which best describes the range of the function f(x) = 2(3)^x?\no y>0\no y≥0\no y>2\no y≥2

Answer:

A. $y>0$

Explanation:

Step1: Analyze the exponential - function property

For the general form of an exponential function $y = a\cdot b^{x}$ ($a\neq0$, $b>0$, $b\neq1$), here $a = 2$ and $b = 3$.

Step2: Consider the value of $3^{x}$

Since $3>1$, for any real - number $x$, $3^{x}>0$.

Step3: Find the range of $y = 2(3)^{x}$

Multiply both sides of $3^{x}>0$ by 2. We get $y=2(3)^{x}>0$. So the range of the function $f(x)=2(3)^{x}$ is $y > 0$.