Please help me solve this problem. Thank you! A function $ f(x) $ is said to have a *removable discontinuity* at $ x = a $ if both of the following conditions hold:1. $ f $ is either not defined or not continuous at $ x = a $.2. $ f(a) $ could either be defined or redefined so that the new function *is* continuous at $ x = a $.---Show that \[ f(x) = \begin{cases} x^2 + 12x + 41 & \text{if } x < -6 \\-2 & \text{if } x = -6 \\-x^2 - 12x - 31 & \text{if } x > -6 \end{cases} \]has a removable discontinuity at $ x = -6 $ by(a) verifying (1) in the definition above, and then(b) verifying (2) in the definition above by determining a value of $ f(-6) $ that would make $ f $ continuous at $ x = -6 $.\[ f(-6) = \boxed{-2} \] would make $ f $ continuous at $ x = -6 $.Now draw a graph of $ f(x) $. It's just a couple of parabolas!### Explanation of the GraphThe graph of $ f(x) $ consists of two parabolas transitioning at $ x = -6 $:- For $ x < -6 $, the graph follows the parabola $ x^2 + 12x + 41 $.- At $ x = -6 $, the function is defined as $-2$.- For $ x > -6 $, the graph follows the inverted parabola $-x^2 - 12x - 31$.The removable discontinuity occurs because the limit of the function from both sides of $ x = -6 $ can be made equal by appropriately defining the value at $ x = -6 $.

Answered: Show that f(x) = x² + 12x +41 -2 -x²…

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Please help me solve this problem. Thank you!

A function $ f(x) $ is said to have a *removable discontinuity* at $ x = a $ if both of the following conditions hold:

1. $ f $ is either not defined or not continuous at $ x = a $.

2. $ f(a) $ could either be defined or redefined so that the new function *is* continuous at $ x = a $.

---

Show that

\[
f(x) =
\begin{cases}
x^2 + 12x + 41 & \text{if } x < -6 \\
-2 & \text{if } x = -6 \\
-x^2 - 12x - 31 & \text{if } x > -6
\end{cases}
\]

has a removable discontinuity at $ x = -6 $ by

(a) verifying (1) in the definition above, and then

(b) verifying (2) in the definition above by determining a value of $ f(-6) $ that would make $ f $ continuous at $ x = -6 $.

\[ f(-6) = \boxed{-2} \] would make $ f $ continuous at $ x = -6 $.

Now draw a graph of $ f(x) $. It's just a couple of parabolas!

### Explanation of the Graph

The graph of $ f(x) $ consists of two parabolas transitioning at $ x = -6 $:

- For $ x < -6 $, the graph follows the parabola $ x^2 + 12x + 41 $.
- At $ x = -6 $, the function is defined as $-2$.
- For $ x > -6 $, the graph follows the inverted parabola $-x^2 - 12x - 31$.

The removable discontinuity occurs because the limit of the function from both sides of $ x = -6 $ can be made equal by appropriately defining the value at $ x = -6 $.

Expert Solution

Step 1: Given the information

The given function,

$f open parentheses x close parentheses equals open curly brackets table row cell x squared plus 12 x plus 41 end cell cell i f space x less than negative 6 end cell row cell negative 2 end cell cell i f space x equals negative 6 end cell row cell negative x squared minus 12 x minus 31 end cell cell i f space x greater than negative 6 end cell end table close$

The aim is to prove the function has a removable discontinuous at $x equals negative 6$ and redefine the function at $x equals negative 6$ that makes the function continuous at the point.