Let y=f(x)=x^2-4 x a. Find the average rate of change of y...

Let $y=f(x)=x^{2}-4 x$ a. Find the average rate of change of $y$ with respect to $x$ in the interval from $x=3$ to $x=4$, from $x=3$ to $x=3.5,$ and from $x=3$ to $x=3.1$ b. Find the (instantaneous) rate of change of $y$ at $x=3$ c. Compare the results obtained in part (a) with the result of part (b).

Key Concepts

Derivative

The derivative is a fundamental tool in calculus that provides an efficient way to compute the instantaneous rate of change. It is obtained by applying the limit process to the difference quotient. For functions where the derivative exists, it encapsulates the local linear behavior of the function and is used extensively in problems involving rates of change and curve analysis.

Secant vs Tangent Relationship

This concept illustrates the transition from the average rate of change (secant slope) to the instantaneous rate of change (tangent slope). As the interval used in computing the secant line reduces in size, the secant line approaches the tangent line at a point; hence, averages over smaller intervals better approximate the true instantaneous rate of change represented by the derivative.

Average Rate of Change

This concept refers to the ratio of the change in the function value to the change in the independent variable over a specified interval. It is calculated as the difference between the function values at two distinct points divided by the difference in the corresponding x-values, essentially finding the slope of the secant line connecting the two points on the graph.

Instantaneous Rate of Change

This represents the derivative of a function at a specific point, which is the limit of the average rate of change as the interval between the points approaches zero. It gives the exact slope of the tangent line to the function’s curve at that point, capturing the behavior of the function at an infinitesimally small scale.

Transcript

00:01 Okay, so we're going to find the average rate of change of this function on three intervals.

00:06 The first one is just from three to four.

00:10 So from x is equal to three to x is equal to four.

00:13 So how we do this is it's going to be equal to our value of our function at x is equal to four minus the value of our function at x is equal to three divided by four minus three.

00:24 So this is really just going to be equal to f of four minus f of three.

00:29 Since the interval length in this case is equal to 1.

00:33 So if we plug in 4 into our equation, this is equal to 4 squared, which is 16, minus 4 times 4 again is 4 squared, which is 16, so this is equal to 0.

00:45 If we plug in f of 3, we get 3 squared, which is equal to 9, minus 3 times 4 is equal to 12, so this is negative 3.

00:57 So now if we look at f of 4, it was 0 minus negative 3 so our average rate of change was 3 again i found this from this equation where i just took f of 4 minus f of 3 the next interval we're going to look at is from 3 to 3 .5 so now this is going to be equal to f of 3 .5 minus f of 3 of 3 divided by 3 .5 minus 3 which is equal to f of 3 .5 minus f of 3 divided by 0 .5 and that's the same as multiplying by 2.

01:38 So we already know the value of f of 3 .3.

01:41 Let's try and find the value of f of 3 .5.

01:44 So 3 .5 squared is equal to 12 .25 and we're minusing 4 times 3 .5 which is equal to 14.

01:58 So this is equal to 1 .75.

02:03 So now we know, and this is actually negative 1 .75, so now we know the two values of f of 3 and f of 3 .5, so we can plug them into our equation that we had up here.

02:14 So it would be negative 1 .75 minus negative 3, which is f of 3, divided by 0 .5.

02:24 So this is equal to negative 1 .75 plus 3, which is equal to 1 .25 divided by .5, which is equal to 2 .5.

02:34 So the average rate of change from 3 to 3 .5 is 2 .5.

02:41 And then the last one we're looking at is from 3 to 3 .1.

02:46 So we're just going to need to find f of 3 .1...

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