Problem 36 - Calculus for Biology and Medicine

Question

Suppose \(f(x)=1 / x, x \in[1,2]\). (a) Find the slope of the secant line connecting the points \((1,1)\) and \((2,1 / 2)\). (b) Find a number \(c \in(1,2)\) 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 \((1,2)\).

Step-by-Step Solution

Answer

The slope of the secant line is \(-\frac{1}{2}\). The value \(c = \sqrt{2}\) such that \(f'(c) = -\frac{1}{2}\) exists by the Mean Value Theorem.

Step 1: Calculate the Slope of the Secant Line

To find the slope of the secant line connecting the points \((1, 1)\) and \((2, \frac{1}{2})\), use the slope formula: \( m = \frac{f(x_2) - f(x_1)}{x_2 - x_1} \), where \((x_1, f(x_1)) = (1, 1)\) and \((x_2, f(x_2)) = (2, \frac{1}{2})\). This gives:\[ m = \frac{\frac{1}{2} - 1}{2 - 1} = \frac{-\frac{1}{2}}{1} = -\frac{1}{2} \] Thus, the slope of the secant line is \(-\frac{1}{2}\).

Step 2: Differentiate the Function

The function is given by \(f(x) = \frac{1}{x}\). To find the derivative \(f'(x)\), use the power rule for derivatives:\[ f'(x) = \frac{d}{dx} \left( x^{-1} \right) = -1 \cdot x^{-2} = -\frac{1}{x^2} \]This derivative expresses the rate of change of \(f(x)\).

Step 3: Solve for \(c\) where \(f'(c) = m\)

Using the derivative \(f'(x) = -\frac{1}{x^2}\), set it equal to the slope of the secant line \(-\frac{1}{2}\):\[ -\frac{1}{c^2} = -\frac{1}{2} \]To solve for \(c\), \[ \frac{1}{c^2} = \frac{1}{2} \]Taking the reciprocal gives:\[ c^2 = 2 \]Thus, taking the positive square root gives \(c = \sqrt{2}\). Since \(\sqrt{2} \approx 1.414\), it is between 1 and 2.

Step 4: Justify Existence of \(c\) using the Mean Value Theorem

By the Mean Value Theorem (MVT), if \(f(x)\) is continuous on the closed interval \([1, 2]\) and differentiable on the open interval \((1, 2)\), then there exists some \(c \in (1, 2)\) such that \(f'(c)\) equals the slope of the secant line, \(-\frac{1}{2}\). Since \(f(x) = \frac{1}{x}\) is both continuous and differentiable on these intervals, \(c = \sqrt{2}\) satisfies the MVT.

Key Concepts

Derivatives

The derivative of a function provides insight into how the function behaves as its input changes. It essentially measures the rate at which the function's value changes for a small change in the input.
For the function given by the exercise, \(f(x) = \frac{1}{x}\), its derivative is found using the power rule.

• The power rule states that for a function of the form \(x^n\), the derivative is \(nx^{n-1}\).
• For \(f(x) = x^{-1}\), the derivative is calculated as \(f'(x) = -1 \cdot x^{-2} = -\frac{1}{x^2}\).

This derivative tells us how \(f(x)\) changes as \(x\) changes. More specifically, it demonstrates that as \(x\) increases, \(-\frac{1}{x^2}\) becomes less negative, indicating that the function's rate of change slows down.

Secant Line

A secant line is a straight line that passes through two points on a curve, giving the average rate of change of the function between those two points. In this exercise, we're looking at the function \(f(x) = \frac{1}{x}\), and the secant line connects the points \((1, 1)\) and \((2, \frac{1}{2})\).
To find the slope of this line, the formula used is:

• \( m = \frac{f(x_2) - f(x_1)}{x_2 - x_1} \)

Here, \((x_2, f(x_2)) = (2, \frac{1}{2})\) and \((x_1, f(x_1)) = (1, 1)\).
By substituting these values, we find the slope \(m\):

• \(m = \frac{\frac{1}{2} - 1}{2 - 1} = -\frac{1}{2}\)

This slope represents the average rate of change from \(x=1\) to \(x=2\). It's the rate at which \(f(x)\) decreases as \(x\) moves from 1 to 2.

Functions

The concept of functions is fundamental in mathematics, representing a relationship where each input is paired with exactly one output. In this exercise, we work with the function \(f(x) = \frac{1}{x}\) over the interval \([1, 2]\).
This function is a classic example of an inverse relationship, where as \(x\) increases, \(f(x)\) decreases.

• On this specified interval, the function is both continuous and differentiable.
• These properties are crucial for applying the Mean Value Theorem.

According to the theorem, if a function is continuous on a closed interval and differentiable on the open interval, there must be at least one point where the derivative equals the slope of the secant line. Thus, for our function, there exists some \(c\) within \((1, 2)\) such that \(f'(c) = -\frac{1}{2}\). By our calculations, \(c\) is \(\sqrt{2}\), validating the theorem and solidifying the relationship described by the function.