The graph of a function f is shown. (a) Find the average...

The graph of a function $f$ is shown. (a) Find the average rate of change of $f$ on the interval $[20,60] .$ (b) Identify an interval on which the average rate of change of $f$ is $0 .$(c) Which interval gives a larger average rate of change, $[40,60]$ or $[40,70] ?$ (d) Compute $\frac{f(40)-f(10)}{40-10}$ . What does this value repre- sent geometrically?

The graph of a function $f$ is shown. (a) Find the average rate of change of $f$ on the interval
$[20,60] .$
(b) Identify an interval on which the average rate of change
of $f$ is $0 .$(c) Which interval gives a larger average rate of change,
$[40,60]$ or $[40,70] ?$
(d) Compute $\frac{f(40)-f(10)}{40-10}$ . What does this value repre-
sent geometrically?

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Key Concepts

Geometric Interpretation of the Difference Quotient

The difference quotient, such as (f(40)-f(10))/(40-10), represents the slope of the secant line between two points on the graph. This value gives a geometric sense of the overall change in the function from one point to the other and is crucial for understanding the function’s behavior over an interval.

Average Rate of Change

This concept refers to the change in the output of a function divided by the change in the input over a given interval. It is calculated as (f(b) - f(a))/(b - a) and represents the slope of the secant line connecting two points on the function's graph.

Secant Line

A secant line is a straight line that connects two points on a function’s graph. Its slope is equal to the function's average rate of change between those two points, providing a geometric interpretation of changes in the function.

Zero Average Rate of Change

When the average rate of change is zero over an interval, it means the function's value does not change between the two endpoints. Geometrically, this corresponds to a horizontal line segment on the graph, indicating no increase or decrease.

Transcript

00:01 Today we will be finding the average rate of change based on a graph.

00:04 To start, we'll look at the range 20 -60, and in this case, we will use the equation the average rate of change equals f -f -b minus f -a over b minus a, and a for this problem is 20, and b is 60.

00:17 So if we look on the graph, 20 is about here, which is 300, and 60 is about here, which is 700.

00:30 So coming back down here to the example.

00:31 We will have 700 minus 300 over 60 minus 20 which equals 400 over 40 which equals 10.

00:45 For part b of this question we'll be looking where the graph has an average rate of change of 0 and that means that there are two points that have the same y value.

00:57 So if you look these two points have the same y value and they're at 400 and this is at 10 and the other one is at 50.

01:06 So we can say that the interval is from 10 to 50.

01:12 To prove that this is true, we can plug it into the formula and both of them are at 400 and 50 minus 10, which is 0 because 400 minus 400 is 0.

01:25 So the average change is 0 at the interval 1050.

01:31 For the third part, we will be looking at whether the interval 4060 or 4070 is larger.

01:38 So to start, we'll look at 4060.

01:42 For 40, it's about here, which is at 200.

01:44 And for 60, we already looked earlier, is at 700...