Problem 2 - 5 Steps to A 5: AP Calculus AB 2021

Question

Let \(f(x)=e^{x} .\) Show that the hypotheses of the Mean Value Theorem are satisfied on the interval [0,1] and find all values of \(c\) that satisfy the conclusion of the theorem.

Step-by-Step Solution

Answer

Question: Find the value of c that satisfies the conclusion of the Mean Value Theorem for the function f(x) = e^x on the interval [0, 1]. Answer: The value of c that satisfies the conclusion of the Mean Value Theorem is: \(c = ln(e - 1)\).

Step 1: Check continuity on [0,1]

Since \(f(x) = e^x\), it is an exponential function and is continuous on \((-\infty, \infty)\). So, it is continuous on the closed interval \([0,1]\).

Step 2: Check differentiability on (0,1)

The derivative of an exponential function is also an exponential function. Thus, \(f'(x) = e^x\) is also continuous on \((-\infty, \infty)\) and thus it is differentiable on the open interval \((0,1)\).

Step 3: Apply Mean Value Theorem to find c

As the conditions of the Mean Value Theorem are satisfied, we can now find the value(s) of \(c\) that satisfy the theorem. We have to find \(c\) in \((0,1)\) such that \(f'(c) = \frac{f(1) - f(0)}{1-0}\). Let's find \(f(1)\) and \(f(0)\): - \(f(1) = e^1 = e\) - \(f(0) = e^0 = 1\) Now we can calculate \(\frac{f(1) - f(0)}{1-0} = \frac{e - 1}{1}\).

Step 4: Solve for c

Now we need to solve the following equation for \(c\): $$f'(c) = e^c = \frac{e - 1}{1}$$ To find \(c\), simply take the natural logarithm of both sides: $$c = ln\left(\frac{e - 1}{1}\right) = ln(e - 1)$$ Thus, the value of \(c\) that satisfies the conclusion of the Mean Value Theorem is: \(c = ln(e - 1)\).

Key Concepts

Continuity

The concept of continuity is a fundamental part of calculus, which intuitively means that a function has no breaks, jumps, or holes within a specific interval. If you imagine drawing the graph of a function without lifting your pencil, that represents continuity.

For the exponential function \(f(x) = e^x\), this quality is exhibited beautifully since it holds continuously for all real numbers. In the context of the Mean Value Theorem, ensuring a function is continuous over a closed interval \([a, b]\) is essential. For the problem at hand, we check the function on the interval \([0, 1]\), and since exponential functions like \(e^x\) are continuous everywhere, this condition is met.

Overall, continuity guarantees that the function does not behave unexpectedly, allowing us to apply powerful calculus tools like the Mean Value Theorem with confidence.

Differentiability

Differentiability refers to whether a function has a derivative at each point of an interval, indicating it is smooth enough to allow for tangent lines to the curve. A differentiable function does not have sharp corners or cusps within its interval.

For our exponential function \(f(x)=e^x\), differentiability is straightforward. The derivative of \(e^x\) is itself \(e^x\), which remains smooth and continuous over any range. More specifically, we are interested in the open interval \((0, 1)\). As such, differentiability is perfectly satisfied.

Why does this matter? Because for the Mean Value Theorem to be applicable, the function must be both continuous and differentiable in its given interval. In simpler terms, it allows calculus to "unleash its potential" to find meaningful results like the critical point \(c\) inside the interval.

Exponential function

The exponential function \(e^x\) is a fascinating entity. It represents one of the key functions in mathematics due to its unique properties and ubiquitous presence in exponential growth or decay.

This function is continuous, differentiable, and its own derivative, which is rather special. The value \(e\) itself, approximately 2.718, is a mathematical constant that appears consistently in various natural phenomena.

• Exponential functions are important in modeling real-world phenomena such as population growth, radioactive decay, and interest rates.
• They possess distinctive properties like having horizontal asymptotes on the x-axis and constant growth rates.

In our problem, the exponential function elegantly meets the prerequisites of the Mean Value Theorem, thus allowing us to find a suitable \(c\) within the interval.

Calculus problem solving

Solving calculus problems often involves a strategic application of several theorems and properties. The Mean Value Theorem (MVT) is one such powerful tool. It acts as a bridge between the derivative of a function and its overall change over an interval.

The problem implicitly enacts a common calculus method: checking that \(f(x)=e^x\) is continuous and differentiable over \([0, 1]\) and \((0, 1)\) respectively. Once confirmed, we apply the MVT to determine the value \(c\), which satisfies the equation:

• \(f'(c) = \frac{f(b) - f(a)}{b-a}\)
• \(c\) is found by equating \(e^c = \frac{e - 1}{1}\) and solving for \(c = \ln(e - 1)\).

By following these systematic steps, calculus problem solving becomes more practicable, allowing students to tackle seemingly complex problems with clarity and logic.