Question

what is the range of the function ( y = 2e^{x}-1 )?
all real numbers less than -1
all real numbers greater than -1
all real numbers less than 1
all real numbers greater than 1

Explanation:

Step1: Analyze the range of \(e^x\)

The exponential function \(e^x\) has a range of \((0,\infty)\), i.e., \(e^x>0\).

Step2: Analyze the range of \(2e^x\)

Multiply the inequality \(e^x > 0\) by \(2\). Using the property of inequalities (if \(a>b\) and \(c>0\), then \(ca>cb\)), we get \(2e^x>0\).

Step3: Analyze the range of \(y = 2e^x-1\)

Subtract \(1\) from the inequality \(2e^x>0\). Using the property of inequalities (if \(a > b\), then \(a - c>b - c\)), we have \(2e^x-1>0 - 1\). So \(y> - 1\).

Answer:

all real numbers greater than \(-1\)