Problem 25 - College Algebra with Modeling and Visualization

Question

Comparing Growth Which function becomes larger for \(0 \leq x \leq 10: f(x)=2 x+1\) or \(g(x)=2^{-x} ?\)

Step-by-Step Solution

Answer

In the range \(0 \leq x \leq 10\), function \(f(x)=2x+1\) becomes larger.

Step 1: Understand the problem

We need to compare the values of two functions, \(f(x) = 2x + 1\) and \(g(x) = 2^{-x}\), for \(x\) in the range \(0 \leq x \leq 10\). We will find which function has greater values in this interval.

Step 2: Evaluate the function at the endpoints

First, let's calculate \(f(x)\) and \(g(x)\) at the endpoints of our range. For \(x = 0\), \(f(0) = 2(0) + 1 = 1\) and \(g(0) = 2^{-0} = 1\). For \(x = 10\), \(f(10) = 2(10) + 1 = 21\) and \(g(10) = 2^{-10} = \frac{1}{1024}\). At \(x = 10\), \(f(x)\) is much larger than \(g(x)\).

Step 3: Compare growth of functions

Function \(f(x) = 2x + 1\) is a linear function with a constant positive slope, so its value increases steadily as \(x\) increases. Function \(g(x) = 2^{-x}\) is an exponential decay function, meaning it decreases as \(x\) increases. Thus, \(f(x)\) will surpass \(g(x)\) as \(x\) gets larger.

Step 4: Compare values for small values of x

For \(x=0\), both functions are equal. As \(x\) increases, \(f(x) = 2x + 1\) grows linearly, while \(g(x) = 2^{-x}\) quickly diminishes. Even at \(x=1\), \(f(1) = 3\) and \(g(1) = \frac{1}{2}\), \(f(x)\) is already larger.

Step 5: Conclusion

After evaluating both functions across the range, it is clear that \(f(x) = 2x + 1\) becomes larger than \(g(x) = 2^{-x}\) and maintains this for all \(x\) in the range from 0 to 10.

Key Concepts

Linear Functions

A linear function is a type of mathematical function that demonstrates a constant rate of change. The general form of a linear function is given by \( y = mx + b \), where:

• \( m \) represents the slope, or the rate of change, indicating how much y changes for a unit change in x.
• \( b \) represents the y-intercept, which is the value of y when x equals zero.

These functions graph as straight lines, and their predictability makes them one of the simplest to understand. In our exercise, the function \( f(x) = 2x + 1 \) has a slope of 2, meaning it rises steadily as x increases.
This consistency allows us to easily predict its values at any point along the given interval. By understanding that the slope remains constant, we can say that \( f(x) \) will consistently increase by 2 for every increase of 1 in x.

Exponential Functions

Exponential functions exhibit different behavior compared to linear functions, particularly in how they grow or decay. Typically, these functions are represented by the formula \( y = a \, b^x \), where:

• \( a \) is a constant coefficient, affecting the function's initial value or height at \( x = 0 \).
• \( b \) is the base of the exponential, determining growth if \( b > 1 \), or decay if \( 0 < b < 1 \).

In this exercise, our exponential function \( g(x) = 2^{-x} \) is an example of exponential decay. As \( x \) increases, the value of \( g(x) \) rapidly decreases.
This difference in behavior from linear functions can be critical when comparing two function types over an interval, especially as exponential decay can quickly approach zero, significantly affecting its comparative values to a growing linear function.

Interval Evaluation

Interval evaluation involves analyzing the behavior of functions over a specific range of values for the variable x. This process helps in understanding how functions behave relatively and interact over certain intervals.
In the context of our exercise, the interval of interest is \( 0 \leq x \leq 10 \). We start by evaluating both functions \( f(x) = 2x + 1 \) and \( g(x) = 2^{-x} \) at the boundaries, which offers insight into their initial and ending behavior.

• At \( x = 0 \), both functions equal 1, showing they start off equal.
• As \( x \) grows towards 10, \( f(x) \) steadily increases to 21, while \( g(x) \) rapidly diminishes towards \( \frac{1}{1024} \).

By evaluating the functions over the entire interval, it's apparent that the linear function \( f(x) \) overtakes and maintains larger values than the exponential decay function \( g(x) \). This evaluation illustrates how diverse functions respond differently within a fixed range of x values.