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. $$ f(x)=x^{2}-6 x-1 ;[-1,3] $$

Short answer

The average rate of change of \(f(x)=x^{2}-6x-1\) on the interval [-1,3] is 3.5, the instantaneous rate of change at -1 is -8 and at 3 is 0. The average rate is not equal to the instantaneous rates at the endpoints.

Step-by-step solution

Graph the function

Using a graphing utility, plot the function \(f(x)=x^{2}-6x-1\). This will give us a visual representation of the function on the interval [-1,3].

Find the average rate of change

We calculate the average rate of change over the interval as the difference in the function values at the endpoints divided by the length of the interval, which means using the formula \(\frac{f(b) - f(a)}{b - a}\), we get \(\frac{f(3) - f(-1)}{3 - (-1)}\). After substituting \(f(3)\) and \(f(-1)\) by respectively 8 and -6, we have \(\frac{8 - (-6)}{3 - (-1)}\) which equals 3.5.

Find the instantaneous rates of change at the endpoints of the interval

Calculate the derivative of the function \(f(x)=x^{2}-6x-1\) which is \(f'(x)=2x-6\). Then substitute the endpoints of the interval into the derivative to find the instantaneous rates of change at the endpoints. The instantaneous rate of change at x=-1 is \(f'(-1)=2*(-1)-6=-8\) and at x=3 is \(f'(3)=2*3-6=0\).

Compare the average rate of change with the instantaneous rates of change at the endpoints

The instantaneous rate of change is -8 at x=-1, 0 at x=3 while the average rate of change is 3.5. Thus, the average rate of change is not equal to the instantaneous rates at the endpoints. We can visually confirm this from the graph where the slope of the secant line (representing the average rate of change) is not equal to the slopes of the tangent lines at the endpoints (representing the instantaneous rates of change).