Problem 12 - Calculus

Question

show that there is a number \(c,\) with \(0 \leq c \leq 1,\) such that \(f(c)=0\) $$f(x)=e^{x}-3 x$$

Step-by-Step Solution

Answer

There is a number \( c \in (0, 1) \) such that \( f(c) = 0 \) by the Intermediate Value Theorem.

Step 1: Determine Continuity

The function \( f(x) = e^x - 3x \) is continuous because both the exponential function \( e^x \) and the linear function \( -3x \) are continuous for all real numbers.

Step 2: Evaluate the Function at Boundaries

Evaluate \( f(x) \) at \( x=0 \) and \( x=1 \): - \( f(0) = e^0 - 3 \times 0 = 1 \)- \( f(1) = e^1 - 3 \times 1 = e - 3 \) Approximate \( e \approx 2.718 \) to get \( f(1) \approx 2.718 - 3 = -0.282 \). Hence, \( f(0) = 1 > 0 \) and \( f(1) < 0 \).

Step 3: Apply the Intermediate Value Theorem

The Intermediate Value Theorem states that if a function \( f \) is continuous on the interval \([a, b]\) and \( f(a) \) and \( f(b) \) have opposite signs, there exists at least one \( c \) in \( (a, b) \) such that \( f(c) = 0 \).Since \( f(0) = 1 > 0 \) and \( f(1) \approx -0.282 < 0 \), there exists at least one \( c \in (0, 1) \) where \( f(c) = 0 \).

Key Concepts

Understanding Continuous Functions

Continuous functions are a cornerstone of calculus and many areas of mathematics. A function is considered continuous if you can draw it on a graph without picking up your pen — essentially, the graph has no breaks or gaps.
In more technical terms, a function \( f(x) \) is continuous at a point \( x = a \) if three conditions are satisfied:

• The function \( f(a) \) is defined.
• The limit of \( f(x) \) as \( x \) approaches \( a \) exists.
• The limit of \( f(x) \) as \( x \) approaches \( a \) is equal to \( f(a) \).

If a function meets these conditions for all points within a certain interval, it is said to be continuous over that interval. In our exercise, we dealt with \( f(x) = e^x - 3x \), which is continuous everywhere because both the exponential function \( e^x \) and the linear function \( -3x \) are continuous across all real numbers.
This property is crucial when applying the Intermediate Value Theorem, as it assures us there's no disruption in the function's behavior over the interval of interest.

Exploring Exponential Functions

Exponential functions have a unique characteristic -- their growth rate. They grow more rapidly than any polynomial function, which makes them interesting and powerful.
An exponential function is typically written as \( e^x \), where \( e \) is Euler's number, approximately equal to 2.718. This special number \( e \) arises naturally in the process of continuous growth and compounded interest applications.
In the function \( f(x) = e^x - 3x \), the exponential part \( e^x \) grows rapidly as \( x \) increases. This rapid growth contributes significantly to the behavior of the overall function. When we say \( e^x \) is continuous, it means its value changes smoothly as \( x \) changes without any jumps or interruptions.
Understanding the characteristics of exponential growth helps explain why the output of \( f(x) \) changes so much between the values \( x=0 \) and \( x=1 \), moving from a positive to a negative value.

Finding Roots of Equations

Finding the roots of an equation, simply put, is finding the values at which the function equals zero. This is important for various applications, such as solving real-world problems where a function describes a certain system, and its roots represent key points where the system changes.
In our exercise, we've employed the Intermediate Value Theorem to find the roots of the function \( f(x) = e^x - 3x \). The theorem asserts that for a continuous function over an interval, if the function takes on different signs at two endpoints, then there must be at least one point \( c \) within the interval where the function is zero.
In practical terms:

• We examined the function at \( x=0 \) and found \( f(0) = 1 \), which is greater than zero.
• We also examined \( x=1 \) and found \( f(1) \approx -0.282 \), which is less than zero.

Since the function changes from positive to negative over the interval \([0, 1]\), there must be at least one \( c \) where \( f(c) = 0 \), confirming the function has at least one root in that interval.
This theorem is a fundamental property in calculus as it guarantees the existence of roots in a given continuous interval, thus providing a powerful tool for analysis and problem-solving.