Question

which best describes the range of the function $f(x) = 2(3)^x$? options: $y \geq 2$, $y > 2$, $y > 0$, $y \geq 0$

Explanation:

Step1: Analyze the exponential function \(3^x\)

The exponential function \(a^x\) (where \(a>1\)) has a range of \(y > 0\) for all real numbers \(x\). For \(y = 3^x\), since the base \(3>1\), the range of \(3^x\) is \(y>0\).

Step2: Analyze the function \(f(x)=2(3)^x\)

When we multiply the function \(3^x\) by \(2\) (a positive constant), we are vertically stretching the graph of \(3^x\) by a factor of \(2\). Since multiplying a positive number (the output of \(3^x\)) by \(2\) still results in a positive number, and as \(x\) varies over all real numbers:

• If \(x\) approaches \(-\infty\), \(3^x\) approaches \(0\), so \(2(3)^x\) approaches \(0\) (but never actually reaches \(0\) because \(3^x>0\) for all \(x\), so \(2(3)^x>0\)).
• If \(x\) increases, \(3^x\) increases, so \(2(3)^x\) also increases without bound.

So the range of \(f(x) = 2(3)^x\) is \(y>0\).

Answer:

\(y > 0\) (the option with \(y > 0\))