Problem 9 - Calculus Early Transcendentals: Pearson New International Edition

Question

In each of the Problems 1-21, a function is defined and a closed interval is given. Decide whether the Mean Value Theorem applies to the given function on the given interval. If it does, find all possible values of c; if not, state the reason. In each problem, sketch the graph of the given function on the given interval. $$ h(x)=e^{-x} ;[0,3] $$

Step-by-Step Solution

Answer

The Mean Value Theorem applies, and the value of \( c \) is \(-\ln \left(-\frac{e^{-3} - 1}{3} \right)\).

Step 1: Verify Continuity

The first step is to check if the function \( h(x) = e^{-x} \) is continuous on the closed interval \([0, 3]\). The function \(e^{-x}\) is an exponential function, which is continuous everywhere on its domain. Therefore, \( h(x) \) is continuous on \([0, 3]\).

Step 2: Verify Differentiability

Next, we check if the function \( h(x) = e^{-x} \) is differentiable on the open interval \((0, 3)\). Exponential functions are differentiable everywhere on their domain, so \( h(x) \) is differentiable on \((0, 3)\).

Step 3: Apply the Mean Value Theorem

Since \( h(x) \) is continuous on \([0, 3]\) and differentiable on \((0, 3)\), the Mean Value Theorem applies. The theorem states that there exists at least one \( c \in (0, 3) \) such that \( h'(c) = \frac{h(3) - h(0)}{3 - 0} \).

Step 4: Calculate the Derivative

Find the derivative of \( h(x) = e^{-x} \). Using the chain rule, the derivative is \( h'(x) = -e^{-x} \).

Step 5: Evaluate at Interval Endpoints

Compute \( h(3) \) and \( h(0) \): \[ h(3) = e^{-3} \] \[ h(0) = e^{0} = 1 \] The difference is: \[ h(3) - h(0) = e^{-3} - 1 \]

Step 6: Calculate the Mean Value

Substitute the values from Step 5 into the Mean Value Theorem formula: \[ h'(c) = \frac{e^{-3} - 1}{3} \]

Step 7: Solve for c

Set \( h'(x) = -e^{-x} \) equal to \( \frac{e^{-3} - 1}{3} \) and solve for \( x \): \[ -e^{-c} = \frac{e^{-3} - 1}{3} \] Multiplying both sides by -1 gives: \[ e^{-c} = -\frac{e^{-3} - 1}{3} \]Taking the natural logarithm of both sides and solving for \( c \), we have:\[ c = -\ln \left(-\frac{e^{-3} - 1}{3} \right) \]

Step 8: Consider Validity of Natural Logarithm

The natural log function is only valid for positive arguments. Verify that the expression inside the logarithm, \(-\frac{e^{-3} - 1}{3} \), is positive. Note that since 1 > e^{-3}, the numerator is negative, making the entire expression positive when multiplied by -1.

Key Concepts

Continuity

One of the crucial aspects of applying the Mean Value Theorem is verifying the continuity of the function over the closed interval. A function is continuous on a closed interval \([a, b]\) if there are no breaks, jumps, or holes over the entire interval.
For the function \(h(x) = e^{-x}\), which is an exponential function, we should know that exponential functions are continuous everywhere in their domain.

• This means there are no sudden changes in the value of \(h(x)\).
• Thus, on the closed interval \([0, 3]\), the function is smooth and uninterrupted.

Verifying continuity is logically the first step because it checks if every point on the curve connects smoothly.
This property ensures a valid application of the Mean Value Theorem in the subsequent steps.

Differentiability

To use the Mean Value Theorem, it is not enough for a function to be continuous; it must also be differentiable on the open interval. Differentiability implies that the function has a defined slope or derivative at every point in the open interval.
For \(h(x) = e^{-x}\), differentiability implies that we can calculate the rate of change of the function at any point between 0 and 3.

• Exponential functions like \(e^{-x}\) are differentiable everywhere they are continuous.
• Hence, \(h(x)\) is differentiable on the interval \( (0, 3) \).

Having determined both continuity and differentiability, the conditions necessary for the Mean Value Theorem are met.
This allows us to find a point \(c\) where the tangent to the curve is parallel to the secant line joining \( (0, h(0)) \) and \( (3, h(3)) \).

Exponential Function

Exponential functions are an important family of functions in mathematics. They are commonly written in the form \( f(x) = a^x \), where \(a\) is a positive constant. These functions are characterized by constant growth or decay rates.
In the context of \(h(x) = e^{-x}\), this represents an exponential decay function.

• The base \(e\), approximately equal to 2.71828, is a fundamental mathematical constant.
• The negative exponent indicates that as \(x\) increases, \(h(x)\) decreases, illustrating the decay behavior.

Exponential functions like \(e^{-x}\) are crucial in modeling a wide range of real-world phenomena, such as radioactive decay and population growth, due to their continuous and smooth nature.
Through \(h(x) = e^{-x}\), we demonstrate how exponential functions maintain core properties like continuity and differentiability across any interval on their domain.

Graph Sketching

Graph sketching is a valuable tool in visualizing how a function behaves over a particular interval.
For \(h(x) = e^{-x}\), sketching the graph on the interval \([0, 3]\) allows us to easily observe its decreasing nature.

• At \(x=0\), the graph starts at \(h(0) = 1\).
• As \(x\) approaches 3, the graph moves downward towards \(h(3) = e^{-3}\).
• The curve is steep initially and gradually flattens out.

With the graph, we can visualize how the slope of the tangent line at some point \(c\) parallels the slope of the secant line as per the Mean Value Theorem.
Graph sketching reinforces the understanding of function behavior, specifically illustrating continuity and differentiability through visual depiction.