Problem 10 - Calculus Early Transcendentals

Question

Let \(f(x)=x^{3}-4 x\) (a) Find the equation of the secant line through the points \((-2, f(-2))\) and \((1, f(1))\). (b) Show that there is only one point \(c\) in the interval \((-2,1)\) that satisfies the conclusion of the Mean-Value Theorem for the secant line in part (a). (c) Find the equation of the tangent line to the graph of \(f\) at the point \((c, f(c)) .\) (d) Use a graphing utility to generate the secant line in part (a) and the tangent line in part (c) in the same coordinate system, and confirm visually that the two lines seem parallel.

Step-by-Step Solution

Answer

The secant line is \(y = -x - 2\). The Mean-Value Theorem applies at \(c = -1\). The tangent line is also \(y = -x - 2\). The lines are parallel.

Step 1: Evaluate f(-2) and f(1)

First, determine the function values at the given points. For \(x = -2\), calculate \(f(-2) = (-2)^3 - 4(-2) = -8 + 8 = 0\). Similarly, for \(x = 1\), we have \(f(1) = (1)^3 - 4(1) = 1 - 4 = -3\). Thus, the points are \((-2, 0)\) and \((1, -3)\).

Step 2: Find the Secant Line

The slope of the secant line through points \((-2, 0)\) and \((1, -3)\) is calculated using the formula \(m = \frac{y_2 - y_1}{x_2 - x_1}\). This gives \(m = \frac{-3 - 0}{1 - (-2)} = \frac{-3}{3} = -1\). The equation of the secant line is then given by \(y - y_1 = m(x - x_1)\). Substitute \(y_1 = 0\), \(x_1 = -2\), and \(m = -1\) to find \(y = -1(x + 2) = -x - 2\).

Step 3: Apply the Mean-Value Theorem

The Mean-Value Theorem states that there exists at least one \(c\) in the interval \((-2, 1)\) such that \(f'(c) = \frac{f(b) - f(a)}{b - a}\). We've already calculated \(f(b) - f(a) = -3 - 0 = -3\) and \(b - a = 1 - (-2) = 3\), so \(f'(c) = -1\).

Step 4: Calculate the Derivative

Find the derivative of \(f(x) = x^3 - 4x\). Derivative \(f'(x) = 3x^2 - 4\). Set \(f'(x) = -1\) to solve for \(c\). This gives the equation \(3c^2 - 4 = -1\), or \(3c^2 = 3\), resulting in \(c^2 = 1\). Thus, \(c = 1\) or \(c = -1\). Since \(-1\) is in the interval \((-2, 1)\), \(c = -1\).

Step 5: Tangent Line at c

Substitute \(c = -1\) into the derivative to find the slope of the tangent line: \(f'(-1) = 3(-1)^2 - 4 = -1\). With point \((-1, f(-1)) = (-1,-3)\), the tangent line equation is \(y - (-3) = -1(x + 1)\), which simplifies to \(y = -x - 2\).

Step 6: Graphing Utility Visualization

Using a graphing tool, plot the secant line \(y = -x - 2\) and the tangent line \(y = -x - 2\) to observe that they are indeed parallel since both have the slope of \(-1\).

Key Concepts

Secant Line

A secant line is an important concept in calculus that connects two points on a curve. In this exercise, the secant line is derived from the points

• \((-2, f(-2))\)
• \((1, f(1))\)

These points are on the graph of the function \(f(x) = x^3 - 4x\). First, calculate the values of the function at these points:
- At \(x = -2\): \(f(-2) = (-2)^3 - 4(-2) = -8 + 8 = 0\)
- At \(x = 1\): \(f(1) = (1)^3 - 4(1) = 1 - 4 = -3\)
With points

• \((-2, 0)\)
• \((1, -3)\)

, we can find the slope \(m\) of the secant line using the formula: \[ m = \frac{y_2 - y_1}{x_2 - x_1} \]Plugging in the numbers: \[ m = \frac{-3 - 0}{1 - (-2)} = \frac{-3}{3} = -1 \]Now, use point-slope form to find the equation of the line: \[ y - y_1 = m(x - x_1) \]Substituting for the first point \((-2, 0)\): \[ y - 0 = -1(x + 2) \]Simplifying gives \[ y = -x - 2 \]This equation represents the secant line between the two points.

Tangent Line

A tangent line touches a curve at exactly one point. It's an essential concept for understanding the behavior of functions. In part (c) of the exercise, it is requested to find the tangent line at the point \((c, f(c))\), where \(c\) is a specific point guaranteed by the Mean Value Theorem.
For the function \(f(x) = x^3 - 4x\), the Mean Value Theorem indicates that there exists a point \(c\) in the interval

• \((-2, 1)\)

where the slope of the tangent line \(f'(c) = -1\), which matches the slope of the secant line.
We previously calculated: - The derivative of the function is \(f'(x) = 3x^2 - 4\)- Set \(f'(c)\) equal to the slope \(-1\): \[ 3c^2 - 4 = -1 \]Solving yields:

• \(3c^2 = 3\)
• \(c^2 = 1\)
• \(c = 1\) or \(c = -1\)

Since we seek \(c\) in

• \((-2, 1)\)

, select \(c = -1\).
The tangent line at \(c = -1\) uses the point

• \((-1, f(-1)) = (-1, -3)\)

And slope \(f'(-1) = -1\):
\[ y + 3 = -1(x + 1) \]Simplifying: \[ y = -x - 2 \]Notice the tangent line and the secant line have the same slope, confirming they are parallel.

Derivative

The derivative of a function provides crucial insights into its behavior, describing how it changes at any given point. For the function \(f(x) = x^3 - 4x\):

• The derivative is \(f'(x) = 3x^2 - 4\).

This derivative helps us find slopes of tangent lines at various points on the curve. In this context, the derivative is directly used for applying the Mean Value Theorem, which asserts there is a point where the tangent slope \(f'(c)\) equates to the secant line's slope.
For this function, we found:

• Set \(f'(x) = -1\) to match the secant line's slope, resulting in the equation: \[ 3x^2 - 4 = -1 \]
• Solve the equation: \[ 3x^2 = 3 \] \[ x^2 = 1 \]
• Possible values for \(x\) are \(x = 1\) or \(x = -1\).

In the interval \((-2, 1)\), only \(x = -1\) fits. At this point, the tangent line is parallel to the secant line, confirming the tangent slope equals the average or "overall" slope between distinct points on the curve.