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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? $\newline$ Choose $1$ answer: $\newline$ (A) $f(c)=0$ for at least one $c$ between $-2$ and $1$ $\newline$ (B) $f(c)=4$ for at least one $c$ between $-2$ and $1$ $\newline$ (C) $f(c)=0$ for at least one $c$ between $[-2,1]$ $4$ and $[-2,1]$ $5$ $\newline$ (D) $f(c)=4$ for at least one $c$ between $[-2,1]$ $4$ and $[-2,1]$ $5$

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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

. 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)=4

for at least one

c

between

-2

and

1

\newline

(C)

f(c)=0

for at least one

c

between

[-2,1]

4

and

[-2,1]

5

\newline

(D)

f(c)=4

for at least one

c

between

[-2,1]

4

and

[-2,1]

5

IVT Explanation: 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 $f(c) = N$ . We are given that $f$ is continuous on the interval $[-2, 1]$ , $[a, b]$ $0$ , and $[a, b]$ $1$ .
Option Analysis (A): We need to determine which of the given options is guaranteed by the Intermediate Value Theorem. Let's analyze each option: $\newline$ (A) $f(c)=0$ for at least one $c$ between $-2$ and $1$ : Since $0$ is not between $f(-2) = 3$ and $f(1) = 6$ , the theorem does not guarantee a $c$ such that $f(c) = 0$ in the interval $[-2, 1]$ .
Option Analysis (B): (B) $f(c)=4$ for at least one $c$ between $-2$ and $1$ : Since $4$ is between $f(-2) = 3$ and $f(1) = 6$ , the theorem guarantees that there is at least one $c$ in the interval $[-2, 1]$ such that $f(c) = 4$ .
Option Analysis (C): $(C) f(c)=0$ for at least one $c$ between $3$ and $6$ : This option is not relevant because the interval between $3$ and $6$ does not correspond to the $x$ -values but to the $y$ -values (function values). The theorem applies to the $x$ -values in the interval $[-2, 1]$ .
Option Analysis (D): (D) $f(c)=4$ for at least one $c$ between $3$ and $6$ : Similar to option (C), this option is not relevant because it refers to the $y$ -values. The theorem guarantees the existence of a $c$ for a value between $f(-2)$ and $f(1)$ , not between $3$ and $6$ .