Problem 4 - College Algebra and Calculus: An Applied Approach

Question

Use a graphing utility to graph the function and find its average rate of change on the interval. Compare this rate with the instantaneous rates of change at the endpoints of the interval. $$ h(x)=2-x ;[0,2] $$

Step-by-Step Solution

Answer

The average rate of change of the function \(h(x) = 2 - x\) over the interval [0,2] is -1. The instantaneous rates of change at the endpoints of this interval are both -1. Therefore, the average rate of change and the instantaneous rates of change are the same on this interval.

Step 1: Find the Average Rate of Change

To calculate the average rate of change of the function \(h(x)=2-x\) over the interval [0,2], use the formula: \[\frac{f(b) - f(a)}{b - a}\] Substitute \(a = 0\) and \(b = 2\) into the equation, with \(h(x) = 2-x\), to determine the average rate. Hence, \[\frac{h(2) - h(0)}{2 - 0} = \frac{(2-2) - (2-0)}{2} = -1\]

Step 2: Find the Instantaneous Rates of Change

The instantaneous rate of change at \(x = a\) is the derivative of \(h\) evaluated at \(x = a\). Thus, the derivative of \(h(x) = 2 - x\) is \(h'(x) = -1\). For \(x = 0\) and \(x = 2\), \(h'(0) = -1\) and \(h'(2) = -1\). Since the derivative \(h'(x) = -1\) remains constant, the function \(h(x) = 2 - x\) is a straight line with a slope of -1.

Step 3: Compare Rates of Change

The average rate of change found in step 1 is -1 and the instantaneous rates of change at \(x = 0\) and \(x = 2\) found in step 2 are also both -1. In conclusion, the average rate of change is equal to the instantaneous rate of change at all points in the interval [0,2]. This is consistent with the Second Mean Value Theorem for Integrals as the function \(h(x) = 2 - x\) is continuous on the interval [0,2] and differentiable on the open interval (0,2).