Problem 38
Question
Suppose that \(f(x)=\ln x, x \in[1, e]\). (a) Find the slope of the secant line connecting the points \((x, y)=(1,0)\) and \((e, 1)\) (b) Find a number \(c \in(1, e)\) such that \(f^{\prime}(c)\) is equal to the slope of the secant line you computed in (a), and explain why such a number must exist in \((1, e)\).
Step-by-Step Solution
Verified Answer
(a) The slope of the secant line is \(\frac{1}{e-1}\). (b) The value of \(c\) is \(e - 1\), which exists by the Mean Value Theorem.
1Step 1: Identify function values for secant line
Compute the function values at the endpoints of the interval. At \(x = 1\), \(f(1) = \ln(1) = 0\). At \(x = e\), \(f(e) = \ln(e) = 1\). Thus, the points on the function \(f(x) = \ln x\) in the interval are \((1, 0)\) and \((e, 1)\).
2Step 2: Compute the slope of the secant line
The formula for the slope \(m\) of a secant line through points \((x_1, y_1)\) and \((x_2, y_2)\) is \[ m = \frac{y_2 - y_1}{x_2 - x_1} \]Substitute the coordinates \((1, 0)\) and \((e, 1)\):\[ m = \frac{1 - 0}{e - 1} = \frac{1}{e - 1} \]
3Step 3: Compute the derivative of the function
The derivative \(f'(x)\) of the function \(f(x) = \ln x\) is calculated as follows: \[ f'(x) = \frac{d}{dx}(\ln x) = \frac{1}{x} \]
4Step 4: Find the value of \(c\) for which \(f'(c)\) equals the secant line slope
We need to find \(c\) such that \(f'(c) = \frac{1}{e-1}\). \(f'(c) = \frac{1}{c}\), so set \(\frac{1}{c} = \frac{1}{e-1}\). Thus, \(c = e - 1\). Since \(e \approx 2.718\), \(c \approx 1.718\), which lies in the interval \((1, e)\).
5Step 5: Justification by Mean Value Theorem
The Mean Value Theorem states that if a function \(f(x)\) is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists at least one number \(c\) in \((a, b)\) such that \[ f'(c) = \frac{f(b) - f(a)}{b - a} \]Since \(f(x) = \ln x\) is continuous and differentiable on the given interval \([1, e]\), there must exist such a \(c\) where \(f'(c)\) matches the secant line slope as we found earlier.
Key Concepts
Secant LineDerivativeNatural Logarithm
Secant Line
A secant line is an important concept in calculus, often used when discussing the difference between average and instantaneous rates of change. Imagine a curve, like the graph of a function, and then picture a straight line that touches this curve at two points. This straight line is our secant line. The slope of a secant line, denoted usually by \( m \), is a measure of the average rate of change of the function over that interval.
For our case, with the function \( f(x) = \ln x \) over the interval \([1, e]\), the secant line connects the points \((1, 0)\) and \((e, 1)\). Using the formula for the slope of a secant line \( m = \frac{y_2 - y_1}{x_2 - x_1} \), and plugging in these points, we calculate:
For our case, with the function \( f(x) = \ln x \) over the interval \([1, e]\), the secant line connects the points \((1, 0)\) and \((e, 1)\). Using the formula for the slope of a secant line \( m = \frac{y_2 - y_1}{x_2 - x_1} \), and plugging in these points, we calculate:
- \((x_1, y_1) = (1, 0)\)
- \((x_2, y_2) = (e, 1)\)
Derivative
The derivative is a key concept in calculus, and it's all about understanding the instantaneous rate of change. Whereas the secant line gives us the average rate of change between two points, a derivative gives us the slope of the tangent line at a single point on the curve. For a function \( f(x) \), its derivative \( f'(x) \) at any point gives us the slope of the function at that point.
In our example, the function \( f(x) = \ln x \) has as its derivative \( f'(x) = \frac{1}{x} \). This means at any point \( x \), the slope of the tangent to the curve is given by \( \frac{1}{x} \). This derivative tells us how quickly or slowly the function \( \ln x \) changes as \( x \) changes.
To find a specific point where the derivative equals the slope of our secant line, we set \( f'(c) = \frac{1}{e - 1} \) or \( \frac{1}{c} = \frac{1}{e - 1} \). Solving this, we get \( c = e - 1 \). This \( c \) represents a specific point in the interval where the instantaneous rate of change equals our previously calculated average rate of change.
In our example, the function \( f(x) = \ln x \) has as its derivative \( f'(x) = \frac{1}{x} \). This means at any point \( x \), the slope of the tangent to the curve is given by \( \frac{1}{x} \). This derivative tells us how quickly or slowly the function \( \ln x \) changes as \( x \) changes.
To find a specific point where the derivative equals the slope of our secant line, we set \( f'(c) = \frac{1}{e - 1} \) or \( \frac{1}{c} = \frac{1}{e - 1} \). Solving this, we get \( c = e - 1 \). This \( c \) represents a specific point in the interval where the instantaneous rate of change equals our previously calculated average rate of change.
Natural Logarithm
Natural logarithms are logarithms to the base \( e \), an important mathematical constant approximately equal to 2.718. Natural logarithms are denoted as \( \ln(x) \). This mathematical function comes into play in various fields such as calculus, mathematical modeling, and compound interest calculations.
The function \( f(x) = \ln x \) is a common example where natural logarithms are applied. The property of this function is that it is continuous and differentiable over its entire domain, which is necessary for certain theorems in calculus like the Mean Value Theorem to be applicable.
The key takeaway about natural logarithms in the context of this exercise is how they allow us to calculate rates of change via their derivative. With \( f(x) = \ln x \), the derivative \( f'(x) = \frac{1}{x} \) illustrates the rate at which the function \( \ln x \) is changing at any point \( x \). Understanding natural logarithms offers a foundation for more advanced calculus concepts such as exponential growth and decay.
The function \( f(x) = \ln x \) is a common example where natural logarithms are applied. The property of this function is that it is continuous and differentiable over its entire domain, which is necessary for certain theorems in calculus like the Mean Value Theorem to be applicable.
The key takeaway about natural logarithms in the context of this exercise is how they allow us to calculate rates of change via their derivative. With \( f(x) = \ln x \), the derivative \( f'(x) = \frac{1}{x} \) illustrates the rate at which the function \( \ln x \) is changing at any point \( x \). Understanding natural logarithms offers a foundation for more advanced calculus concepts such as exponential growth and decay.
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