Problem 30 - Calculus with Concepts in Calculus

Question

Determine at which points the graphs of the given pair of functions intersect. $$ f(x)=e^{x} \text { and } g(x)=e^{-x^{2}} $$

Step-by-Step Solution

Answer

The graphs intersect at points (0, 1) and (-1, 0.368).

Step 1: Set the Functions Equal

To find the points of intersection, we set the two functions equal to each other: \[ e^x = e^{-x^2} \]

Step 2: Apply the Natural Logarithm

Take the natural logarithm of both sides to simplify the equation. Since the base of the logarithm and the base of the exponetiated numbers are the same, we find:\[ x = -x^2 \]

Step 3: Rearrange the Equation

Rearrange the equation from Step 2 to form a quadratic equation:\[ x^2 + x = 0 \]

Step 4: Factor the Quadratic Equation

Factor the quadratic equation:\[ x(x + 1) = 0 \]

Step 5: Solve for x

Solve the factors set to zero:\[ x = 0 \] and \[ x = -1 \]

Step 6: Find Corresponding y-values

Substitute \( x = 0 \) into either function to find \( y \):\[ f(0) = e^0 = 1 \]Then substitute \( x = -1 \):\[ f(-1) = e^{-1} \approx 0.368 \]

Step 7: List Intersection Points

The points of intersection are where both functions have the same value. These points are: (0, 1) and (-1, 0.368)

Key Concepts

Understanding Exponential Functions

Exponential functions are a class of mathematical functions where the variable is in the exponent. Typically expressed in the form \( f(x) = a^x \), where \( a \) is a positive constant. In this exercise, we have both \( f(x) = e^x \) and \( g(x) = e^{-x^2} \), showcasing exponential growth and decay correspondingly.

• Exponential Growth: In \( f(x) = e^x \), as \( x \) increases, \( e^x \) increases rapidly, demonstrating exponential growth.
  
• Exponential Decay: In \( g(x) = e^{-x^2} \), the negative exponent induces decay, meaning as \( x \) moves away from zero in both directions, the value of \( g(x) \) approaches zero.
  

Visualizing these functions by plotting them can assist in identifying their behavior and understanding where they might intersect.

Exploring Quadratic Equations

Quadratic equations are polynomial equations of degree two. They are typically expressed as \( ax^2 + bx + c = 0 \). Here in our exercise, we derive a quadratic equation: \( x^2 + x = 0 \).

• Forms of Quadratic Equations: The standard form is \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are coefficients.
  
• Solutions: The solutions to a quadratic equation are the points where the parabola intersects the x-axis, also known as the roots or zeros of the function.
  

This quadratic equation set up by rearranging the intersection condition shows us potential x-values where the functions \( f(x) \) and \( g(x) \) might be equal.

The Role of Natural Logarithm

A natural logarithm is a logarithm with the base \( e \), where \( e \) is approximately 2.718. Applying the natural logarithm simplifies equations involving exponentials, because the logarithm and its power function, \( e \), are inverse functions.

• Simplification: For \( e^x = e^{-x^2} \), taking the natural logarithm of both sides simplifies directly to \( x = -x^2 \).
  
• Inverse Operation: The natural logarithm effectively "undoes" the exponential on both sides, simplifying the equation significantly.
  

Understanding this concept aids in the manipulation and solution of equations involving exponential functions.

Factoring Equations

Factoring is a method used to break down equations into simple products, making them easier to solve. For quadratic equations, this involves finding numbers or expressions that multiply together to give the original quadratic.

• Factoring Method: The method involves expressing \( x^2 + x = 0 \) as \( x(x + 1) = 0 \), which helps identify the roots or solutions.
  
• Solving: Setting each factor from \( x(x + 1) \) to zero gives \( x = 0 \) and \( x = -1 \), the solutions where the quadratic intersected the x-axis.
  

Factoring not only simplifies the process of solving quadratic equations but also provides insight into the structure and solutions of the equations.