MTH What is a Difference Quotient?

Transcription

1 MTH What is a Difference Quotient? Another way to think about the difference quotient is that it yields a new function that gives the average rate of change of the original function, for two points that are h units apart horizontally. Let s describe this with a picture: Suppose that the graph shown at right represents a function, f ( x) 0.05x 0.1x 0.95x 1, and we want to look at an average rate of change (the slope between two points) for a pair of points that are units apart. This second graph shows a place where two points along the function are exactly two units apart. Their x-coordinates are -4 and -. We could calculate the average rate of change if we had the y values. We could find them by substituting these input values into the function and finding the output values. units Negative average rate of change (slope) units Similarly, we could find the average rate change for several places along the function that were exactly two units apart. Some of these average rates of change (slopes) would be positive, and some of the average rates of change would be negative. Some average rates of change could actually be zero. Positive average rate of change (slope) units Positive average rate of change (slope) units

2 But what if we wanted a quick way to find ALL of the average rates of change for points that were horizontally two units apart? First, we could calculate the difference quotient (bear with me on the calculations here): Original function f ( x) 0.05x 0.1x 0.95x 1 Function with x + h as input f ( x h) 0.05( x h) 0.1( x h) 0.95( x h) 1 Expanded f ( x h) 0.05x 0.15x h 0.15xh 0.05h 0.1x 0.xh 0.1h 0.95x 0.95h 1 Subtracting original function f ( x h) f ( x) 0.15x h 0.15xh 0.05h 0.xh 0.1h 0. 95h The difference quotient itself is a function, but it really has three inputs: 1. x (a starting point). h (How far away horizontally the second point should be). the function that you re taking the difference quotient of (in our case, the function f ) So, the difference quotient might be written like this: x, h, f ) f ( x h) h f ( x) 0.15x h 0.15xh 0.05h h 0.xh 0.1h 0.95h ** The notation above for Difference Quotient written as a function of three inputs is not in the textbook. You are not required to know it, and it will not be on any assessment in this course. Simplifying x, h, f ) f ( x h) h f ( x) 0.15x 0.15xh 0.05h 0.x 0.1h 0.95 So, we ve calculated the difference quotient. But what do we do with it? There are quite a few letters. The h in the difference quotient represents how far apart horizontally your two points are. We wanted our points to be exactly units apart horizontally, so let s let h =.

3 Substituting for h x,, f ) f ( x ) f ( x) 0.15x 0.15x() 0.05() 0.x 0.1() 0.95 f ( x ) f ( x) Simplified Difference Quotient( x,, f ) 0.15x 0.1x Hmm that s not so bad. It s a function; the right side is in the correct form for a quadratic equation (I know that officially in class we haven t gotten to quadratic equations, but most people already know what it is from a previous math experience). Let s graph it: Okay, so it s a parabola, but what does it mean? Well, the output (the height) of this function gives the average rate of change (slope between two points) of two points that are exactly horizontal units apart in our original function. So, what does it mean when the parabola has a height of zero? like here and here Now remember that the Difference Quotient output represents an average rate of change (slope) between two points that are horizontally h units apart. Since the output of the new function represents the average rate of change (slope) of the original function, it means that these x values are talking about places on the graph that have an average rate of change (or slope) of zero (like a horizontal line) with a point that is horizontally exactly two units apart from it.

4 There is a method to find these places where the height of a quadratic equation is zero (the quadratic formula, for those who remember it; again, we haven t officially talked about this yet, but many people probably remember it from previous math experience): x,, f ) 0.15x x 0.1x x 0.95 It is important to note WHY we are setting the y value equal to zero. We are looking for the x-values when the height of the parabola (the y-coordinate) is zero. Using the quadratic formula (bear with me on the math if you don t remember how this part works) a 0.15 b 0.10 c 0.95 b b 4ac b b 4ac x or x a a Just to verify that these values do actually have an output of zero on the parabola: ,, f ) 0.15( ).4548* ,, f ) 0.15( ) 4.7* ( ) ( ) 0.95 So yes, both x values that we found have an output of zero for the parabola, which was the output of our difference quotient.

5 Therefore, these are the x values at which the original function will have an average rate of change of zero when paired with the point that is two units horizontally to the right of it. (See diagram) x = units y = y = units x = The calculations here show that each of the inputs on the left produce an output that is equal to the height of the green horizontal line: f ( ) 0.05( ) f ( ) 0.05( ) ( ) 0.1( ) 0.95( ) ( ) 1 The second input is exactly units more than the first input, verifying that the two inputs are both exactly two units apart horizontally, and have the same height. To calculate average rate of change, we use the formula for slope between two points: y x y x 1 1 ( ) ( ) ( ) ( ) 0 0

6 The calculations here show that each of the inputs on the right produce an output that is equal to the height of the blue horizontal line: f ( ) 0.05( ) f ( ) 0.05( ) ( ) 0.1( ) 0.95( ) ( ) 1 Again, the second input is exactly units more than the first input, verifying that the two inputs are both exactly two units apart horizontally, and have the same height. Since they have the same height, it again follows that the average rate of change between those two points is zero. Why might this be useful? Suppose that a new airport was being built and they want to build it away from any obstructions (they don t want any mountains or hills right next to it). They want a runway miles long, but the land is full of hills and valleys. Where should they flatten the land for miles? If the original function, f ( x) 0.05x 0.1x 0.95x 1 represented the elevation of the land, and if x represented distance in miles, we just found two places where we could build a mile airstrip that would be level. The place to the right of the origin is down low and has hills/obstructions on both sides, whereas the place to the left of the origin would be on top of a hill, without obstruction. This is not the only example of where the difference quotient might be useful, but it is one example of a reasonable application.