The average rate of change over a given interval is the slope of the secant line through the endpoints of the interval.
![m = \dfrac{f(b)-f(a)}{b-a} = \dfrac{f(103)-f(100)}{103-100}\\\\ = \dfrac{(103)^2-2500-[(100)^2-2500]}{103-100} = \dfrac{609}{3}=203](https://tex.z-dn.net/?f=m%20%3D%20%5Cdfrac%7Bf%28b%29-f%28a%29%7D%7Bb-a%7D%20%3D%20%20%5Cdfrac%7Bf%28103%29-f%28100%29%7D%7B103-100%7D%5C%5C%5C%5C%20%3D%20%5Cdfrac%7B%28103%29%5E2-2500-%5B%28100%29%5E2-2500%5D%7D%7B103-100%7D%20%3D%20%20%5Cdfrac%7B609%7D%7B3%7D%3D203)
$203/unit.
The instantaneous rate of change at a certain point is the slope of the tangent line.
We differentiate C to get 2x. We substitute 100 for x to get that the instantaneous rate of change at 100 is $200/unit.
$203/unit.
The instantaneous rate of change at a certain point is the slope of the tangent line.
We differentiate C to get 2x. We substitute 100 for x to get that the instantaneous rate of change at 100 is $200/unit.