Continuity and One Sided Limits Worksheet

**Page 1**

1.

If $f$($x$) = $\frac{1}{{x}^{2}+9}$, then which of the following statement(s) is/are correct?

I. $f$($x$) is continuous on the interval [- 1, 3]

II. $f$($x$) is not right continuous at - 1

III. $f$($x$) is continuous on the interval[- 4, 0]

IV. $f$($x$) is not left continuous at 0

III. $f$($x$) is continuous on the interval

a. | I, II only | ||

b. | I only | ||

c. | II only | ||

d. | II, IV only | ||

e. | I, III only |

=

=

Draw the graph of

[

=

=

Draw the graph of

[

The correct choice is E.

Correct answer : (5)

2.

Which of the following statements are correct for the function $f$($x$) = $x$^{2} - 3$x$ - 10 using the Intermediate Value Theorem?

I. $f$($x$) has a zero on [- 3, - 1]

II. $f$($x$) has no zero on [4, 6]

III. $f$($x$) has no zero on [-1, 0]

IV. $f$($x$) has no zero on [- 3, - 1]

I. $f$($x$) has a zero on [- 3, - 1]

II. $f$($x$) has no zero on [4, 6]

III. $f$($x$) has no zero on [-1, 0]

IV. $f$($x$) has no zero on [- 3, - 1]

a. | I, III only | ||

b. | II, IV only | ||

c. | IV only | ||

d. | II, III only | ||

e. | III, IV only |

[Write the function.]

[Polynomial functions are continuous on R.]

[Find the values of

As

[Use the Intermediate Value Theorem in [- 3, -1].]

As

[Use the Intermediate Value Theorem in [4, 6].]

As

[Use the Intermediate Value Theorem in [-1, 0].]

The correct choice is A.

Correct answer : (1)

3.

Use Intermediate Value Theorem to choose the correct statements from the following for the function $f$($x$) = | 2$x$ + 3 |.

I. $f$($x$) has a zero on [1, 2]

II. $f$($x$) has no zero on [- 3, - 2]

III. $f$($x$) has no zero on [- 2, - 1]

IV. $f$($x$) has a zero on [-1, 0]

I. $f$($x$) has a zero on [1, 2]

II. $f$($x$) has no zero on [- 3, - 2]

III. $f$($x$) has no zero on [- 2, - 1]

IV. $f$($x$) has a zero on [-1, 0]

a. | I, II only | ||

b. | II, III only | ||

c. | III, IV only | ||

d. | II, IV only | ||

e. | I, IV only |

[Write the function.]

[Find the values of

As

[Use Intermediate Value Theorem in [1, 2].]

As

[Use Intermediate Value Theorem in [- 3, - 2].]

As

[Use Intermediate Value Theorem in [- 2, - 1].]

As

[Use Intermediate Value Theorem in [- 1, 0].]

The correct choice is B.

Correct answer : (2)

4.

Use Intermediate Value Theorem to choose the correct statements from the following for the function $f$($x$) = $\frac{2x}{{x}^{2}+4}$.

I. $f$($x$) has a zero on [- 1, 1]

II. $f$($x$) has no zero on [- 2, - 1]

III. $f$($x$) has a zero on [1, 2]

IV. $f$($x$) has a zero on [- 3, - 2]

I. $f$($x$) has a zero on [- 1, 1]

II. $f$($x$) has no zero on [- 2, - 1]

III. $f$($x$) has a zero on [1, 2]

IV. $f$($x$) has a zero on [- 3, - 2]

a. | I, II only | ||

b. | II, III only | ||

c. | I, IV only | ||

d. | II, IV only | ||

e. | III, IV only |

[Write the function.]

[Find

As

[Use Intermediate Value Theorem in [- 1, 1].]

As

[Use Intermediate Value Theorem in [- 2, -1].]

As

[Use Intermediate Value Theorem in [1, 2].]

As

[Use Intermediate Value Theorem in [- 3, - 2].]

The correct choice is A.

Correct answer : (1)

5.

Use Intermediate Value Theorem to choose the correct statement from the following for the function $f$($x$) = $\frac{{4x}^{2}+3x}{x-1}$.

I. $f$($x$) has a zero on [- $\frac{1}{2}$, $\frac{1}{2}$]

II. $f$($x$) has no zero on [- 2, -1]

III. $f$($x$) has a zero on [2, 3]

IV. $f$($x$) has a zero on [3, 4]

I. $f$($x$) has a zero on [- $\frac{1}{2}$, $\frac{1}{2}$]

II. $f$($x$) has no zero on [- 2, -1]

III. $f$($x$) has a zero on [2, 3]

IV. $f$($x$) has a zero on [3, 4]

a. | I, II only | ||

b. | II, III only | ||

c. | III, IV only | ||

d. | I, IV only | ||

e. | I, III only |

[Write the function.]

[Find the values of

As

[Use Intermediate Value Theorem in [-

As

[Use Intermediate Value Theorem in [- 2 , -1].]

As

[Use Intermediate Value Theorem in [2, 3].]

As

[Use Intermediate Value Theorem in [3, 4].]

The correct choice is A.

Correct answer : (1)

6.

Use Intermediate Value Theorem to choose the correct statements from the following for the function $f$($x$) = $e$^{2$x$} - 2$x$.

I. $f$($x$) has a zero on [- 2, - 1]

II. $f$($x$) has a zero on [0, $\frac{1}{2}$]

III. $f$($x$) has no zero on [$\frac{1}{2}$, 1]

IV. $f$($x$) has no zero on [- 1, 0]

I. $f$($x$) has a zero on [- 2, - 1]

II. $f$($x$) has a zero on [0, $\frac{1}{2}$]

III. $f$($x$) has no zero on [$\frac{1}{2}$, 1]

IV. $f$($x$) has no zero on [- 1, 0]

a. | III, IV only | ||

b. | I, III only | ||

c. | I, II only | ||

d. | II, IV only | ||

e. | II, III only |

[Write the function.]

[Find the values of

As

[Use Intermediate Value Theorem in [- 2, - 1].]

As

[Use Intermediate Value Theorem in [0,

As

[Use Intermediate Value Theorem in [

As

[Use Intermediate Value Theorem in [- 1, 0].]

The correct choice is A.

Correct answer : (1)

7.

Use Intermediate Value Theorem to choose the correct statements for the function $f$($x$) = sin $x$.

a. | III, IV only | ||

b. | II, IV only | ||

c. | I, III only | ||

d. | II, III only | ||

e. | I, IV only |

[Write the function.]

[Find the values of

As

[Use Intermediate Value Theorem in [

As

[Use Intermediate Value Theorem in [

As

[Use Intermediate Value Theorem in [-

As

[Use Intermediate Value Theorem in [-

The correct choice is A.

Correct answer : (1)

8.

Use Intermediate Value Theorem to choose the correct statements from the following for the function $f$($x$) = - $x$ + ln($x$^{2} + 1).

I. $f$($x$) has a zero on [- 1, 1]

II. $f$($x$) has no zero on [1, 2]

III. $f$($x$) has a zero on [- 2, - 1]

Iv. $f$($x$) has a zero on [2, 3]

I. $f$($x$) has a zero on [- 1, 1]

II. $f$($x$) has no zero on [1, 2]

III. $f$($x$) has a zero on [- 2, - 1]

Iv. $f$($x$) has a zero on [2, 3]

a. | I, II only | ||

b. | II, III only | ||

c. | II, IV only | ||

d. | III, IV only | ||

e. | I, IV only |

[Write the function.]

[Find the values of

As

[Use Intermediate Value Theorem in [- 1, 1].]

As

[Use Intermediate Value Theorem in [1, 2].]

As

[Use Intermediate Value Theorem in [- 2, - 1].]

As

[Use Intermediate Value Theorem in [2, 3].]

The correct choice is A.

Correct answer : (1)

9.

Use Intermediate Value Theorem to choose the correct statements from the following for the function $f$($x$) = 2$x$ - $\sqrt{25-{x}^{2}}$.

I. $f$($x$) has a zero on [2, 3]

II. $f$($x$) has no zero on [1, 2]

III. $f$($x$) has a zero on [3, 4]

IV. $f$($x$) has a zero on [- 3, - 2]

I. $f$($x$) has a zero on [2, 3]

II. $f$($x$) has no zero on [1, 2]

III. $f$($x$) has a zero on [3, 4]

IV. $f$($x$) has a zero on [- 3, - 2]

a. | I, II only | ||

b. | II, III only | ||

c. | III, IV only | ||

d. | II, IV only | ||

e. | I, IV only |

[Write the function.]

[Find the values of

As

[Use Intermediate Value Theorem in [2, 3].]

As

[Use Intermediate Value Theorem in [1, 2].]

As

[Use Intermediate Value Theorem in [3, 4].]

As

[Use Intermediate Value Theorem in [- 3, - 2].]

The correct choice is A.

Correct answer : (1)

10.

By using Intermediate Value Theorem, choose the correct statements from the following for the function $f$($x$) = 2$x$$\sqrt{{x}^{2}+5}$.

I. $f$($x$) has a zero on [- $\frac{1}{2}$, $\frac{1}{2}$]

II. $f$($x$) has a zero on [$\frac{1}{2}$, 1]

III. $f$($x$) has a zero on [1, 2]

IV. $f$($x$) has no zero on [- 2, - $\frac{1}{2}$]

I. $f$($x$) has a zero on [- $\frac{1}{2}$, $\frac{1}{2}$]

II. $f$($x$) has a zero on [$\frac{1}{2}$, 1]

III. $f$($x$) has a zero on [1, 2]

IV. $f$($x$) has no zero on [- 2, - $\frac{1}{2}$]

a. | I, IV only | ||

b. | II, IV only | ||

c. | III, IV only | ||

d. | I, III only | ||

e. | II, III only |

[Write the function.]

[Find the values of

As

[Use Intermediate Value Theorem in [-

As

[Use Intermediate Value Theorem in [

As

[Use Intermediate Value Theorem in [1, 2].]

As

[Use Intermediate Value Theorem in [- 2, -

The correct choice is A.

Correct answer : (1)