Problem 29 - Calculus with Concepts in Calculus

Question

Determine at which points the graphs of the given pair of functions intersect. $$ f(x)=e^{x} \text { and } g(x)=e^{1-x} \text { (Hint: Take natural logarithms.) } $$

Step-by-Step Solution

Answer

The graphs intersect at the point \( \left( \frac{1}{2}, e^{1/2} \right) \).

Step 1: Set the Functions Equal

To find the points of intersection of the graphs of the functions, we must set \( f(x) = g(x) \). Thus, we have \( e^x = e^{1-x} \).

Step 2: Apply Natural Logarithms

To solve the equation \( e^x = e^{1-x} \), take the natural logarithm of both sides. This yields \( \ln(e^x) = \ln(e^{1-x}) \).

Step 3: Simplify Using Logarithm Properties

Use the property \( \ln(e^a) = a \) to simplify both sides of the equation. This gives us \( x = 1-x \).

Step 4: Solve the Linear Equation

Solve the equation \( x = 1 - x \). Add \( x \) to both sides to obtain \( 2x = 1 \). Now, divide both sides by 2 to find \( x = \frac{1}{2} \).

Step 5: Find the Intersection Point

Substitute \( x = \frac{1}{2} \) into either original function to find the y-coordinate. Using \( f(x) = e^x \), we get \( f\left(\frac{1}{2}\right) = e^{1/2} \). Thus, the point of intersection is \( \left( \frac{1}{2}, e^{1/2} \right) \).

Key Concepts

Natural Logarithms

Natural logarithms are logarithms with the base of the mathematical constant \( e \), which is approximately 2.71828. The natural logarithm of a number \( x \) is denoted as \( \ln(x) \). One of the main properties of natural logarithms is that \( \ln(e^a) = a \). This means that taking the logarithm of a number with base \( e \) simply gives us the exponent back.

Natural logarithms are the inverse operation of exponentiation when the base is \( e \). This makes them extremely useful in solving equations where the exponential function is involved. For instance, if you have an equation \( e^y = x \), then \( y = \ln(x) \).

• Natural logarithms help convert exponential equations to a linear form, making them easier to solve.
• The natural log function \( \ln(x) \) is undefined for \( x \leq 0 \) since exponentials do not produce these values.

Exponential Functions

Exponential functions are a type of mathematical function where a constant base is raised to a variable exponent, commonly denoted as \( e^x \), where \( e \) is Euler's number (~2.718). These functions are characterized by their rapid growth or decay and have unique properties that differentiate them from mere polynomials or linear functions.

The most important property of exponential functions is the rule \( e^{a+b} = e^a \cdot e^b \). This allows for the addition of exponents in multiplication scenarios and simplifies many complex calculations.

• Exponential growth means that the function increases by an increasing rate.
• Exponential decay describes a function that decreases rapidly at first and then slowly as it approaches zero.

In many real-world applications, exponential functions model phenomena such as population growth, radioactive decay, and interest compounding, thanks to their unique rate of change. In the context of our original problem, the use of exponential functions allows us to set up equations in a way that can easily be solved for their intersection points.

Solving Equations

Solving equations involves finding the values of variables that make the equation true. When dealing with functions, finding their intersection involves setting them equal and solving for the variable. This means determining where their values overlap for some input.

In the given problem, we need to determine at which values of \( x \) the equations \( e^x = e^{1-x} \) hold true. The process is as follows:

• First, you set the equations equal to each other, treating them like an equality rather than two separate equations.
• Use natural logarithms to simplify the equation. Since \( \ln(e^x) = x \), the natural logarithms help in bringing down the exponent. This simplifies the process to a linear equation.
• Solve the resulting simpler equation, in this case, \( x = 1 - x \), which easily reduces to \( x = \frac{1}{2} \) by algebraic manipulation.

This method of solving equations is not only specific to this problem but is widely applicable. By taking known properties of exponential functions and logarithms, one can transform and solve seemingly complex equations into more manageable forms. The principles of equation solving help clarify many mathematical concepts by finding specific values that satisfy given conditions.