Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

$Which of the following functions are continuous at x=1 ? {:[f(x)=e^(x)-1],[g(x)=ln(e^(x)-1)]:} Choose 1 answer: (A) f only (B) g only (C) Both f and g (D) Neither f nor g$

Which of the following functions are continuous at $x=1$ ? $\newline$ $\begin{array}{l} f(x)=e^{x}-1 \\ g(x)=\ln \left(e^{x}-1\right) \end{array}$ $\newline$ Choose $1$ answer: $\newline$ (A) $f$ only $\newline$ (B) $g$ only $\newline$ (C) Both $f$ and $g$ $\newline$ (D) Neither $f$ nor $g$

View step-by-step help

Home

Math Problems

Algebra 2

Domain and range of absolute value functions: equations

Full solution

Q. Which of the following functions are continuous at

x=1

\newline

\begin{array}{l} f(x)=e^{x}-1 \\ g(x)=\ln \left(e^{x}-1\right) \end{array}

\newline

Choose

1

answer:

\newline

(A)

f

only

\newline

(B)

g

only

\newline

f

and

g

\newline

(D) Neither

f

nor

g

Consider function $f(x)$ : Let's first consider the function $f(x) = e^x - 1$ . To determine if $f$ is continuous at $x=1$ , we need to check if the limit of $f(x)$ as $x$ approaches $1$ is equal to $f(1)$ . Calculate the limit of $f(x)$ as $x$ approaches $1$ : $f(x) = e^x - 1$ $1$ . Now, calculate $f(1)$ : $f(x) = e^x - 1$ $3$ . Since the limit of $f(x)$ as $x$ approaches $1$ is equal to $f(1)$ , $f$ is continuous at $x=1$ .
Check continuity at $x=1$ : Now let's consider the function $g(x) = \ln(e^x - 1)$ . To determine if $g$ is continuous at $x=1$ , we need to check if the limit of $g(x)$ as $x$ approaches $1$ is equal to $g(1)$ . Calculate the limit of $g(x)$ as $x$ approaches $1$ : $g(x) = \ln(e^x - 1)$ $1$ . Since $g(x) = \ln(e^x - 1)$ $2$ is always positive, $g(x) = \ln(e^x - 1)$ $3$ is positive for $g(x) = \ln(e^x - 1)$ $4$ . Therefore, the natural logarithm is defined for $g(x) = \ln(e^x - 1)$ $3$ when $x=1$ . Now, calculate $g(1)$ : $g(x) = \ln(e^x - 1)$ $8$ . Since $g(x) = \ln(e^x - 1)$ $9$ is positive, $g$ $0$ is defined, and the limit of $g(x)$ as $x$ approaches $1$ is equal to $g(1)$ , $g$ is continuous at $x=1$ .