Problem 14 - Essential Calculus Early Transcendentals

Question

Find the number \(c\) that satisfies the conclusion of the Mean Value Theorem on the given interval. Graph the function, the secant line through the endpoints, and the tangent line at \((c, f(c))\) . Are the secant line and the tangent line parallel? \(f(x)=e^{-x}, \quad[0,2]\)

Step-by-Step Solution

Answer

The value of \(c\) is approximately 0.79.

Step 1: Understand the Mean Value Theorem

The Mean Value Theorem states that for a function \( f \) that is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), there is at least one number \( c \) in \((a, b)\) such that \[ f'(c) = \frac{f(b) - f(a)}{b - a}. \]

Step 2: Calculate f(a) and f(b)

For the function \( f(x) = e^{-x} \), calculate \( f(0) \) and \( f(2) \): 1. \( f(0) = e^{0} = 1 \) 2. \( f(2) = e^{-2} \) which is approximately 0.135.

Step 3: Compute the Secant Line Slope

Using the values of \( f(a) \) and \( f(b) \), calculate the slope of the secant line: \[ \text{slope} = \frac{f(2) - f(0)}{2 - 0} = \frac{e^{-2} - 1}{2}. \]

Step 4: Find the Derivative of f(x)

Differentiate \( f(x) = e^{-x} \) to find \( f'(x) \): \[ f'(x) = -e^{-x}. \]

Step 5: Solve for c

Set \( f'(c) = \frac{e^{-2} - 1}{2} \) and solve for \( c \): \[ -e^{-c} = \frac{e^{-2} - 1}{2}. \] Solving yields: \[ e^{-c} = -\frac{e^{-2} - 1}{2}. \] Taking the natural logarithm of both sides:\[ -c = \ln\left(-\frac{e^{-2} - 1}{2}\right), \] then \[ c \approx 0.79. \]

Step 6: Graph and Compare Lines

Graph \( f(x) = e^{-x} \), the secant line through \((0, f(0))\) and \((2, f(2))\), and the tangent line at \((c, f(c))\) using the slope found in Step 3. Compare to ensure that the tangent line at that \( c \) is indeed parallel to the secant line.

Step 7: Conclusion

The value of \( c \) that satisfies the Mean Value Theorem on the interval \([0, 2]\) is approximately 0.79. The secant line and the tangent line at \((c, f(c))\) are parallel.

Key Concepts

Secant Line

A secant line is a straight line that connects two points on the graph of a function. In the Mean Value Theorem (MVT), the secant line serves as a bridge between the endpoints of the function's interval. For example, with the function \( f(x) = e^{-x} \) on
the interval \([0,2]\), the secant line passes through the points \((0, f(0))\) and \((2, f(2))\). To determine its slope, which is crucial for applying the MVT, we use the formula:

• \( \text{slope} = \frac{f(b) - f(a)}{b - a} \)

The slope indicates how steep the secant line is between these two points. This slope is critical because the MVT guarantees that there is at least one point \(c\) in the interval where the tangent line's slope equals the secant line's slope. This forms the heart of the theorem's statement.

Tangent Line

A tangent line touches a curve at exactly one point and represents the curve's instantaneous rate of change at that precise point. In the context of the Mean Value Theorem, we're interested in finding a tangent line parallel to the secant line. For our function \(f(x) = e^{-x}\), the tangent line at a point \(c\) on the interval \((0,2)\) should have a slope equal to the secant line's slope. This is what ensures parallelism. Calculating this involves finding the derivative of the function:

• \( f'(x) = -e^{-x} \)

The derivative gives us a tool to calculate the slope of the tangent line at any point \(x\). By setting the derivative equal to the secant line's slope, we can solve for \(c\). This \(c\) gives us the exact point where the tangent line is parallel to the secant line.

Function Derivative

The derivative of a function, denoted \( f'(x) \), is the cornerstone of calculus. It provides us with the rate of change or the slope of the tangent line at any given point on a function's graph. For the function \( f(x) = e^{-x} \), the derivative is:

• \( f'(x) = -e^{-x} \)

This derivative function helps us determine how the original function is changing at any point. Within the context of the Mean Value Theorem, finding \( f'(c) \) allows us to determine whether there is a tangent line parallel to the secant line, confirming the theorem's conclusion. So, when applying the theorem, the derivative is paramount in locating the specific \(c\) that makes \( f'(c) = \frac{f(b) - f(a)}{b - a} \).

Parallel Lines

When two lines are parallel, they have the same slope and will never intersect. In the scenario given by the Mean Value Theorem for the function \(f(x) = e^{-x}\), we want the tangent line at point \((c, f(c))\) and the secant line to be parallel. This occurs when their slopes are equal.

• The slope of the secant line is found using: \(\frac{f(2) - f(0)}{2 - 0}\).
• The slope of the tangent line at \(c\) is given by \(f'(c)\).

To ensure the lines are parallel, set \(f'(c)\) equal to the slope of the secant line. Solving this equation provides the value of \(c\), showcasing the unique synergy between the geometry of the lines and calculus. When they are parallel, it confirms the Mean Value Theorem's prediction that such a \(c\) exists.