lus/x9e81a4f98389efdf:limits-and-continuity/x9e81a4f98389efdf:removing-discontinuities/e/continuity?classCode=ED3AZEZWI A schematic reacti... Google Drive Dashboard North Wake Hall P... Blackboard Learn khan academy AP Khan Academy Get Al Tutoring NEW 주 Donate [ד jtolentino-1 Removable discontinuities Google Classroom Let f(x)=(x+1)/(xe^(x)+e^(x)) when x!=-1. f is continuous for all real numbers. Find f(-1). Choose 1 answer: (A) 1 (B) e (c) 2 Start over 2 of 4

Check Function Continuity: To find the value of f(-1), we need to determine if the function f(x) has a removable discontinuity at x = -1. If it does, we can find the limit of f(x) as x approaches -1 to find f(-1).Plug in x = -1: First, let's plug in x = -1 into the function to see if it is defined at that point. f(-1) = (-1 + 1) / ((-1)e^(-1) + e^(-1)) f(-1) = 0 / ((-1)/e + 1/e) f(-1) = 0 / ((-1 + 1)/e) f(-1) = 0 / 0 This is an indeterminate form, which means the function is not defined at x = -1, and there is a discontinuity at that point.Apply L'Hôpital's Rule: Since f(x) is continuous for all real numbers and there is a discontinuity at x = -1, it must be a removable discontinuity. To find the value of f(-1), we need to calculate the limit of f(x) as x approaches -1.Calculate Limit: We can use L'Hôpital's Rule to evaluate the limit since we have an indeterminate form of 0/0. We will differentiate the numerator and the denominator separately. The derivative of the numerator with respect to x is: d/dx (x + 1) = 1 The derivative of the denominator with respect to x is: d/dx (xe^(x) + e^(x)) = e^(x) + xe^(x) + e^(x) = e^(x) + xe^(x) + e^(x)Final Value of f(-1): Now we apply L'Hôpital's Rule by taking the limit of the derivatives as x approaches -1: limit as x -> -1 of (1) / (e^(x) + xe^(x) + e^(x)) Plugging in x = -1, we get: limit = 1 / (e^(-1) + (-1)e^(-1) + e^(-1)) limit = 1 / ((1/e) - (1/e) + (1/e)) limit = 1 / (1/e) limit = eTherefore, the value of f(-1) is e, which means the correct answer is (B) e.

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lus/x9e81a4f98389efdf:limits-and-continuity/x9e81a4f98389efdf:removing-discontinuities/e/continuity?classCode=ED3AZEZWI
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Donate [ד
jtolentino-1
Removable discontinuities
Google Classroom
Let
f(x)=(x+1)/(xe^(x)+e^(x)) when
x!=-1.

f is continuous for all real numbers.
Find
f(-1).
Choose 1 answer:
(A) 1
(B)
e
(c) 2
Start over 2 of 4

Let $f(x)=\frac{x+1}{xe^{x}+e^{x}}$ when $x\neq-1$ . $\newline$ $f$ is continuous for all real numbers. $\newline$ Find $f(-1)$ . $\newline$ Choose $1$ answer: $\newline$ (A) $1$ $\newline$ (B) $e$ $\newline$ (C) $2$

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Full solution

Q. Let

f(x)=\frac{x+1}{xe^{x}+e^{x}}

when

x\neq-1

\newline

f

is continuous for all real numbers.

\newline

Find

f(-1)

\newline

Choose

1

answer:

\newline

(A)

1

\newline

(B)

e

\newline

(C)

2

Check Function Continuity: To find the value of $f(-1)$ , we need to determine if the function $f(x)$ has a removable discontinuity at $x = -1$ . If it does, we can find the limit of $f(x)$ as $x$ approaches $-1$ to find $f(-1)$ .
Plug in $x = -1$ : First, let's plug in $x = -1$ into the function to see if it is defined at that point. $\newline$ $f(-1) = (-1 + 1) / ((-1)e^{-1} + e^{-1})$ $\newline$ $f(-1) = 0 / ((-1)/e + 1/e)$ $\newline$ $f(-1) = 0 / ((-1 + 1)/e)$ $\newline$ $f(-1) = 0 / 0$ $\newline$ This is an indeterminate form, which means the function is not defined at $x = -1$ , and there is a discontinuity at that point.
Apply L'Hôpital's Rule: Since $f(x)$ is continuous for all real numbers and there is a discontinuity at $x = -1$ , it must be a removable discontinuity. To find the value of $f(-1)$ , we need to calculate the limit of $f(x)$ as $x$ approaches $-1$ .
Calculate Limit: We can use L'Hôpital's Rule to evaluate the limit since we have an indeterminate form of $0/0$ . We will differentiate the numerator and the denominator separately. $\newline$ The derivative of the numerator with respect to $x$ is: $\newline$ $\frac{d}{dx} (x + 1) = 1$ $\newline$ The derivative of the denominator with respect to $x$ is: $\newline$ $\frac{d}{dx} (xe^{x} + e^{x}) = e^{x} + xe^{x} + e^{x} = 2e^{x} + xe^{x}$
Final Value of $f(-1)$ : Now we apply L'Hôpital's Rule by taking the limit of the derivatives as $x$ approaches $-1$ :
$\lim_{x \to -1} \frac{1}{e^{x} + xe^{x} + e^{x}}$
Plugging in $x = -1$ , we get:
$\lim = \frac{1}{e^{-1} + (-1)e^{-1} + e^{-1}}$
$\lim = \frac{1}{(\frac{1}{e}) - (\frac{1}{e}) + (\frac{1}{e})}$
$\lim = \frac{1}{\frac{1}{e}}$
$\lim = e$
Final Value of $f(-1)$ : Now we apply L'Hôpital's Rule by taking the limit of the derivatives as $x$ approaches $-1$ :
$\lim_{x \to -1} \frac{1}{e^{x} + xe^{x} + e^{x}}$
Plugging in $x = -1$ , we get:
$\lim = \frac{1}{e^{-1} + (-1)e^{-1} + e^{-1}}$
$\lim = \frac{1}{(\frac{1}{e}) - (\frac{1}{e}) + (\frac{1}{e})}$
$\lim = \frac{1}{\frac{1}{e}}$
$\lim = e$ Therefore, the value of $f(-1)$ is $x$ $0$ , which means the correct answer is (B) $x$ $0$ .