Question

what are the domain and range of the function $f(x) = 4(\sqrt3{81})^x$?
○ {x| x is a real number}; {y| y > 0}
○ {x| x > 4}; {y| y > 0}
○ {x| x is a real number}; {y| y > 4}
○ {x| x > 4}; {y| y > 4}

Explanation:

Step1: Simplify the base

First, simplify $\sqrt[3]{81}$. We know $81=3^4$, so $\sqrt[3]{81}=3^{\frac{4}{3}}$, which is a positive real number greater than 1.

Step2: Determine the domain

For exponential functions $a \cdot b^x$ where $b>0, b
eq1$, $x$ can be any real number. So the domain is all real numbers.

Step3: Determine the range

Since $3^{\frac{4}{3}}>0$, $(3^{\frac{4}{3}})^x>0$ for all real $x$. Multiply by 4: $4\cdot(3^{\frac{4}{3}})^x>0$. So the range is all positive real numbers.

Answer:

$\{x| x \text{ is a real number}\}; \{y| y > 0\}$