Problem 4 - College Algebra

Question

The graph of \(f(x)=\left(\frac{1}{2}\right)^{-x}\) is refl cted about the \(y\) -axis and compressed vertically by a factor of \(\frac{1}{5}\). What is the equation of the new function, \(g(x) ?\) State its \(y\) -intercept, domain, and range.

Step-by-Step Solution

Answer

The new function is \(g(x) = \frac{1}{5} \left(\frac{1}{2}\right)^x\); the \(y\)-intercept is \(\frac{1}{5}\), domain is \((-\infty, \infty)\), and range is \((0, \infty)\).

Step 1: Understand the Transformations

The original function is \(f(x)=\left(\frac{1}{2}\right)^{-x}\). To reflect about the \(y\)-axis, we replace \(x\) with \(-x\), so it becomes \(\left(\frac{1}{2}\right)^{x}\). Then, to compress this function vertically by a factor of \(\frac{1}{5}\), multiply the entire function by \(\frac{1}{5}\).

Step 2: Formulate the New Function

Apply the transformations to obtain the equation of the new function: \(g(x) = \frac{1}{5} \cdot \left(\frac{1}{2}\right)^{x}\).

Step 3: Identify the y-intercept

The \(y\)-intercept of a function is found by evaluating the function at \(x=0\). Substitute \(x=0\) into \(g(x)\): \(g(0) = \frac{1}{5} \cdot \left(\frac{1}{2}\right)^{0} = \frac{1}{5} \cdot 1 = \frac{1}{5}\). The \(y\)-intercept is \(\frac{1}{5}\).

Step 4: Determine the Domain

The domain of \(g(x)\) is the set of all possible \(x\)-values. Since \(\left(\frac{1}{2}\right)^{x}\) is defined for all real numbers, the domain of \(g(x)\) is \((-\infty, \infty)\).

Step 5: Determine the Range

The range is the set of all possible \(g(x)\)-values. Since \(\left(\frac{1}{2}\right)^{x}\) is always positive and the smallest value \(\left(\frac{1}{2}\right)^{x}\) approaches is 0 as \(x\to\infty\), the range of \(\frac{1}{5}\times\left(\frac{1}{2}\right)^{x}\) is \((0, \infty)\).

Key Concepts

Exponential Function

An exponential function is one where a constant base is raised to a variable exponent. In the given exercise, we start with the exponential function \( f(x) = \left(\frac{1}{2}\right)^{-x} \). Exponential functions like this are important in various fields due to their unique growth or decay curves.
Understanding transformations is key to manipulating these functions.

• **Base of the Exponent:** The number \(\frac{1}{2}\) in this case is the base. It is crucial because it determines the behavior of the function as \(x\) changes.
• **Negative Exponent:** Initially, the exponent is negative \(-x\), which reflects the function across the \(y\)-axis. It reverses the usual direction of growth to decay.
• **Transformation:** By replacing \(x\) with \(-x\), the function becomes \(\left(\frac{1}{2}\right)^{x} \), changing it from decreasing to increasing as \(x\) rises.

Once transformed, we compress it vertically by multiplying the function by \( \frac{1}{5}\). This reduces the slope and makes the growth less steep.

Y-Intercept

The \(y\)-intercept of a function is the point where the graph crosses the \(y\)-axis. For any function \(f(x)\), the \(y\)-intercept is found by evaluating \(f(x)\) at \(x = 0\).
Let’s consider our final transformed function: \(g(x) = \frac{1}{5} \cdot \left(\frac{1}{2}\right)^{x}\).
Here's how you find the \(y\)-intercept:

• **Substitute \(x = 0\):** Calculate \(g(0) = \frac{1}{5} \cdot \left(\frac{1}{2}\right)^{0}\).
• **Simplify:** Since any number to the power of 0 is 1, \(\left(\frac{1}{2}\right)^{0} = 1\).
• **Result:** Multiply to get \( \frac{1}{5}\), so \(g(0) = \frac{1}{5}\). This means the graph intersects the \(y\)-axis at \( \frac{1}{5}\).

The simplicity of calculating the \(y\)-intercept helps you quickly pinpoint characteristics of a graph.

Domain and Range

Domain and range describe the set of inputs (domain) and outputs (range) that a function can generate. For the exponential function \(g(x) = \frac{1}{5} \cdot \left(\frac{1}{2}\right)^{x}\), these sets tell us the nature of the function rigorously.
**Domain**
An exponential function has a domain consisting of all real numbers, \(( -\infty, \infty )\). This means there is no restriction on the \(x\)-values; every real number can be substituted into the function.
**Range**

• Since \(\left(\frac{1}{2}\right)^{x}\) always produces positive values and gets closer to zero as \(x\) gets larger, \(g(x)\) itself is always positive as well.
• This is multiplied by \(\frac{1}{5}\), proportionally compressing the values between \(0\) and positive infinity.

Hence, the range of \(g(x)\) is \((0, \infty)\), signifying \(g(x)\) never reaches zero or becomes negative.