Problem 35 - Calculus for Biology and Medicine

Question

Suppose \(f(x)=x^{2}, x \in[0,2]\). (a) Find the slope of the secant line connecting the points \((0,0)\) and \((2,4)\). (b) Find a number \(c \in(0,2)\) 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 \((0,2)\).

Step-by-Step Solution

Answer

(a) Slope of secant line is 2. (b) The number is \(c=1\), which exists due to the Mean Value Theorem.

Step 1: Calculate the Change in Function Values

To find the slope of the secant line, we need to determine the change in the function values between the two points. Given points are \((0,0)\) and \((2,4)\). Compute the difference in the \(y\)-values: \(f(2) - f(0) = 4 - 0 = 4\).

Step 2: Calculate the Change in x-values

Next, determine the difference in the \(x\)-values of points \((0,0)\) and \((2,4)\). This is calculated as \(x_2 - x_1 = 2 - 0 = 2\).

Step 3: Calculate the Slope of the Secant Line

The slope of the secant line is given by the formula \(m = \frac{f(x_2) - f(x_1)}{x_2-x_1}\). Substituting the values from the previous steps gives \(m = \frac{4}{2} = 2\).

Step 4: Determine the Derivative of f(x)

Find the derivative \(f'(x)\) to check where it equals the slope of the secant line. Since \(f(x) = x^2\), the derivative \(f'(x) = 2x\).

Step 5: Solve for c in the Interval (0, 2)

Set the derivative equal to the slope of the secant line to find \(c\): \(2x = 2\). Solving for \(x\), we get \(x = 1\). Since \(1 \in (0, 2)\), \(c = 1\) is the required value.

Step 6: Verify Using the Mean Value Theorem

According to the Mean Value Theorem, if a function is continuous on \([0, 2]\) and differentiable on \((0, 2)\), there must be a \(c\) such that \(f'(c)\) equals the slope of the secant line. Since \(f(x) = x^2\) is both continuous and differentiable on the given interval, \(c = 1\) satisfies this condition.

Key Concepts

Understanding Secant Lines

In mathematics, the secant line is a straight line that connects two points on a curve. For the function \(f(x)=x^2\) on the interval \([0,2]\), the secant line connects the points \((0,0)\) and \((2,4)\). To find the slope of this line, we need to calculate the change in the function's values and the corresponding change in the \(x\)-values. Here:

• The change in \(y\)-values: \(f(2) - f(0) = 4 - 0 = 4\)
• The change in \(x\)-values: \(2 - 0 = 2\)

The slope \(m\) is given by \(m = \frac{f(x_2) - f(x_1)}{x_2 - x_1}\). Substituting the known values gives us \(m = \frac{4}{2} = 2\). This slope represents how steeply the secant line rises as you move from \(x=0\) to \(x=2\).
Knowing how to calculate the slope of a secant line is crucial for understanding changes between two points on a curve.

Exploring Differentiation

Differentiation is the process of finding the derivative of a function. The derivative gives us the rate at which a function is changing at any given point, essentially providing a "dynamic" understanding of its behavior. For our quadratic function \(f(x) = x^2\), the derivative is \(f'(x) = 2x\).
This derivative, \(f'(x)\), tells us the slope of the tangent line at any particular point \(x\). While a secant line connects two points, a tangent line just touches the curve at a single point and shows the immediate rate of change at that point.
Differentiation allows us to identify how rapidly the curve changes direction, making it a powerful tool for various applications in math and sciences.

Slope Calculation: Connecting Secants and Tangents

Calculating the slope of both secant and tangent lines helps us understand the different ways a curve can be approximated by a straight line. For a function continuous and differentiable like \(f(x) = x^2\), the Mean Value Theorem assists us by asserting that there exists a special point \(c\) in \((0, 2)\) where the slope of the tangent (found by differentiating) matches that of the secant line.
Given our derivative \(f'(x) = 2x\), setting it equal to the secant slope (\(m = 2\)) gives \(2x = 2\). Solving for \(x\), we find \(x = 1\). This means that at \(x = 1\), the slope of the tangent also equals 2, just like the secant. Identifying such a point helps link average rates of change over an interval (represented by secants) to instantaneous change at a precise point (tangents), providing a deeper understanding of the curve's geometry.