- Start with the property of the exponential function \(e^x > 0\).
- Apply the scalar transformation: \(2e^x > 0\).
- Apply the vertical shift: \(2e^x - 1 > -1\).
- Identify that the range is all real numbers greater than \(-1\), which corresponds to option \(\boxed{a}\).
Explanation
- Analysis of the Parent Function
The given function is \(y = 2e^x - 1\). To find its range, we start by analyzing the properties of the parent exponential function, \(f(x) = e^x\). The function \(f(x) = e^x\) is defined for all real numbers \(x\), and its output is always strictly positive, meaning \(e^x > 0\) for all \(x \in \mathbb{R}\).
- Transformations and Range Determination
We apply transformations to the inequality \(e^x > 0\) to match the form of the given function \(y = 2e^x - 1\). First, we multiply the inequality by the positive constant 2:
\[2e^x > 2 \times 0\]
\[2e^x > 0\]
Next, we subtract 1 from both sides of the inequality:
\[2e^x - 1 > 0 - 1\]
\[2e^x - 1 > -1\]
Since \(y = 2e^x - 1\), this inequality shows that \(y > -1\).
- Conclusion and Final Answer Selection
The range of the function \(y = 2e^x - 1\) consists of all real numbers greater than \(-1\). Comparing this result to the given options:
Aall real numbers greater than \(-1\)
Ball real numbers less than \(-1\)
Call real numbers less than \(1\)
Dall real numbers greater than \(1\)
The correct choice is (a). The final answer is \(\boxed{a}\).
Examples
Understanding the range of functions is essential in fields like finance, where you might model the growth of an investment. If an account balance follows the function \(V(t) = 2e^t - 1\) (where \(V\) is value in thousands of dollars and \(t\) is time), knowing that the range is greater than -1 tells you that while the account balance can theoretically start just below zero (due to fees, for instance), it grows rapidly as time progresses. This helps in predicting potential financial outcomes and planning investments.