Problem 35 - Calculus for Biology and Medicine

Question

\(f(x)=x^{3}, x \in[0,1]\). (a) Find the slope of the secant line connecting the points \((x, y)=(0,0)\) and \((1,1)\) (b) Find a number \(c \in(0,1)\) 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 \((0,1)\).

Step-by-Step Solution

Answer

(a) Slope of secant line is 1. (b) \(c = \frac{1}{\sqrt{3}}\) and it exists due to the Mean Value Theorem.

Step 1: Identify the Points

We are given two points on the graph of the function \(f(x) = x^3\): the first point is \((0, f(0)) = (0, 0^3) = (0, 0)\) and the second point is \((1, f(1)) = (1, 1^3) = (1, 1)\). These are the points we will use to find the slope of the secant line.

Step 2: Calculate the Slope of the Secant Line

The slope of the secant line connecting two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by the formula \(m = \frac{y_2 - y_1}{x_2 - x_1}\). Using the points \((0, 0)\) and \((1, 1)\), the slope is \(m = \frac{1 - 0}{1 - 0} = 1\).

Step 3: Find the Derivative of the Function

The derivative of the function \(f(x) = x^3\) is \(f'(x) = 3x^2\). This derivative represents the slope of the tangent line at any point \(x\) on \(f(x)\).

Step 4: Solve for a Number \(c\) Where the Derivative Equals Secant Slope

We need to find \(c\) such that \(f'(c) = 1\). Setting the derivative equal to the slope of the secant line, we have \(3c^2 = 1\). Solving for \(c\), we get \(c^2 = \frac{1}{3}\) and therefore \(c = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}}\).

Step 5: Verify Existence with the Mean Value Theorem

The Mean Value Theorem states that if a function is continuous on \([a, b]\) and differentiable on \((a, b)\), then there exists some \(c\) in \((a, b)\) such that \(f'(c)\) equals the slope of the secant line. Since \(f(x) = x^3\) is continuous and differentiable on \([0, 1]\), there must exist a \(c \in (0, 1)\) such that \(f'(c) = 1\). We found this \(c\) to be \(\frac{1}{\sqrt{3}}\).

Key Concepts

Slope of the Secant Line

The first concept we'll explore is the slope of the secant line. Imagine you have a curve and you draw a straight line that connects two distinct points on this curve. This line is called the secant line, and its slope represents how steep or flat this line is between those two points.

To find the slope of a secant line, we use the formula:

• \(m = \frac{y_2 - y_1}{x_2 - x_1}\)

This formula calculates the "rise" over the "run," or the vertical change over the horizontal change between the two points.

In our exercise, the function given is \(f(x) = x^3\), and the points of interest are \((0, 0)\) and \((1, 1)\). Plug these into the formula to find the slope of the secant line:

• \(m = \frac{1 - 0}{1 - 0} = 1\)

Thus, the slope of the secant line connecting these two points is 1.

This understanding of the secant line is crucial for applying the Mean Value Theorem and finding tangents to curves.

Differentiability

Differentiability refers to a function's ability to have a derivative at each point within a given interval. A function is differentiable if you can draw a tangent line at any point on its graph without any jumps or sharp turns.

When a function is differentiable, it means that small changes in the function's input (the \(x\)-values) result in smooth and predictable changes in the output (the \(y\)-values).

For the function \(f(x) = x^3\), its derivative is \(f'(x) = 3x^2\). The presence of this derivative across the entire interval \((0, 1)\) confirms that \(f(x) = x^3\) is smoothly changing without any sharp angles or discontinuities.

This property is what allows us to apply the Mean Value Theorem. We're assured that somewhere in the interval \((0, 1)\), there is a point \(c\) where the tangent line is parallel to the secant line, matching its slope.

Continuity

Continuity means that a function does not "break" or have any holes over its interval. It’s one of the basic requirements for applying the Mean Value Theorem, which needs a function to be continuous on a closed interval \([a, b]\).

For the function \(f(x) = x^3\), imagine traveling along its curve from \(x = 0\) to \(x = 1\) without lifting your pencil. This means that the function is continuous because there are no breaks or jumps along \(f(x)\) from start to finish.

This seamless flow from point to point ensures that we can trust the function's behavior across the interval. When we confirm that a function is both continuous and differentiable, it sets the stage perfectly for tools like the Mean Value Theorem, which require both of these properties to make reliable predictions about tangent and secant line relationships.