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)=0 for at least one c between -2 and 1 (B) f(c)=0 for at least one c between 3 and 6 (C) f(c)=4 for at least one c between -2 and 1 (D) f(c)=4 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 and Interval: We are given that f is continuous on the closed interval [-2, 1], f(-2) = 3, and f(1) = 6. We need to determine which statement is guaranteed by the Intermediate Value Theorem.Analysis of Option (A): Option (A) suggests that f(c) = 0 for some c between -2 and 1. However, since f(-2) = 3 and f(1) = 6, and 0 is not between 3 and 6, the Intermediate Value Theorem does not guarantee that f(c) will equal 0 on the interval [-2, 1].Analysis of Option (B): Option (B) suggests that f(c) = 0 for some c between 3 and 6. This is not relevant to the Intermediate Value Theorem as it applies to the values of c, not the values of f(c), and the interval given is [-2, 1], not [3, 6].Analysis of Option (C): Option (C) suggests that f(c) = 4 for some c between -2 and 1. Since 4 is between f(-2) = 3 and f(1) = 6, the Intermediate Value Theorem guarantees that there is at least one c in the interval [-2, 1] such that f(c) = 4.Analysis of Option (D): Option (D) suggests that f(c) = 4 for some c between 3 and 6. This is again not relevant to the Intermediate Value Theorem as it applies to the values of c in the interval [-2, 1], not the values of f(c) in the interval [3, 6].

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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)=0 for at least one
c between -2 and 1
(B)
f(c)=0 for at least one
c between 3 and 6
(C)
f(c)=4 for at least one
c between -2 and 1
(D)
f(c)=4 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) $f(c)=0$ for at least one $c$ between $-2$ and $1$ $\newline$ (B) $f(c)=0$ for at least one $c$ between $3$ and $6$ $\newline$ (C) $f(c)=4$ for at least one $c$ between $-2$ and $1$ $\newline$ (D) $f(c)=4$ 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)=0

for at least one

c

between

-2

and

1

\newline

(B)

f(c)=0

for at least one

c

between

3

and

6

\newline

(C)

f(c)=4

for at least one

c

between

-2

and

1

\newline

(D)

f(c)=4

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 and Interval: We are given that $f$ is continuous on the closed interval $[-2, 1]$ , $f(-2) = 3$ , and $f(1) = 6$ . We need to determine which statement is guaranteed by the Intermediate Value Theorem.
Analysis of Option (A): Option (A) suggests that $f(c) = 0$ for some $c$ between $-2$ and $1$ . However, since $f(-2) = 3$ and $f(1) = 6$ , and $0$ is not between $3$ and $6$ , the Intermediate Value Theorem does not guarantee that $f(c)$ will equal $0$ on the interval $c$ $1$ .
Analysis of Option (B): Option (B) suggests that $f(c) = 0$ for some $c$ between $3$ and $6$ . This is not relevant to the Intermediate Value Theorem as it applies to the values of $c$ , not the values of $f(c)$ , and the interval given is $[-2, 1]$ , not $[3, 6]$ .
Analysis of Option (C): Option (C) suggests that $f(c) = 4$ for some $c$ between $-2$ and $1$ . Since $4$ is between $f(-2) = 3$ and $f(1) = 6$ , the Intermediate Value Theorem guarantees that there is at least one $c$ in the interval $[-2, 1]$ such that $f(c) = 4$ .
Analysis of Option (D): Option (D) suggests that $f(c) = 4$ for some $c$ between $3$ and $6$ . This is again not relevant to the Intermediate Value Theorem as it applies to the values of $c$ in the interval $[-2, 1]$ , not the values of $f(c)$ in the interval $[3, 6]$ .

More problems from Intermediate Value Theorem

Question

Let

f

be a continuous function on the closed interval

[1,5]

, where

f(1)=1