between $3$ and $6$ $\newline$ (C) $f(c)=0$ for at least one $c$ between $-2$ and $1$ $\newline$ (D) $f(c)=0$ for at least one $c$ between $3$ and $6$

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Calculus

Intermediate Value Theorem

Full solution

Q. Let

f

be a continuous function on the closed interval

[-2,1]

, where

f(-2)=3

and

f(1)=6

\newline

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

\newline

Choose

1

answer:

\newline

(A)

f(c)=4

for at least one

c

between

-2

and

1

\newline

(B)

f(c)=4

for at least one

c

between

3

and

6

\newline

(C)

f(c)=0

for at least one

c

between

-2

and

1

\newline

(D)

f(c)=0

for at least one

c

between

3

and

6

The Intermediate Value Theorem: The Intermediate Value Theorem states that if a function $f$ 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 number $c$ in the interval $(a, b)$ such that $f(c) = N$ .
Given Function Values: We are given that $f(-2) = 3$ and $f(1) = 6$ . This means that the function $f$ takes on the value $3$ at $x = -2$ and the value $6$ at $x = 1$ .
Checking Answer Choices: We need to determine if there is a value $c$ in the interval $[-2, 1]$ such that $f(c)$ is equal to a certain value. We will check each answer choice to see which one is guaranteed by the Intermediate Value Theorem.
Choice (A) Analysis: For choice (A), we are looking for a value $c$ between $-2$ and $1$ such that $f(c) = 4$ . Since $4$ is between $f(-2) = 3$ and $f(1) = 6$ , the Intermediate Value Theorem guarantees that there is at least one such $c$ in the interval $[-2, 1]$ .
Choice (B) Analysis: For choice (B), we are looking for a value $c$ between $3$ and $6$ such that $f(c) = 4$ . This choice is not relevant to the Intermediate Value Theorem because it refers to values of $f(c)$ , not values of $c$ .
Choice (C) Analysis: For choice (C), we are looking for a value $c$ between $-2$ and $1$ such that $f(c) = 0$ . Since $0$ is not between $f(-2) = 3$ and $f(1) = 6$ , the Intermediate Value Theorem does not guarantee that there is a $c$ such that $f(c) = 0$ in the interval $[-2, 1]$ .
Choice (D) Analysis: For choice (D), we are looking for a value $c$ between $3$ and $6$ such that $f(c) = 0$ . This choice is also not relevant to the Intermediate Value Theorem because it refers to values of $f(c)$ , not values of $c$ .

More problems from Intermediate Value Theorem

Question

Let

f

be a continuous function on the closed interval

[1,5]

, where

f(1)=1