Problem 37 - Calculus for Biology and Medicine

Question

. Suppose that \(f(x)=x^{2}, x \in[-1,1]\). (a) Find the slope of the secant line connecting the points \((x, y)=(-1,1)\) and \((1,1)\) (b) Find a number \(c \in(-1,1)\) 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 \((-1,1)\).

Step-by-Step Solution

Answer

The secant slope is 0. The number is \(c = 0\), by the Mean Value Theorem.

Step 1: Calculate f(x) at the Given Points

Given points are \((-1, f(-1))\) and \((1, f(1))\). Evaluate the function at these points.\(f(-1) = (-1)^2 = 1\)\(f(1) = 1^2 = 1\).So, the points are \((-1, 1)\) and \((1, 1)\).

Step 2: Find the Slope of the Secant Line

The formula for the slope \(m\) of the secant line connecting the points \((x_1, y_1)\) and \((x_2, y_2)\) is:\[m = \frac{y_2 - y_1}{x_2 - x_1}\]Substitute the points \((-1, 1)\) and \((1, 1)\):\[m = \frac{1 - 1}{1 - (-1)} = \frac{0}{2} = 0\]So, the slope of the secant line is 0.

Step 3: Calculate the Derivative f'(x)

The function is \(f(x) = x^2\). Using differentiation, the derivative \(f'(x)\) is:\[f'(x) = 2x\].

Step 4: Set f'(c) Equal to the Secant Slope

We want to find \(c\) such that \(f'(c) = 0\), matching the secant slope.From \(f'(x) = 2x\), set:\[2c = 0\]Solving this gives \(c = 0\).

Step 5: Application of the Mean Value Theorem

By the Mean Value Theorem, since \(f(x)\) is continuous and differentiable on \([-1, 1]\), there exists at least one \(c\) in \((-1, 1)\) such that \(f'(c)\) equals the slope of the secant line.In this case, \(c = 0\) meets this condition.

Key Concepts

Secant Line

A secant line is a straight line that touches a curve at two distinct points. In mathematical analysis, it's often used to approximate a curve by a simpler form, which is especially helpful for understanding changes between two values. In the given problem, our function is defined as \( f(x) = x^2 \). The points of interest are \((-1, 1)\) and \((1, 1)\), both of which lie on the parabola. The secant line that connects these two points describes a straight path between them. To find its slope, which represents the average rate of change between these points, we use the formula for the slope:

• \( m = \frac{y_2 - y_1}{x_2 - x_1} \)

Substituting the given points into the formula gives \( m = \frac{1 - 1}{1 - (-1)} = 0 \). The slope being zero indicates that the secant line is horizontal. This concept is a fundamental stepping stone in understanding dynamic changes in calculus.

Differentiation

Differentiation is the process of finding the derivative of a function, which represents the rate at which the function is changing at any given point. For the function \( f(x) = x^2 \), the process of differentiation involves applying the power rule. This rule states that for \( f(x) = x^n \), the derivative \( f'(x) \) is \( n \times x^{n-1} \).Applying this rule to \( f(x) = x^2 \), we find the derivative:

• \( f'(x) = 2x \)

This derivative function, \( f'(x) = 2x \), provides the slope of the tangent line to any point \( x \) on the curve \( f(x) = x^2 \).In relation to the problem, we use differentiation to derive the formula \( f'(x) = 2x \). We then solve for \( c \) by setting the equation equal to the slope of the secant line, which was found to be zero. Solving \( 2c = 0 \), we get \( c = 0 \). Differentiation thus helps connect the slope of a secant line with the instantaneous rate of change at a given point.

Slope

The slope of a line is a measurement of its steepness. In this exercise, you encounter the slope in both the secant line and the tangent line concepts. The slope is crucial because it offers insights into how one variable changes relative to another.In the context of the secant line connecting points \((-1, 1)\) and \((1, 1)\), the slope is zero. This implies no vertical change between the points, hence a horizontal line.Furthermore, the derivative \( f'(x) = 2x \) represents the slope of the tangent line to the function \( f(x) = x^2 \) at any point \( x \). For example, reflecting on \( f'(x) = 0 \) at \( x = 0 \) shows that, at this particular point, the tangent line is horizontal, matching our secant line's slope.By employing the Mean Value Theorem, we use this slope equality concept. The theorem states that for a continuous and differentiable function like \( x^2 \), there is a point \( c \) where the instantaneous rate of change (or slope of the tangent) matches the average rate of change (slope of the secant). This is a powerful indicator of how calculus bridges conceptual gaps between average and instantaneous changes.