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$Which of the following functions are continuous at x=3 ? {:[g(x)=ln(x-3)],[f(x)=e^(x-3)]:} Choose 1 answer: (A) g only (B) f only (C) Both g and f (D) Neither g nor f$

Which of the following functions are continuous at $x=3$ ? $\newline$ $\begin{array}{l} g(x)=\ln (x-3) \\ f(x)=e^{x-3} \end{array}$ $\newline$ Choose $1$ answer: $\newline$ (A) $g$ only $\newline$ (B) $f$ only $\newline$ (C) Both $g$ and $f$ $\newline$ (D) Neither $g$ nor $f$

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Math Problems

Algebra 2

Domain and range of absolute value functions: equations

Full solution

Q. Which of the following functions are continuous at

x=3

\newline

\begin{array}{l} g(x)=\ln (x-3) \\ f(x)=e^{x-3} \end{array}

\newline

Choose

1

answer:

\newline

(A)

g

only

\newline

(B)

f

only

\newline

g

and

f

\newline

(D) Neither

g

nor

f

Checking for Continuity: To determine if the functions are continuous at $x=3$ , we need to check if they are defined at that point and if there are any discontinuities such as holes, jumps, or vertical asymptotes.
Analysis of $g(x)$ : Let's first consider $g(x) = \ln(x-3)$ . The natural logarithm function $\ln(x)$ is continuous for all x > 0. However, for $g(x)$ , the argument of the logarithm is $(x-3)$ , which means $g(x)$ is only defined for x > 3. At $x = 3$ , the function $g(x)$ is not defined because $g(x) = \ln(x-3)$ $0$ is undefined. Therefore, $g(x)$ has a discontinuity at $x = 3$ .
Analysis of $f(x)$ : Now let's consider $f(x) = e^{(x-3)}$ . The exponential function $e^x$ is continuous for all real numbers. Since the transformation $x-3$ simply shifts the graph horizontally, it does not introduce any discontinuities. Therefore, $f(x)$ is continuous at $x = 3$ because $e^{(3-3)} = e^0 = 1$ , which is defined.
Conclusion: Based on the analysis, $g(x)$ is not continuous at $x = 3$ because it is not defined at that point, while $f(x)$ is continuous at $x = 3$ .