Let g(x)={[e^(x)," for "x -1]:} Is g continuous at x=-1 ? Choose 1 answer: (A) Yes (B) No

Check Left-Hand Limit: To determine if g(x) is continuous at x=-1, we need to check if the left-hand limit, the right-hand limit, and the value of the function at x=-1 are all equal.Find Right-Hand Limit: First, let's find the left-hand limit as x approaches -1 from the left. Since for x -1, g(x) = -e^(-x), so the right-hand limit is -e^(-(-1)) which simplifies to -e^(1).Compare Limits and Value: Next, we need to find the value of the function at x=-1. Since x=-1 falls in the first case of the piecewise function where x <= -1, g(-1) = e^(-1).Conclusion: Now we compare the left-hand limit, right-hand limit, and the value of the function at x=-1. The left-hand limit is e^(-1), the right-hand limit is -e^(1), and the value of the function at x=-1 is e^(-1).Since the left-hand limit e^(-1) is not equal to the right-hand limit -e^(1), the function g(x) is not continuous at x=-1.

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

$Let g(x)={[e^(x)," for "x <= -1],[-e^(-x)," for "x > -1]:} Is g continuous at x=-1 ? Choose 1 answer: (A) Yes (B) No$

Let $\newline$ $g(x)=\left\{\begin{array}{ll} e^{x} & \text { for } x \leq-1 \\ -e^{-x} & \text { for } x>-1 \end{array}\right.$ $\newline$ Is $g$ continuous at $x=-1$ ? $\newline$ Choose $1$ answer: $\newline$ (A) Yes $\newline$ (B) No

View step-by-step help

Home

Math Problems

Calculus

Intermediate Value Theorem

Full solution

Q. Let

\newline

g(x)=\left\{\begin{array}{ll} e^{x} & \text { for } x \leq-1 \\ -e^{-x} & \text { for } x>-1 \end{array}\right.

\newline

g

continuous at

x=-1

\newline

Choose

1

answer:

\newline

(A) Yes

\newline

(B) No

Check Left-Hand Limit: To determine if $g(x)$ is continuous at $x=-1$ , we need to check if the left-hand limit, the right-hand limit, and the value of the function at $x=-1$ are all equal.
Find Right-Hand Limit: First, let's find the left-hand limit as $x$ approaches $-1$ from the left. Since for $x \leq -1$ , $g(x) = e^{x}$ , the left-hand limit is $e^{-1}$ .
Calculate Function Value: Now, let's find the right-hand limit as $x$ approaches $-1$ from the right. For x > -1, $g(x) = -e^{-x}$ , so the right-hand limit is $-e^{-(-1)}$ which simplifies to $-e^{1}$ .
Compare Limits and Value: Next, we need to find the value of the function at $x=-1$ . Since $x=-1$ falls in the first case of the piecewise function where $x \leq -1$ , $g(-1) = e^{-1}$ .
Conclusion: Now we compare the left-hand limit, right-hand limit, and the value of the function at $x=-1$ . The left-hand limit is $e^{-1}$ , the right-hand limit is $-e^{1}$ , and the value of the function at $x=-1$ is $e^{-1}$ .
Conclusion: Now we compare the left-hand limit, right-hand limit, and the value of the function at $x=-1$ . The left-hand limit is $e^{-1}$ , the right-hand limit is $-e^{1}$ , and the value of the function at $x=-1$ is $e^{-1}$ .Since the left-hand limit $e^{-1}$ is not equal to the right-hand limit $-e^{1}$ , the function $g(x)$ is not continuous at $x=-1$ .