Problem 12 - 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. $$ g(x)=x^{3}-1 ;[-1,1] $$

Step-by-Step Solution

Answer

After computing the function values, the average rate of change over the interval [-1, 1] is 0. The instantaneous rates of change at the endpoints -1 and 1 are 3 and 3 respectively. The instantaneous rate of change at the endpoints is higher when compared to the average rate of change on the interval. This shows how the function behaves differently at different points in the interval [-1,1].

Step 1: Finding the average rate

The average rate of change of \(g(x)\) over the interval \([-1,1]\) is given by the formula \(\frac{g(1)-g(-1)}{1-(-1)}\). Substitute \(g(-1)\) and \(g(1)\) with the respective function values to find the rate.

Step 2: Finding the instantaneous rate of change at the endpoints

The instantaneous rate of change of \(g(x)\) at a point \(x=a\) is given by the derivative \(g'(a)\). First, compute the derivative of the function \(g'(x)=3x^{2}\). Then, substitute \(x=-1\) and \(x=1\) to find the instantaneous rates of change at the ends of the interval.

Step 3: Comparing the rates

After obtaining the average rate of change and instantaneous rates of change, you can now compare them. Remember, the function will be increasing if the rate of change is positive and decreasing if the rate of change is negative. The rate of change tells us the steepness or the slope of the curve at a particular point.