Question 1

Answer true or false. The Intermediate-Value Theorem can be used to approximate the locations of all discontinuities for  .

Accepted Answer

The answer of Answer true or false. The Intermediate-Value Theorem...

Question 2

Show that  is not a continuous function.

Accepted Answer

The answer of Show that  is not a continuous...

Question 3

Answer true or false. The function  is continuous everywhere.

Accepted Answer

Without knowing the exact function, it is impossible to say for certain whether it is continuous everywhere. However, if we assume that this function is well-defined and has been proven to be continuous, then the answer would be true.

Question 4

Answer true or false. The function  has a removable discontinuity at x = 5.

Accepted Answer

Without knowing the actual function, it is impossible to determine if there is a removable discontinuity at x = 5.

Question 5

Answer true or false. f(x) = x⁴ - 2x² + 11 = 0 has at least one solution on the interval [0, 9].

Accepted Answer

The answer of Answer true or false. f(x) = x⁴...

Question 6

Answer true or false. f(x) = x² - 6x + 5 = 0 has at least one solution on the interval [0, 2].

Accepted Answer

The function is continuous on the interval [0, 2] and changes sign (from positive to negative) at some point between x=1 and x=2. By the Intermediate Value Theorem, it has at least one root on that interval.

Question 7

Find the x-coordinates for all points of discontinuity for  .
A) 6 and 2
B) 6
C) -2 and -6
D) -2
E) 2

Accepted Answer

The answer of Find the x-coordinates for all points of...

Question 8

Answer true or false. f(x) = tan (x⁴ - 1) has no point of discontinuity.

Accepted Answer

Since tangent has a vertical asymptote at odd multiples of pi/2, and x⁴ - 1 can take on such values, there exists points of discontinuity.

Question 9

Find any points of discontinuity for  .

Accepted Answer

The answer of Find any points of discontinuity for ...

Question 10

Find the x-coordinates for all points of discontinuity for  .
A) 5
B) 7
C) -5
D) 0 and 7
E) None exists.

Accepted Answer

The answer of Find the x-coordinates for all points of...

Question 11

A point of discontinuity of  is at
A) 
B) 
C) 
D) 
E)

Accepted Answer

The answer of A point of discontinuity of  is...

Question 12

Find the value of k, if possible, that will make the function continuous. 
A) -2
B) 6
C) -6
D) 2
E) None exists.

Accepted Answer

The answer of Find the value of k, if possible,...

Question 13

Define  so that it will be continuous everywhere.

Accepted Answer

The answer of Define  so that it will be...

Question 14

Answer true or false. f(x) = x⁵ - 5x⁴ + 9 has no point of discontinuity.

Accepted Answer

The function is a polynomial and all polynomials are continuous on their domains, therefore there is no point of discontinuity.

Question 15

Redefine  so that it will be continuous everywhere.

Accepted Answer

The answer of Redefine  so that it will be...

Question 16

Use the fact that is a solution of x⁴ - 2 = 0 to approximate with an error of at most 0.005. A) 1.169 B) 1.179 C) 1.189 D) 1.199 E) 1.209

Accepted Answer

We can use the fact that 11ead0bc_8e83_e9c2_99a0_4b72a0e2f7d2_TB6988_11 is a solution of x⁴ - 2 = 0 to find an approximation for 11ead0bc_8e84_10d3_99a0_6d98660c1b31_TB6988_11. Let a be the decimal approximation of 11ead0bc_8e83_e9c2_99a0_4b72a0e2f7d2_TB6988_11 with an error of at most 0.005. Then we have: a⁴ - 2 = 0 +- 0.005 Adding 2 to both sides and using the fact that ⁴ = 2^(52), we get: a = 2/(2^(52)) +- 0.005 Simplifying, we get: a = 1.179 +- 0.005 Therefore, the best choice is B, 1.179.

Question 17

Answer true or false. f(x) = |x² - 4| has points of discontinuity at x = -2 and x = 2.

Accepted Answer

The function f(x) = |x^2 - 4| is continuous everywhere because the absolute value function and polynomial functions are continuous everywhere, and the composition of continuous functions is also continuous.

Question 18

Find the x-coordinates for all points of discontinuity for  .
A) 0
B) -2 and 2
C) -2
D) 2
E) 5

Accepted Answer

The answer of Find the x-coordinates for all points of...

Question 19

Answer true or false. If f and g are each continuous at c,  may be discontinuous at c.

Accepted Answer

This statement is known as the "limit of a product rule" and states that if f and g are each continuous at c, then their product (f*g) may still be discontinuous at c.

Question 20

Find any points of discontinuity for  .

Accepted Answer

The answer of Find any points of discontinuity for ...

Quiz 1: Limits and Continuity

Answer true or false. The Intermediate-Value Theorem can be used to approximate the locations of all discontinuities for .

Show that is not a continuous function.

Answer true or false. The function is continuous everywhere.

Answer true or false. The function has a removable discontinuity at x = 5.

Answer true or false. f(x) = x4 - 2x2 + 11 = 0 has at least one solution on the interval [0, 9].

Answer true or false. f(x) = x2 - 6x + 5 = 0 has at least one solution on the interval [0, 2].

Find the x-coordinates for all points of discontinuity for .

Answer true or false. f(x) = tan (x4 - 1) has no point of discontinuity.

Find any points of discontinuity for .

Find the x-coordinates for all points of discontinuity for .

A point of discontinuity of is at

Find the value of k, if possible, that will make the function continuous.

Define so that it will be continuous everywhere.

Answer true or false. f(x) = x5 - 5x4 + 9 has no point of discontinuity.

Redefine so that it will be continuous everywhere.

Use the fact that is a solution of x4 - 2 = 0 to approximate with an error of at most 0.005.

Answer true or false. f(x) = |x2 - 4| has points of discontinuity at x = -2 and x = 2.

Find the x-coordinates for all points of discontinuity for .

Answer true or false. If f and g are each continuous at c, may be discontinuous at c.

Find any points of discontinuity for .

Answer true or false. f(x) = x⁴ - 2x² + 11 = 0 has at least one solution on the interval [0, 9].

Answer true or false. f(x) = x² - 6x + 5 = 0 has at least one solution on the interval [0, 2].

Answer true or false. f(x) = tan (x⁴ - 1) has no point of discontinuity.

Answer true or false. f(x) = x⁵ - 5x⁴ + 9 has no point of discontinuity.

Use the fact that is a solution of x⁴ - 2 = 0 to approximate with an error of at most 0.005.

Answer true or false. f(x) = |x² - 4| has points of discontinuity at x = -2 and x = 2.