Question

what is the range of the function ( y = e^{4x} )?
( ylt0 )
( ygt0 )
( ylt4 )
( ygt4 )

Explanation:

Step1: Analyze the exponential function properties

The general form of an exponential function is \(y = a^{x}\), where \(a>0,a
eq1\). For the function \(y = e^{u}\), the base \(e\approx2.718>0\). The value of \(e^{u}\) is always positive for any real - valued \(u\). Here \(u = 4x\), and \(x\in R\).

Step2: Determine the range

Since \(4x\) can take any real value (because \(x\in R\)), and \(y = e^{4x}\), and \(e^{t}>0\) for all \(t\in R\) (let \(t = 4x\)). As \(x\to-\infty\), \(4x\to-\infty\) and \(\lim_{x\to-\infty}e^{4x}=0\) (approaching \(0\) but never reaching \(0\)), and as \(x\to+\infty\), \(4x\to+\infty\) and \(e^{4x}\to+\infty\).

Answer:

\(y>0\)