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Let
h(x)=(1)/(x).
Can we use the intermediate value theorem to say the equation
h(x)=0.5 has a solution where
-1 <= x <= 1 ?
Choose 1 answer:
(A) No, since the function is not continuous on that interval.
(B) No, since 0.5 is not between
h(-1) and
h(1).
(C) Yes, both conditions for using the intermediate value theorem have been met.

Let $h(x)=\frac{1}{x}$ . $\newline$ Can we use the intermediate value theorem to say the equation $h(x)=0.5$ has a solution where $-1 \leq x \leq 1$ ? $\newline$ Choose $1$ answer: $\newline$ (A) No, since the function is not continuous on that interval. $\newline$ (B) No, since $0$ . $5$ is not between $h(-1)$ and $h(1)$ . $\newline$ (C) Yes, both conditions for using the intermediate value theorem have been met.

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Intermediate Value Theorem

Full solution

Q. Let

h(x)=\frac{1}{x}

.

\newline

Can we use the intermediate value theorem to say the equation

h(x)=0.5

has a solution where

-1 \leq x \leq 1

?

\newline

Choose

1

answer:

\newline

(A) No, since the function is not continuous on that interval.

\newline

(B) No, since

0

.

5

is not between

h(-1)

and

h(1)

.

\newline

(C) Yes, both conditions for using the intermediate value theorem have been met.

Check Conditions: To use the intermediate value theorem, we need to check two conditions: the function must be continuous on the closed interval $a, b$ , and the value we are looking for (in this case, $0.5$ ) must be between the values of the function at the endpoints of the interval.
Function Continuity: First, let's check if the function $h(x) = \frac{1}{x}$ is continuous on the interval $[-1, 1]$ . We know that the function $\frac{1}{x}$ is not defined at $x = 0$ , which lies within the interval $[-1, 1]$ . Therefore, the function is not continuous on the entire interval.
Use of Theorem: Since the function is not continuous on the interval $[-1, 1]$ due to the discontinuity at $x = 0$ , we cannot use the intermediate value theorem to guarantee that there is a solution to the equation $h(x) = 0.5$ on this interval.

More problems from Intermediate Value Theorem

Question

f

be a continuous function on the closed interval

[1,5]

f(1)=1

f(5)=-3

\newline

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

\newline

1

\newline

f(c)=2

for at least one

c

-3

1

\newline

f(c)=-2

for at least one

c

-3

1

\newline

f(c)=2

for at least one

c

1

5

\newline

f(c)=-2

for at least one

c

1

5

Posted 9 months ago

Question

h

be a continuous function on the closed interval

[-3,4]

h(-3)=-1

h(4)=2

\newline

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

\newline

1

\newline

h(c)=1

for at least one

c

-3

4

\newline

h(c)=-2

for at least one

c

-3

4

\newline

h(c)=-2

for at least one

c

-1

2

\newline

h(c)=1

for at least one

c

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2

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Question

g

be a continuous function on the closed interval

[-3,3]

g(-3)=0

g(3)=6

\newline

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

\newline

1

\newline

g(c)=-2

for at least one

c

-3

3

\newline

g(c)=-2

for at least one

c

0

6

\newline

g(c)=5

for at least one

c

-3

3

\newline

g(c)=5

for at least one

c

0

6

Posted 9 months ago

Question

g

be a continuous function on the closed interval

[-1,4]

g(-1)=-4

g(4)=1

\newline

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

\newline

1

\newline

g(c)=3

for at least one

c

-1

4

\newline

g(c)=-3

for at least one

c

-4

1

\newline

g(c)=3

for at least one

c

-4

1

\newline

g(c)=-3

for at least one

c

-1

4

Posted 9 months ago

Question

g

be a continuous function on the closed interval

[-1,3]

g(-1)=-2

g(3)=-5

\newline

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

\newline

1

\newline

g(c)=-3

for at least one

c

-5

-2

\newline

g(c)=0

for at least one

c

-5

-2

\newline

g(c)=0

for at least one

c

-1

3

\newline

g(c)=-3

for at least one

c

-1

3

Posted 9 months ago

Question

f

be a continuous function on the closed interval

[-2,1]

f(-2)=3

f(1)=6

\newline

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

\newline

1

\newline

f(c)=0

for at least one

c

-2

1

\newline

f(c)=0

for at least one

c

3

6

\newline

f(c)=4

for at least one

c

-2

1

\newline

f(c)=4

for at least one

c

3

6

Posted 9 months ago

Question

f

be a continuous function on the closed interval

[-5,0]

f(-5)=0

f(0)=5

\newline

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

\newline

1

\newline

f(c)=-2

for at least one

c

-5

0

\newline

f(c)=2

for at least one

c

0

5

\newline

f(c)=2

for at least one

c

-5

0

\newline

f(c)=-2

for at least one

c

0

5

Posted 9 months ago

Question

f

be a continuous function on the closed interval

[1,5]

f(1)=1

f(5)=-3

\newline

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

\newline

1

\newline

f(c)=2

for at least one

c

1

5

\newline

f(c)=-2

for at least one

c

-3

1

\newline

f(c)=-2

for at least one

c

1

5

\newline

f(c)=2

for at least one

c

-3

1

Posted 9 months ago

Question

h

be a continuous function on the closed interval

[0,4]

h(0)=2

h(4)=-2

\newline

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

\newline

1

\newline

h(c)=3

for at least one

c

0

4

\newline

h(c)=-1

for at least one

c

0

4

\newline

h(c)=-1

for at least one

c

-2

2

\newline

h(c)=3

for at least one

c

-2

2

Posted 9 months ago

Question

You pick a card at random, put it back, and then pick another card at random.

\newline

4

\newline

5

\newline

6

\newline

7

\newline

What is the probability of picking a number greater than

5

and then picking a

4

\newline

Write your answer as a percentage.

Posted 1 month ago

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