Problem 62 - Precalculus with Calculus Previews

Question

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

Step-by-Step Solution

Answer

The graphs intersect at approximately \(x \approx 0.585\).

Step 1: Graph the Functions

To start, graph the functions based on their equations. For \(f(x) = 2^x\), it is an exponential function with a base of 2, which means it will grow rapidly as \(x\) increases. For \(g(x) = 3 - 2^x\), notice this is a horizontal reflection of \(f(x)\) shifted up by 3 units. Graph both functions on the same set of axes.

Step 2: Estimate Intersection Points

Observe the graphs of both functions. Identify the points where the graphs intersect. These points are where \(f(x) = g(x)\). Approximately, this seems to occur at \(x \approx 1\).

Step 3: Solve Algebraically for Intersection

Set the functions equal to each other to find the exact intersection point: \(2^x = 3 - 2^x\). Solve for \(x\) by first rearranging to have \(2 \times 2^x = 3\), leading to \(2^x = \frac{3}{2}\).

Step 4: Solve for \(x\)

To solve \(2^x = \frac{3}{2}\), take the logarithm of both sides. This gives \(x \cdot \ln(2) = \ln\left(\frac{3}{2}\right)\). Finally, solve for \(x\) by dividing both sides by \(\ln(2)\): \[ x = \frac{\ln(\frac{3}{2})}{\ln(2)} \approx 0.585 \].

Step 5: Verification

Plug \(x = 0.585\) into both original functions to ensure they yield approximately the same value. This confirms the x-coordinate of the point of intersection.

Key Concepts

Intersection points

When graphing functions, finding the intersection points is crucial as they represent the values where both functions are equal.
For the given functions, the intersections signify the exact intersections of their graphs on a coordinate plane. These points show where the equation of one function equals the other.
In this exercise, we focus on the functions:

• \( f(x) = 2^x \): a basic exponential curve growing as \(x\) increases,
• \( g(x) = 3 - 2^x \): this function mirrors \( f(x) \) and then moves it upwards by 3 units.

The intersection is found by solving where \( f(x) = g(x) \) algebraically, giving us the approximate \( x \)-value of 0.585.
This point shows that at \( x = 0.585 \), both functions produce the same output, confirming the intersection's location.

Exponential Functions

Exponential functions are powerful tools in mathematics, describing growth with a consistent base raised to the power of the variable.
They show rapid changes and are often used to model real-life phenomena such as population growth, compound interest, or radioactive decay.
For example, the function \( f(x) = 2^x \) in our exercise illustrates:

• A base of 2 which indicates that for every increase in \(x\), \( f(x) \) doubles.
• A graph that is always increasing, reflecting its rapid growth.

Understanding the behavior and graphing of such functions allows us to appreciate their impact and solve various mathematical problems effectively.

Logarithmic Equations

Logarithmic equations are the inverses of exponential functions. They are essential for solving equations involving exponential terms.
In the context of this exercise, solving \( 2^x = \frac{3}{2} \) involves taking the logarithm of both sides:
Use the natural logarithm (ln) to simplify the equation as follows:

• \( x \cdot \ln(2) = \ln\left(\frac{3}{2}\right) \): Here, \(x\) is isolated by using the property of logarithms which states that \( \ln(a^b) = b \cdot \ln(a) \).
• Solutions are derived by dividing both sides by \( \ln(2) \), leading to:\[ x = \frac{\ln\left(\frac{3}{2}\right)}{\ln(2)} \approx 0.585 \].

This value is the \( x \) coordinate where the graphs of \( f(x) \) and \( g(x) \) intersect.
Logarithms make it possible to simplify and solve exponential equations, offering a clear path to understanding the points of intersection.