Problem 160 - Calculus Volume 1

Question

Graph the functions on a calculator and draw the secant line that connects the endpoints. Estimate the number of points \(c\) such that \(f^{\prime}(c)(b-a)=f(b)-f(a)\). \(y=x^{6}-\frac{3}{4} x^{5}-\frac{9}{8} x^{4}+\frac{15}{16} x^{3}+\frac{3}{32} x^{2}+\frac{3}{16} x+\frac{1}{32} \quad\) over[-1,1]

Step-by-Step Solution

Answer

There is at least one value of \( c \) such that \( f'(c) = -\frac{43}{64} \) in the interval \(-1, 1\).

Step 1: Understand the Mean Value Theorem

The Mean Value Theorem (MVT) for derivatives states that for a function \( f(x) \) which is continuous on \([a, b]\) and differentiable on \((a, b)\), there exists at least one number \( c \) in \((a, b)\) such that \( f'(c) = \frac{f(b) - f(a)}{b-a} \). For this problem, \( f(x) = x^{6} - \frac{3}{4}x^{5} - \frac{9}{8}x^{4} + \frac{15}{16}x^{3} + \frac{3}{32}x^{2} + \frac{3}{16}x + \frac{1}{32} \) on the interval \([-1, 1]\).

Step 2: Calculate \(f(a)\) and \(f(b)\)

Set \( a = -1 \) and \( b = 1 \). Calculate \( f(-1) \) and \( f(1) \).\[ f(-1) = (-1)^{6} - \frac{3}{4}(-1)^{5} - \frac{9}{8}(-1)^{4} + \frac{15}{16}(-1)^{3} + \frac{3}{32}(-1)^{2} + \frac{3}{16}(-1) + \frac{1}{32} = 1 + \frac{3}{4} - \frac{9}{8} - \frac{15}{16} + \frac{3}{32} - \frac{3}{16} + \frac{1}{32} \]\[ = \frac{89}{64} \]\[ f(1) = 1^{6} - \frac{3}{4} \cdot 1^{5} - \frac{9}{8} \cdot 1^{4} + \frac{15}{16} \cdot 1^{3} + \frac{3}{32} \cdot 1^{2} + \frac{3}{16} \cdot 1 + \frac{1}{32} \]\[ = 1 - \frac{3}{4} - \frac{9}{8} + \frac{15}{16} + \frac{3}{32} + \frac{3}{16} + \frac{1}{32} = \frac{3}{64} \].

Step 3: Find the Average Rate of Change

The average rate of change over \([-1,1]\) is calculated using the formula \( \frac{f(b) - f(a)}{b-a} \).\[ \frac{f(1) - f(-1)}{1 - (-1)} = \frac{\frac{3}{64} - \frac{89}{64}}{2} = \frac{-\frac{86}{64}}{2} = \frac{-86}{128} = -\frac{43}{64} \].

Step 4: Graph the Function and Secant Line

Graph the function \( y = x^{6} - \frac{3}{4}x^{5} - \frac{9}{8}x^{4} + \frac{15}{16}x^{3} + \frac{3}{32}x^{2} + \frac{3}{16}x + \frac{1}{32} \) on a calculator. Draw the secant line that connects the points \((-1, f(-1))\) and \((1, f(1))\) with a slope of \(-\frac{43}{64}\).

Step 5: Estimate Values of \(c\)

The slope of the tangent line at \( x = c \) is given by \( f'(c) \). To find \( c \), calculate the derivative and set it equal to the slope of the secant line:\[ f'(x) = 6x^5 - \frac{15}{4}x^4 - \frac{9}{2}x^3 + \frac{45}{16}x^2 + \frac{3}{16}x + \frac{3}{16} \]Solve \( f'(x) = -\frac{43}{64} \) to estimate value(s) of \( c \) within \(-1 < c < 1\). This typically requires numerical estimation using a calculator.

Key Concepts

Derivative Estimation

The concept of derivative estimation revolves around finding the instantaneous rate of change of a function at a certain point by using its derivative. The derivative, denoted as \( f'(x) \), gives you the slope of the tangent line to the function at a specific point. To estimate points where this slope equals the average rate of change across an interval, we often rely on the Mean Value Theorem.

According to the Mean Value Theorem, there is at least one point \( c \) in the interval \((a, b)\) where the derivative equals the overall average rate of change. In mathematical terms, \( f'(c) = \frac{f(b) - f(a)}{b-a} \). To find this point, you compute the derivative of the function, set it equal to the average rate of change, and solve for \( c \). This often entails solving algebraic equations, which can be done analytically or with the help of numerical estimation techniques.

Average Rate of Change

The average rate of change of a function is a simple yet powerful concept. It helps you understand how much a function, like the one you've graphed, changes on average between two points. For a given function \( f(x) \) over an interval \([a, b]\), you calculate it as \( \frac{f(b) - f(a)}{b-a} \). This formula essentially finds the slope of the secant line connecting the endpoints of the interval.

In practical terms, this provides insight into how the function behaves as you move from point \( a \) to point \( b \). In our example, the average rate of change, \(-\frac{43}{64}\), represents how much the function's output decreases per unit increase in \( x \) within the interval \([-1, 1]\). The calculation of this rate is fundamental in many areas like physics or economics, as it gives a straightforward view of overall change without dealing with complexities of the function's derivative.

Numerical Estimation

Numerical estimation is essential when dealing with complex equations, particularly derivatives, that might be challenging to solve algebraically. When the polynomial derivative derived from the function is equated to the average slope, solving for \( c \) might be difficult due to the equation's complexity. This is where numerical methods step in.

By employing numerical techniques, such as graphing calculators or computer software, you can approximate solutions for expressions like \( f'(x) = -\frac{43}{64} \). These tools iterate through possible \( x \) values within the stipulated interval and compute the derivative until the correct approximation is found.

This powerful approach allows for determining precise estimates of \( c \), as per the Mean Value Theorem, even for intricate polynomials or transcendental equations. Utilizing such methods offers a great deal of flexibility and accuracy for many mathematical applications.