Question
Find the domain of the given function. a) f(x) = 1/6^x b) f(x) = 1/2^3x -2
Answers
GPT-4o mini
To find the domain of the given functions, we need to determine the set of all possible input values (x values) for which the functions are defined.
### a) \( f(x) = \frac{1}{6^x} \)
The function \( f(x) \) consists of a fraction where the denominator is \( 6^x \).
The base \( 6 \) raised to any real number \( x \) is always positive and never equals zero. Thus, \( 6^x \) is defined for all real numbers and can never be zero. So, the function \( \frac{1}{6^x} \) is also defined for all real numbers.
**Domain of \( f(x) = \frac{1}{6^x} \):**
\[
\text{All real numbers} \quad (-\infty, \infty)
\]
### b) \( f(x) = \frac{1}{2^{3x} - 2} \)
In this function, the denominator is \( 2^{3x} - 2 \). For this function to be defined, the denominator must not equal zero.
To find where the denominator is zero, we set it equal to zero and solve for \( x \):
\[
2^{3x} - 2 = 0
\]
\[
2^{3x} = 2
\]
\[
3x = 1 \quad \text{(taking logarithm base 2 on both sides)}
\]
\[
x = \frac{1}{3}
\]
Thus, \( f(x) \) is undefined at \( x = \frac{1}{3} \).
The function is defined for all other real numbers, so the domain can be expressed as all real numbers except \( \frac{1}{3} \).
**Domain of \( f(x) = \frac{1}{2^{3x} - 2} \):**
\[
(-\infty, \frac{1}{3}) \cup (\frac{1}{3}, \infty)
\]
### Summary of Domains:
- a) \( f(x) = \frac{1}{6^x} \): \( (-\infty, \infty) \)
- b) \( f(x) = \frac{1}{2^{3x} - 2} \): \( (-\infty, \frac{1}{3}) \cup (\frac{1}{3}, \infty) \)
### a) \( f(x) = \frac{1}{6^x} \)
The function \( f(x) \) consists of a fraction where the denominator is \( 6^x \).
The base \( 6 \) raised to any real number \( x \) is always positive and never equals zero. Thus, \( 6^x \) is defined for all real numbers and can never be zero. So, the function \( \frac{1}{6^x} \) is also defined for all real numbers.
**Domain of \( f(x) = \frac{1}{6^x} \):**
\[
\text{All real numbers} \quad (-\infty, \infty)
\]
### b) \( f(x) = \frac{1}{2^{3x} - 2} \)
In this function, the denominator is \( 2^{3x} - 2 \). For this function to be defined, the denominator must not equal zero.
To find where the denominator is zero, we set it equal to zero and solve for \( x \):
\[
2^{3x} - 2 = 0
\]
\[
2^{3x} = 2
\]
\[
3x = 1 \quad \text{(taking logarithm base 2 on both sides)}
\]
\[
x = \frac{1}{3}
\]
Thus, \( f(x) \) is undefined at \( x = \frac{1}{3} \).
The function is defined for all other real numbers, so the domain can be expressed as all real numbers except \( \frac{1}{3} \).
**Domain of \( f(x) = \frac{1}{2^{3x} - 2} \):**
\[
(-\infty, \frac{1}{3}) \cup (\frac{1}{3}, \infty)
\]
### Summary of Domains:
- a) \( f(x) = \frac{1}{6^x} \): \( (-\infty, \infty) \)
- b) \( f(x) = \frac{1}{2^{3x} - 2} \): \( (-\infty, \frac{1}{3}) \cup (\frac{1}{3}, \infty) \)