Problem 30 - Essential Calculus

Question

Consider the graph of the function \(f(x)=-x^{2}-x+6 .\) (a) Find the equation of the secant line joining the points (-2,4) and (2,0) (b) Use the Mean Value Theorem to determine a point \(c\) in the interval (-2,2) such that the tangent line at \(c\) is parallel to the secant line. (c) Find the equation of the tangent line through \(c\). (d) Use a graphing utility to graph \(f,\) the secant line, and the tangent line.

Step-by-Step Solution

Answer

The equation of the secant line is \(y = -x + 2\). There exists a point \(c = 0\) in the interval (-2,2) such that the tangent line at \(c\) is parallel to the secant line. The equation of the tangent line through \(c\) is \(y = -x + 6\).

Step 1: Find the equation of the secant line

The equation of the secant line that passes through two points can be found by using the formula for the slope of a line: \(m = (y_2 - y_1) / (x_2 - x_1)\). Here \(x_1=-2, x_2=2, y_1=4, y_2=0\). So, \( m = (4-0)/(-2-2) = -1 \). The secant line equation is \(y - y_1 = m(x - x_1)\). Plugging in \(m=-1\) and one of the points, say \((-2,4)\), we get \(y - 4 = -1(x + 2)\). Simplifying gives \(y = -x + 2\).

Step 2: Use the Mean Value Theorem to find point \(c\)

The Mean Value Theorem states that if a function is continuous on the closed interval [a, b], and differentiable on the open interval (a, b), then there exists some number \(c\) in (a, b) such that the derivative of the function at point \(c\) equals the slope of the secant line over the interval [a, b]. We have \(f'(x) = -2x - 1\), and we equate it to the slope of the secant line, -1. This gives us \( -2x - 1 = -1 \). Solving for \(x\), we get \(x = 0\), which lies in the interval (-2, 2). So, \(c = 0\).

Step 3: Find the equation of the tangent line through \(c\)

The slope of the tangent line at point \(c\) is equal to \(f'(c)\) which is -1 (from the previous step). So, the equation of the tangent line is given by \(y - y_c = m(x - c)\), where \(m\) is the slope and \((c,y_c)\) is the point of tangency. Plugging in \(m = -1\) and using \(c = 0, y_c = f(0) = 6\), we have \(y - 6 = -1(x - 0)\). This simplifies to \(y = -x + 6\).

Step 4: Graphing the function, secant line and tangent line

To graph the function and the lines, a graphing utility such as Desmos or a graphing calculator can be used. Plot \(f(x) = -x^2 - x + 6\), the secant line \(y = -x + 2\), and the tangent line \(y = -x + 6\), on the same graph. You will observe that the secant and tangent lines intersect the function at the specific points \((-2, 4)\), \((2, 0)\) and \((0, 6)\), respectively.

Key Concepts

Secant Line

A secant line is a line that intersects a curve at two distinct points. It provides a straight-line approximation of the curve over an interval. For a function, you can find the equation of the secant line between two points using the formula for the slope of a line:

• Slope, \( m = (y_2 - y_1) / (x_2 - x_1) \), where \((x_1, y_1)\) and \((x_2, y_2)\) are the coordinates of the points.
• The formula for the line then follows: \( y - y_1 = m(x - x_1) \).

Given the points \((-2, 4)\) and \((2, 0)\) in our function \( f(x) = -x^2 - x + 6 \), the slope \( m \) works out to be -1, leading to the equation \( y = -x + 2 \). This secant line summarizes how the function behaves on the interval (-2, 2). The line is a key concept in calculus, often used in conjunction with the Mean Value Theorem.

Tangent Line

A tangent line touches a curve at exactly one point and represents the instantaneous rate of change (slope) of the function at that point. Unlike the secant line, which averages the rate of change over an interval, a tangent line gives us an exact 'snapshot' of the function's behavior at a specific point. To find the equation of a tangent line:

• Calculate the derivative, \( f'(x) \), which indicates the slope of the tangent.
• Determine the point of tangency, say \( (c, f(c)) \).
• Apply the point-slope form: \( y - y_c = m(x - c) \).

In our task, once we used the Mean Value Theorem to find the point \( c = 0 \) in the interval (-2, 2), we calculated the corresponding tangent line slope, \( f'(0) = -1 \). Therefore, the tangent line's equation is \( y = -x + 6 \), illustrating the behavior of our function precisely at \( x = 0 \).

Graphing Utility

A graphing utility helps visualize mathematical functions and their related lines, like secant and tangent lines. It enables clear understanding by representing equations graphically, which is useful for interpreting complex algebraic calculations. Using a tool such as Desmos or a graphing calculator:

• Input the original function \( f(x) = -x^2 - x + 6 \).
• Add the equations of the secant line \( y = -x + 2 \) and the tangent line \( y = -x + 6 \).
• Observe how these lines interact with the graph of the function at given points.

Graphing all three together reveals intersections at \((-2, 4)\) and \( (2, 0) \) for the secant line, and at \( (0, 6) \) for the tangent line. Visual tools simplify comprehension by linking the algebraic and graphical representations, reinforcing the understanding of how secant lines approximate interval behavior and tangent lines represent specific point behaviors.