Problem 64 - Precalculus with Calculus Previews

Question

Graph the given functions. Determine the approximate \(x\) -coordinates of the points of intersection of their graphs. $$ f(x)=\frac{1}{3} \cdot 2^{x^{2}}, \quad g(x)=2^{x^{2}}-1 $$

Step-by-Step Solution

Answer

The graphs intersect at approximately \( x \approx \pm 1 \).

Step 1: Graph Preparation

To start, understand the problem by identifying the functions: \( f(x) = \frac{1}{3} \cdot 2^{x^2} \) and \( g(x) = 2^{x^2} - 1 \). These are both exponential functions with a base of 2 and depend on \( x^2 \), making them symmetric about the y-axis. We'll plot them on the same set of axes for comparison.

Step 2: Plot Function f(x)

To graph \( f(x) = \frac{1}{3} \cdot 2^{x^2} \), calculate a few points. For example, when \( x = 0 \), \( f(0) = \frac{1}{3} \cdot 2^0 = \frac{1}{3} \). When \( x = 1 \), \( f(1) = \frac{1}{3} \cdot 2^1 = \frac{2}{3} \). Plot several points, including negative and positive values of \( x \), to capture the shape of the curve.

Step 3: Plot Function g(x)

To graph \( g(x) = 2^{x^2} - 1 \), again, calculate points. When \( x = 0 \), \( g(0) = 2^0 - 1 = 0 \). When \( x = 1 \), \( g(1) = 2^1 - 1 = 1 \). Plot additional points for negative and positive \( x \). Notice this graph shifts the exponential 1 unit downward.

Step 4: Identify Intersection Points

With the graphs plotted, look for the intersection points by comparing the graphs visually. These will occur where \( f(x) = g(x) \). If plotted accurately, intersections will visually become evident. Alternatively, solve \( \frac{1}{3} \cdot 2^{x^2} = 2^{x^2} - 1 \) to confirm graphically identified points.

Step 5: Approximate Solution

Refine your initial estimate by checking closely plotted points. Each graph should show symmetric behavior about the y-axis. The symmetry helps estimate solutions even if exact points aren't obvious. For example, expect intersection near \( x \approx \pm 1 \) as both calculations for nearby points will reveal similar values.

Key Concepts

Exponential Functions

Exponential functions are an essential concept in mathematics that involve quantities growing or decaying at constant rates. In these functions, a constant base is raised to a variable exponent. The general form of an exponential function is \( y = a \, b^{x} \), where \( a \) is a constant scaling factor, \( b \) is the base that is a positive real number, and \( x \) is the exponent. For the functions given:

• \( f(x) = \frac{1}{3} \cdot 2^{x^{2}} \)
• \( g(x) = 2^{x^{2}} - 1 \)

Both functions have a base of 2, indicating exponential growth behavior. However, note the unique characteristics:

• Scaling factor: The function \( f(x) \) includes a scaling factor of \( \frac{1}{3} \), which compresses the growth rate vertically.
• Shifted graph: The function \( g(x) \) includes a \(-1\) term, shifting the graph downward.

The variable \( x^2 \) is significant—it determines that the base is raised to a non-negative power, further emphasizing the growth symmetry as \( x \) changes. Exponential functions like these are prevalent in applications such as population growth, radioactive decay, and financial calculations.

Graphing

Graphing functions, especially exponential ones, involves plotting points to visualize their behavior on a coordinate plane. Let's focus on the steps to graph these functions.First, choose values for \( x \) within a reasonable range, typically around 0, where initial exponential changes are evident. Calculating a few points, for example:

• For \( f(x) = \frac{1}{3} \cdot 2^{x^{2}} \):
• \( x = 0 \), \( f(0) = \frac{1}{3} \cdot 2^{0} = \frac{1}{3} \)
• \( x = 1 \), \( f(1) = \frac{1}{3} \cdot 2^{1} = \frac{2}{3} \)
• \( x = -1 \), calculate similarly since the result is based on \( x^2 \)
• For \( g(x) = 2^{x^{2}} - 1 \):
• \( x = 0 \), \( g(0) = 2^{0} - 1 = 0 \)
• \( x = 1 \), \( g(1) = 2^{1} - 1 = 1 \)
• \( x = -1 \), calculate knowing the graph's symmetry

Plot these points to sketch each function's curve. Observations while graphing:

• \( f(x) \) will appear more compressed due to \( \frac{1}{3} \) scaling.
• \( g(x) \) appears similar but is shifted down by 1 unit.

Graph intersections indicate where \( f(x) = g(x) \). Since they occur where the graphs cross, visual estimates provide approximation points for intersections. Choose additional points or solve algebraically for better precision.

Symmetry of Functions

When dealing with exponential functions such as \( f(x) = \frac{1}{3} \cdot 2^{x^2} \) and \( g(x) = 2^{x^2} - 1 \), understanding symmetry makes graphing simpler.In our problem, both functions involve \( x^2 \) in their exponent, which results in inherent symmetry about the y-axis. This means:

• Even Function Nature: Both functions have identical values for \( x \) and \( -x \).
• Simplified Plotting: Knowing the right half of the graph automatically provides the left half due to symmetry.

This symmetry also aids in determining intersection points, since you can focus on just the positive \( x \)-axis and reflect your findings to the negative side.Symmetry allows for efficient graph checking and helps confirm calculations. Spotting the symmetry visually on your graphs is an effective way to double-check the correctness of your plots. For the given functions, quantify each intersection estimate by understanding the symmetrical structure, and refine the points through more detailed calculation if necessary.