Question
Find the domains of the functions \( f \) and \( g \) - \[ \begin{array}{r}f(x)=\sqrt[4]{3 x-9} \\ g(x)=\sqrt[3]{2 x+4}\end{array} \] Write your answers using interval notation.
Ask by Klein Chang. in the United States
Mar 08,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
The domain of \( f(x) = \sqrt[4]{3x - 9} \) is \( [3, \infty) \) and the domain of \( g(x) = \sqrt[3]{2x + 4} \) is \( (-\infty, \infty) \).
Solution
1. For the function \( f(x) = \sqrt[4]{3x - 9} \), the expression under the fourth root must be non-negative:
\[
3x - 9 \geq 0
\]
Solving for \( x \):
\[
3x \geq 9 \quad \Rightarrow \quad x \geq 3
\]
Thus, the domain of \( f \) in interval notation is:
\[
[3, \infty)
\]
2. For the function \( g(x) = \sqrt[3]{2x + 4} \), the cube root is defined for all real numbers, so there is no restriction on \( x \). Hence, the domain of \( g \) in interval notation is:
\[
(-\infty, \infty)
\]
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Simplify this solution Mind Expander
To find the domain of the function \( f(x) = \sqrt[4]{3x - 9} \), we need to ensure that the expression inside the fourth root is non-negative. This means we need: \[ 3x - 9 \geq 0 \] Solving this inequality: \[ 3x \geq 9 \] \[ x \geq 3 \] Thus, the domain of \( f \) is \( [3, \infty) \). For the function \( g(x) = \sqrt[3]{2x + 4} \), there are no restrictions on the cube root, as it is defined for all real numbers. Therefore, the domain for \( g \) is: \[ (-\infty, \infty) \] In summary, the domains are: - \( f(x) \): \( [3, \infty) \) - \( g(x) \): \( (-\infty, \infty) \)
