# 2.1 Average Rate Of Changeap Calculus

2.1 Defining Average and Instantaneous Rates of Change at a Point Imagine you have a general function f(x), and you want to find the rate of change of that function over a certain interval a, b. Thatâ€™s pretty basic: you would just find the slope of the line that connects those two points. Construction similar to one used in AP Daily video Unit 2.1 (2020-21 academic year), intended to show how: 'average rate of change' may be interpreted as the slope of a secant line over an interval 'instantaneous rate of change' (a.k.a. Derivative) may be interpreted as the slope of a tangent line at a single input value.

#### Answer

a) $frac{f(x)-f(3)}{x-3}$ b) $f'(3) = limlimits_{h to 0}frac{f(3+h)-f(3)}{h}$

#### Work Step by Step

## 2.1 Average Rate Of Changeap Calculus Equation

## 2.1 Average Rate Of Changeap Calculus Formula

For part a, we know that the secant line is the average rate of change on the interval between the two points from equation 6. To calculate the slope of this line, we need to calculate the change in y divided by the change in x: $frac{f(x_{2})-f(x_{1})}{x_{2}-x_{1}}$ We can then plug our two given points into the equation to get our answer: $frac{f(x)-f(3)}{x-3}$ For part b, we know from equation 4 that the equation to find the slope of the tangent line is: $f'(x) = limlimits_{h to 0}frac{f(a+h)-f(a)}{h}$ The question tells us that point P is located at (3, f(3)). Next we can plug in 3 for a to get the equation for the tangent line: $f'(3) = limlimits_{h to 0}frac{f(3+h)-f(3)}{h}$