Problem 6 - Precalculus : Building Concepts and Connections

Question

Complete them to review topics relevant to the remaining exercises. Let \(f(x)=\left(\frac{1}{2}\right)^{x} .\) As \(x \rightarrow-\infty, f(x) \rightarrow\)__________.

Step-by-Step Solution

Answer

As \(x \rightarrow-\infty, f(x) \rightarrow \infty\).

Step 1: Understanding the Function

The function given to us is an exponential function, more specifically \(f(x)=\left(\frac{1}{2}\right)^{x}\). In this, the base is a fraction (1/2), hence this is a decaying exponential function.

Step 2: Analyzing the Function as \(x\) Approaches Negative Infinity

As \(x\) approaches negative infinity, this means that \(x\) is becoming larger but in the negative direction. For a decaying function like \(f(x)=\left(\frac{1}{2}\right)^{x}\), as \(x\) decreases, the value of the function actually increases.

Step 3: Determining the Limit

So, as \(x\) decreases to negative infinity, the value of the function \(f(x)=\left(\frac{1}{2}\right)^{x}\) will keep increasing. Thus we can say the value of \(f(x)\) approaches positive infinity as \(x\) approaches negative infinity.

Key Concepts

Decay Function

One of the core characteristics of many exponential functions is their behavior as decay functions. In the case of the function \( f(x)=\left(\frac{1}{2}\right)^{x} \), we have a decay function because the base of the exponent is a fraction less than 1. This means the function decreases as \( x \) becomes larger (but positive). However, an important point is how the function behaves for negative values of \( x \).

As \( x \) grows more negative, the exponent effectively turns into a form of multiplication resulting in increasing values of the function. A decay function like \( f(x)=\left(\frac{1}{2}\right)^{x} \) will shrink towards zero for positive \( x \). Conversely, as \( x \) shifts to negative, it causes the function to grow towards higher positive values.

Limit as x Approaches Infinity

Calculating the limit of a function as \( x \) tends towards infinity is crucial for understanding the behavior of the function at its extremes. For exponential functions, this entails examining how the function behaves as \( x \) either grows incredibly large or diminishes into negative infinity.

With our given function \( f(x)=\left(\frac{1}{2}\right)^{x} \), the focus is on \( x \) approaching negative infinity. Here, we see quite a counterintuitive result: as \( x \) grows larger in the negative direction, the fraction \( \left(\frac{1}{2}\right) \) raised to a negative number flips the fraction, effectively turning it into a format that multiplies for larger positive outputs. This leads the function to approach positive infinity as \( x \) goes further into negative infinity.

Negative Exponent Analysis

The concept of negative exponents is critical to understanding why some functions behave the way they do over different domains. A negative exponent essentially indicates the reciprocal of the base raised to the positive of that exponent. For instance, \( \left(\frac{1}{2}\right)^{-x} \) is equivalent to \( 2^x \).

In the function \( f(x)=\left(\frac{1}{2}\right)^{x} \), as \( x \) becomes more negative, it acts as though the base is being raised to an increasing positive power. Each increase (in the negative direction) in \( x \) effectively decreases the power, thus driving the results higher—a concept stabilizing our understanding of decay functions when extended into negative realms. Understanding this principle helps demonstrate why the function escalates rather than diminishes for negative values of \( x \).

• Negative exponents flip the base fraction.
• They result in the reciprocal power multiplication.
• This multiplication leads to increasing function values.