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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)=3 for at least one
c between 0 and 4
(B)
h(c)=-1 for at least one
c between 0 and 4
(C)
h(c)=-1 for at least one
c between -2 and 2
(D)
h(c)=3 for at least one
c between -2 and 2

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

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Intermediate Value Theorem

Full solution

Q. Let

h

be a continuous function on the closed interval

[0,4]

, where

h(0)=2

and

h(4)=-2

.

\newline

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

\newline

Choose

1

answer:

\newline

(A)

h(c)=3

for at least one

c

between

0

and

4

\newline

(B)

h(c)=-1

for at least one

c

between

0

and

4

\newline

(C)

h(c)=-1

for at least one

c

between

-2

and

2

\newline

(D)

h(c)=3

for at least one

c

between

-2

and

2

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 $h$ on the interval $[0, 4]$ .
Given Function Values: We are given that $h(0) = 2$ and $h(4) = -2$ . This means that the function $h$ takes on at least all values between $2$ and $-2$ on the interval $[0, 4]$ because it is continuous.
Option (A) Analysis: Option (A) suggests that $h(c) = 3$ for some $c$ in $[0, 4]$ . However, since $3$ is not between $h(0) = 2$ and $h(4) = -2$ , the Intermediate Value Theorem does not guarantee that there is a $c$ such that $h(c) = 3$ .
Option (B) Analysis: Option (B) suggests that $h(c) = -1$ for some $c$ in $[0, 4]$ . Since $-1$ is between $h(0) = 2$ and $h(4) = -2$ , the Intermediate Value Theorem guarantees that there is at least one $c$ in $[0, 4]$ such that $h(c) = -1$ .
Option (C) Analysis: Option (C) is incorrect because the interval given is $[-2, 2]$ , which is not the interval over which we know the function $h$ is continuous. The function $h$ is defined on the interval $0, 4$ , so we cannot apply the Intermediate Value Theorem to the interval $[-2, 2]$ .
Option (D) Analysis: Option (D) is incorrect for the same reason as option (A). The value $3$ is not between $h(0) = 2$ and $h(4) = -2$ , so the Intermediate Value Theorem does not guarantee a $c$ such that $h(c) = 3$ in any interval.

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