Problem 16 - Single Variable Calculus

Question

In Exercises 9-16, verify that the function satisfies the hypotheses of the Mean Value Theorem on the given interval, and find all values of \(c\) that satisfy the conclusion of the theorem. $$ g(x)=\ln x ; \quad[1.3] $$

Step-by-Step Solution

Answer

The function \(g(x) = \ln x\) is continuous on the interval [1, 3] and differentiable on the interval (1, 3). Applying the Mean Value Theorem, there exists a value \(c\) in the open interval (1, 3) that satisfies the equation \(g'(c) = \frac{g(3) - g(1)}{3 - 1}\). After calculating \(g'(c)\) and \(g(3) - g(1)\) and solving for \(c\), we find that \(c = \frac{2}{\ln(3)}\), which indeed lies within the open interval (1, 3).

Step 1: Verify continuity on [1, 3]

We know that the natural logarithm function, \(g(x) = \ln x\), is continuous for all x > 0. Since the interval [1, 3] is within the domain where the function is continuous, we can say that \(g(x)=\ln x\) is continuous on the interval [1, 3].

Step 2: Verify differentiability on (1, 3)

We also know that the derivative of the natural logarithm function is given by \(g'(x) = \frac{1}{x}\). This derivative is defined for all x > 0. Since the open interval (1, 3) is within the domain where the derivative is defined, we can say that \(g(x)=\ln x\) is differentiable on the interval (1, 3).

Step 3: Apply the Mean Value Theorem

Since \(g(x)=\ln x\) is continuous on [1, 3] and differentiable on (1, 3), we can apply the Mean Value Theorem. By the MVT, there exists at least one \(c\) in the open interval (1, 3) such that: $$ g'(c) = \frac{g(3)-g(1)}{3-1} $$

Step 4: Compute g'(c) and g(3)-g(1)

First let's compute \(g'(c)\) and \(g(3)-g(1)\): $$ g'(c) = \frac{1}{c} $$ $$ g(3)-g(1) = \ln(3) - \ln(1) = \ln(3) $$

Step 5: Solve for c

Now substitute \(g'(c)\) and \(g(3)-g(1)\) back into the MVT equation and solve for \(c\): $$ \frac{1}{c} = \frac{\ln(3)}{3-1} $$ $$ \frac{1}{c} = \frac{\ln(3)}{2} $$ Multiply both sides by \(c\) and by 2 to isolate \(c\): $$ 2 = c\ln(3) $$ Divide both sides by \(\ln(3)\) to get: $$ c = \frac{2}{\ln(3)} $$ Since \(\frac{2}{\ln(3)}\) is within the open interval (1, 3), we have found the value of \(c\) that satisfies the conclusion of the Mean Value Theorem: $$ c = \frac{2}{\ln(3)} $$

Key Concepts

Continuity

In mathematics, continuity refers to the property of a function where small changes in the input lead to small changes in the output. For a function to be continuous at a point, it must be defined at that point, and the limit of the function as it approaches that point must equal the function's value at that point.

The function \( g(x) = \ln x \), the natural logarithm function, is continuous for all \( x > 0 \). This means that within any interval greater than zero, there are no breaks, jumps, or holes in the graph. In the original exercise, the function is evaluated for continuity on the interval \([1, 3]\). Since 1 and 3 are both greater than zero, \( g(x) \) is continuous on this interval.

• Continuous functions have no sudden changes in value.
• The graph of a continuous function can be drawn without lifting the pencil from the paper.

Differentiability

Differentiability is about a function having a derivative at a certain point. If a function is differentiable at a point, it means there is a defined tangent at that point and the function is smooth (no sharp corners) there.

For the function \( g(x) = \ln x \), the derivative is \( g'(x) = \frac{1}{x} \). This derivative exists for all \( x > 0 \), indicating that the function is differentiable across intervals that fit within this domain. For our interval of interest, \((1, 3)\), \( g(x) \) is differentiable because 1 and 3 are within this domain.

• A differentiable function is always continuous, but a continuous function is not necessarily differentiable.
• Being differentiable means having a slope or rate of change at every point in the interval.

Natural Logarithm Function

The natural logarithm, denoted as \( \ln x \), is a logarithm with the base \( e \), where \( e \) is an irrational constant approximately equal to 2.71828. It is the inverse function of the exponential function \( e^x \). The natural logarithm has key properties:

• \( \ln(1) = 0 \) because \( e^0 = 1 \).
• \( \ln(e) = 1 \) because \( e^1 = e \).
• It is defined only for \( x > 0 \).

In calculus, the natural logarithm is used to solve problems involving growth and decay, and it frequently appears when dealing with derivatives and integrals. Its continuousness and differentiability (as seen through the function \( \ln x \) on our interval) make it a vital mathematical tool.

Understanding \( \ln x \) helps in effectively applying calculus concepts, including the Mean Value Theorem.

Calculus

Calculus is the branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. It is used to describe changes and motion, providing tools for understanding and modeling dynamic systems.

**Key Concepts:**

• **Derivative:** Measures how a function changes as its input changes, rarely appears alone in pure mathematics, but is essential in real-world applications involving change rates.
• **Integral:** Involves finding the accumulation of quantities, such as areas under a curve. Used extensively in engineering and physics.

In our exercise, the Mean Value Theorem from calculus is employed to find a specific value \( c \) where the function's instantaneous rate of change matches the average rate of change over an interval. This theorem requires a function to be continuous at the closed interval and differentiable at the open interval. Using the function \( g(x) = \ln x \), these criteria are met, allowing us to successfully find \( c \).

Mastering calculus concepts like continuity, differentiability, and applications like the MVT, opens up a deeper understanding of the mathematical principles that describe the natural world.