### Understanding Continuity and Discontinuity of Functions#### Discontinuity in Function \( h \)Consider the function \( h(x) \):\[ h(x) = \begin{cases} \frac{1}{x + 2} & \text{if } x \ne -2 \\1 & \text{if } x = -2 \end{cases} \]**Discontinuity Explanation:**To determine why the function \( h \) is discontinuous at \( a = -2 \):1. **Limit from the Left and Right:** - As \( x \) approaches \(-2\) from either direction (left or right), \( \frac{1}{x + 2} \) becomes undefined. However, for values very close to but not equal to \(-2\), the function yields values approaching \( \pm \infty \).2. **Value at \( x = -2 \):** - When \( x \) reaches \(-2\), \( h(x) \) is defined as \( 1 \), which is explicitly given to avoid the undefined nature of \( \frac{1}{x + 2} \).Since \( \lim_{{x \to -2}} h(x) \) does not equal \( h(-2) \), the function \( h \) is discontinuous at \( x = -2 \).#### Continuity in Function \( f \)Consider the function \( f(x) \):\[ f(x) = \frac{3v - 1}{v^2 + 2v - 15} \]**Continuity and Domain:**1. **Finding the Domain:** - The denominator \( v^2 + 2v - 15 \) should not be zero to avoid undefined values. - Solving \( v^2 + 2v - 15 = 0 \): - \( v^2 + 5v - 3v - 15 = 0 \) - \((v + 5)(v - 3) = 0\) Thus, the denominator is zero when \( v = -5 \) or \( v = 3 \). Hence, the domain of \( f \) is all real numbers except \( v = -5 \) and \( v = 3 \).2. **Checking for Continu

As per the company's guidelines, we are supposed to solve first question for you, kindly repost…

Answered: Explain why the function h is…

Math

Calculus

Explain why the function h is discontinuous at a = -2. 1 x = -2 h(x) = x + 2 1 x = -2

College Algebra (MindTap Course List)

12th Edition

ISBN:9781305652231

Author:R. David Gustafson, Jeff Hughes

Publisher:R. David Gustafson, Jeff Hughes

Chapter3: Functions

Section3.3: More On Functions; Piecewise-defined Functions

Problem 84E

See similar textbooks

Related questions

Q: is this a function and why y = x2 + 3

Q: Find the point on the graph of the function f(x) = (x + 1)² that is closest to the point (-6, 1).…

A: Concept: The distance formula for the points x1,y1 and x2,y2 D=x2-x12+(y2-y1)2

Q: Find the breakeven point of the following functions: C(x) = 5x - 7 R(x) = 2x Approximately (2,…

Q: Complete the table given the following function: f(x) = -2x^2+1 f(x) - 2x² + 1 = - X f(x) -4 -31 -2…

A: The answer is in step 2.

Q: Given f(X) = 0 Which term or

Q: Find f (5) for the function f (x) = x+4

A: 1) The given function is fx=x+4. Evaluate f5 as follows. f5=5+4=9 Therefore, the value of f5 is 9.

Q: The function V(t)= = - - 3500t+18000, where V is value and t is time in years, can be used to find…

A: Given function:- Vt=-3500t+18000

Q: Determine what x = 5 is (max/min/neither) of the function f (x) = 2x2 − 3x + 9

A: Given f (x) = 2x2 − 3x + 9

Question

100%

### Understanding Continuity and Discontinuity of Functions

#### Discontinuity in Function \( h \)

Consider the function \( h(x) \):

\[ h(x) =
\begin{cases}
\frac{1}{x + 2} & \text{if } x \ne -2 \\
1 & \text{if } x = -2
\end{cases} \]

**Discontinuity Explanation:**

To determine why the function \( h \) is discontinuous at \( a = -2 \):

1. **Limit from the Left and Right:**
- As \( x \) approaches \(-2\) from either direction (left or right), \( \frac{1}{x + 2} \) becomes undefined. However, for values very close to but not equal to \(-2\), the function yields values approaching \( \pm \infty \).

2. **Value at \( x = -2 \):**
- When \( x \) reaches \(-2\), \( h(x) \) is defined as \( 1 \), which is explicitly given to avoid the undefined nature of \( \frac{1}{x + 2} \).

Since \( \lim_{{x \to -2}} h(x) \) does not equal \( h(-2) \), the function \( h \) is discontinuous at \( x = -2 \).

#### Continuity in Function \( f \)

Consider the function \( f(x) \):

\[ f(x) = \frac{3v - 1}{v^2 + 2v - 15} \]

**Continuity and Domain:**

1. **Finding the Domain:**
- The denominator \( v^2 + 2v - 15 \) should not be zero to avoid undefined values.
- Solving \( v^2 + 2v - 15 = 0 \):
- \( v^2 + 5v - 3v - 15 = 0 \)
- \((v + 5)(v - 3) = 0\)

Thus, the denominator is zero when \( v = -5 \) or \( v = 3 \). Hence, the domain of \( f \) is all real numbers except \( v = -5 \) and \( v = 3 \).

2. **Checking for Continu

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Step by step

Solved in 2 steps with 1 images

SEE SOLUTION Check out a sample Q&A here

Follow-up Questions

Read through expert solutions to related follow-up questions below.

Follow-up Question

### Continuity of the Function \( f(v) \)

**Problem Statement:**
Explain why the function \( f \) is continuous at every number in its domain. State the domain.

\[ f(v) = \frac{3v - 1}{v^2 + 2v - 15} \]

**Solution Explanation:**

To determine the continuity of the function \( f \) at every number in its domain, first, we need to identify the domain of the function. The function \( f \) is given as a rational function. Rational functions are continuous everywhere they are defined. The function is defined for all real numbers \( v \) except where the denominator is zero.

1. **Find the Domain:**
To find where the denominator is zero, solve the equation \( v^2 + 2v - 15 = 0 \):
\[
v^2 + 2v - 15 = 0
\]
Factorize the quadratic expression:
\[
(v + 5)(v - 3) = 0
\]
Set each factor equal to zero:
\[
v + 5 = 0 \quad \Rightarrow \quad v = -5
\]
\[
v - 3 = 0 \quad \Rightarrow \quad v = 3
\]
Therefore, the denominator is zero at \( v = -5 \) and \( v = 3 \).

2. **State the Domain:**
Since the function \( f(v) \) is undefined where the denominator is zero, the domain of \( f \) is all real numbers except \( v = -5 \) and \( v = 3 \):
\[
\text{Domain: } v \in \mathbb{R} \setminus \{-5, 3\}
\]

3. **Continuity:**
The function \( f(v) \) is continuous at every point in its domain. This is because for rational functions, continuity is determined by the denominator. As long as the denominator is not zero, the function is continuous. So, for all \( v \) in the domain \( \mathbb{R} \setminus \{-5, 3\} \), the function is continuous.

In conclusion, the function \( f(v) \) is continuous at every

Solution

by Bartleby Expert

SEE SOLUTION

Similar questions

How can you determine whether a function is odd or even from the formula of the function?
How do we determine the domain of a function defined by an equation?
If xis large, which function grows faster, f(x)=2x or g(x)=x2?
During a nine-hour snowstorm, it snows at a rate of 1 inch per hour for the first 2 hours, at a rate of 2 inches per hour for the next 6 hours, and at a rate of 0.5 inch per hour for the final hour. Write and graph a piecewise-defined function that gives the depth of the snow during the snowstorm. How many inches of snow accumulated from the storm?
Rightmost xvalue where f(x) =3
Page > of 10 ZOOM + [6] Use the algebraic test to determine if these functions are odd, even, or neither. [Odd: f(-x) = -f(x); Even: ƒ(-x) = f(x)] x2 a. f(x) = x2 - 5 b. 8(x) = x' + x? c. h(x) = x – 2x* d. q(x) = 6 а. b.
Example Let f (x) = x³ – with domain (-x, 0). Determine where f is increasing and where f is decreasing.
3x / x^2+4x-5
ƒ (x) = − ½-|x| − 14x^8 +7 is this a function if so how and why
where is the function decreasing (interval notetion)
X^3-8/x-2 How do I find interval on which function is continuous and discontinuities?
To make the function one to one we restrict the domain x>-4 d(x) =(x+4)^2-2