Problem 21 - Algebra and Trigonometry Real Mathematics, Real People

Question

Use a graphing utility to graph \(f\) and \(g\) in the same viewing window. Approximate the point of intersection of the graphs of \(f\) and \(g .\) Then solve the equation \(f(x)=g(x)\) algebraically. $$\begin{aligned}&f(x)=\ln e^{x+1}\\\&g(x)=2 x+5\end{aligned}$$

Step-by-Step Solution

Answer

The point of intersection approximated visually from the graph gives a preliminary idea of the solution. The algebraic solution for x is obtained by solving the equation \(x + 1 = 2x + 5\), which yields \(x = -4\).

Step 1 Graphing the functions

Use a graphing software to plot both functions \(f\) and \(g\). The function \(f(x) = \ln e^{x+1}\) and \(g(x) = 2x + 5\). Make sure both these functions are in the same viewing window.

Step 2 Approximating the intersection point

Inspect the graph visually to approximate the intersecting point. This point signifies where the two functions meet, which means the output of both functions is equal at this point (x-coordinate).

Step 3 Solving the equation

Solve the algebraic equation \(f(x) = g(x)\) mathematically by setting \(\ln e^{x+1} = 2x + 5\). Since the natural logarithm and the exponential function are inverse functions, \(\ln e^{x+1}\) simplifies to \(x + 1\). Now the equation \(x + 1 = 2x + 5\) can be solved for \(x\).

Key Concepts

Graphing Functions

Graphing functions is a fundamental aspect of algebra and calculus. It involves plotting a visual representation of equations on a coordinate plane. In this exercise, we graph two functions:

• \( f(x) = \ln e^{x+1} \)
• \( g(x) = 2x + 5 \)

Using graphing software or a calculator, you can input both functions to see how they behave. Both functions should appear in the same viewing window, allowing you to examine their shapes and where they might intersect. Visualizing these graphs helps in understanding the behavior of each function over the range of \(x\) values.

Points of Intersection

When graphing two functions together, the point of intersection is where they cross each other on the graph. This point is significant because it represents the solution to the equation \( f(x) = g(x) \). To find this point:

• Look for where both graphs intersect.
• The \( x \)-coordinate at the intersection provides the solution to the equation \( f(x) = g(x) \).
• The \( y \)-coordinate shows the common value of both functions at that \( x \) value.

In our case, the equation \( \ln e^{x+1} = 2x + 5 \) visualized shows where the two expressions produce the same outcome. Approximating this intersection through graphing is an initial step before solving it algebraically.

Natural Logarithms

Natural logarithms are logarithms with the base \( e \), where \( e \approx 2.718 \). They are denoted as \( \ln \). In the problem, we encounter \( \ln e^{x+1} \). A key property of logarithms is that the logarithm of an exponential with the same base simplifies the expression.

• The property \( \ln e^{k} = k \) allows us to simplify \( \ln e^{x+1} \) to \( x + 1 \).

This property makes complex expressions more manageable, simplifying equation solving. Understanding this property and its application is crucial for solving many algebraic problems involving logarithms and exponentials.

Exponential Functions

Exponential functions involve the expression \( e^{x} \), where \( e \) is a constant approximately equal to 2.718. They are characterized by rapid growth or decay. The function \( f(x) = \ln e^{x+1} \) has such an exponential part.

• Exponential functions are the inverses of logarithmic functions.
• This means they "undo" each other, so \( \ln e^{x+1} \) simplifies to \( x+1 \).
• This simplification is used to equate \( f(x) \) and \( g(x) \), making solving for \( x \) possible.

Understanding exponential functions, especially their inverse relationships with logarithms, is vital for mastering equations that feature both types of expressions. This knowledge enables you to transition from graphing a function to solving its equations algebraically.