Algebra II: Strand 2. Linear Functions; Topic 2. Slope and Rate of Change; Task 2.2.1

Transcription

1 1 TASK 2.2.1: AVERAGE RATES OF CHANGE Solutions One of the ways in which we describe functions is by whether they are increasing, decreasing, or constant on an interval in their domain. If the graph of f consistently rises on the interval then we conclude that the function f is increasing on the interval. If the graph of f consistently falls on the interval then we conclude that the function f is decreasing on the interval. If the graph of f remains constant (horizontal) then we conclude that the function f is constant on the interval. This can be expressed symbolically as: A function f is increasing on an interval if for any x 1 and x 2 in the interval, where x 2 > x 1, then f (x 2 ) > f (x 1 ). A function f is decreasing on an interval if for any x 1 and x 2 in the interval, where x 2 > x 1, then f (x 2 ) < f (x 1 ). A function f is constant on an interval if for any x 1 and x 2 in the interval, where x 2 > x 1, then f (x 2 ) = f (x 1 ). A single function can be increasing on some intervals, decreasing on others, and constant on yet others. Consider the function shown below: f is increasing on the interval (-1,3) because for any x 1 and x 2 in the interval (-1,3), where x 2 > x 1, then f (x 2 ) > f (x 1 ). f is constant on an interval (3,6) because for any x 1 and x 2 in the interval (3,6), where x 2 > x 1, then f (x 2 ) = f (x 1 ). f is decreasing on the interval (6,10) because for any x 1 and x 2 in the interval (6,10), where x 2 > x 1, then f (x 2 ) < f (x 1 ).

2 2 It is useful to not only determine whether or not a function is increasing or decreasing over an interval but also how fast the function is either increasing or decreasing. To investigate how fast a function is increasing or decreasing over an interval, we calculate the change in f(x) relative to the change in x over the interval. This yields the average rate of change over the interval. To make this notion more mathematically precise we define the average rate of change of f from c to x as follows: If c is in the domain of a function y=f(x), the average rate of change of f from c to x is defined as: Average rate of change =!y!x f (x) " f (c) =, x # c x " c This expression is also called the difference quotient of f at c. Graphically, we can see that the average rate of change of f from c to x corresponds to the slope of the secant line passing through (c, f(c)) and (x, f(x)) as in the figure below: For a given value of c, the average rate of change of f from c to x is a function of x. Thus, we can say that for a given value of c, the function r c given by: r c (x) =!y f (x) " f (c) = is a function of f and represents the average rate of change of f!x x " c from c to x for any x not equal to c. If a function is increasing on the interval (c,x) then the sign of r c is positive. Why? When f is increasing x>c implies f(x)>f(c) so both (x-c) and (f(x)-f(c)) are positive.

3 3 If a function is decreasing on the interval (c,x) then the sign of r c is negative. Why? When f is decreasing x>c implies f(x)<f(c) so (x-c) is positive and (f(x)- f(c)) is negative and then their ratio is negative. If a function is constant on the interval (c,x) then r c (x)=0. Why? When f is constant x>c implies f(x)=f(c) so (x-c) is positive and (f(x)-f(c)) is zero and then their ratio is zero. Note that r c (x)>0 does not imply that f is increasing on (c,x). However, if we find that r c (x)>0 for any x in the interval (c,d) then we can conclude that f is increasing on the interval (c,d). The following illustrates this point: Consider the function f with graph f (8)! f (2) If we let c=2 and x = 8, then r 2 (8) = = 5! 4 = 1 is positive, but f is not 8! increasing on the interval (2,8). However, for any x in the interval (2,3), r 2 (x) is positive which would imply that f is increasing on the interval (2,3). Let s Try This! Complete the function tables for each of the following functions by finding f x i ( ) = f for the given values of x. You can use the graphing i calculator to make your calculations by following these steps: ( ) and

4 4 Enter the x values for the given function in L1. Press STAT, then ENTER to access the list editor. Enter the x i values in L1. Cursor to list L2 and move the cursor up to highlight the L2 at the top of the column. Type the function, replacing the x in the function with L1 (2 nd 1). For example, for x + 2, enter L The screen shots are for Problem 1, f (x) = 2x 1: Press ENTER. This will calculate the function values and store them in L2.

5 5 Move the cursor to list L3 and cursor up to highlight the L3 at the top of the column. Press 2 nd STAT (LIST), OPS, ΔList (7) Type L2 (2 nd 2), and close the parenthesis. Press ENTER. This will calculate Δf (x) for each pair of successive function values in list L2. This is also r(x) since the x values given are such that Δx = 1.

6 6 Graph each function on the graphing calculator and sketch the graph on the grid provided. Press Y= and enter Y1= 2x -1. Press ZOOM 6.

7 7 1. f (x) = 2x! 1 i x i f ( x i ) ( ) = f

8 8 2. f (x) =!3x + 2 i x i f ( x i ) ( ) = f

9 9 3. f (x) = 1 2 x! 2 i x i f ( x i ) ( ) = f

10 10 4. f (x) = x 2 i x i f ( x i ) ( ) = f

11 11 5. f (x) = x 3 i x i f ( x i ) ( ) = f

12 12 6. f (x) = 2 x i x i f ( x i ) ( ) = f

13 Reflect and Summarize What observations can you make about the average rate of change of linear functions compared to the average rate of change of non-linear functions? Linear functions have a constant average rate of change. Non-linear functions do not have a constant average rate of change. 2. Using the average rate of change of a function of f from c to x, r c (x) = f (x)! f (c), x! c and the slope-intercept form of the equation of a line, y = mx + b, prove that the slope, m, is the average rate of change of the function on any interval x 1,x 2 ( ). The rate of change of f from x 1 to x 2 is given by ( ) = f ( x 2 )! f ( x 1 ) r x1 x 2 x 2! x 1 for x 1 " x 2 ( ) = mx + b, we can express Since f x f (x 1 ) = mx 1 + b and f (x 2 ) = mx 2 + b, and substitute into r x1 ( x 2 ), giving ( ( ) = mx + b 2 )! ( mx 1 + b) r x1 x 2 x 2! x 1 Simplifying the numerator, we have r x1 ( x 2 ) = mx + b! mx! b 2 1 = mx! mx + b! b 2 1 = mx! mx 2 1 x 2! x 1 x 2! x 1 x 2! x 1 factoring out the m, we have ( ) = m ( x! x 2 1) r x1 x 2 x 2! x 1 = m for x 1 " x 2 Since x 1 and x 2 had no other restrictions other than x 1 " x 2, we have shown that the average rate of change of f from x 1 to x 2, for any x 1 and x 2 is m.

14 14 Math notes This task builds toward the understanding that the average rate of change of a linear function over any interval is constant and is equal to the slope of the linear function. We learn later in calculus that the slope of a linear function also corresponds to the instantaneous rate of change at any given point. Teaching notes It is helpful to introduce this task by sketching several graphs of functions and asking participants to note that parts of the graphs are rising, parts are falling, and parts are horizontal and that we have a mathematical way of describing this. Remind participants that we are examining the change in y over the change in x when we calculate values f (x)! f (c) for r c (x) =. Have participants work in individually initially and then move x! c them into groups to compare findings. Assign each group one of the questions 1-6 to present on chart paper. After group presentations of questions 1-6, have the participants work in groups on Reflect and Summarize task. After hearing participant responses to question 2 of the Reflect and Summarize task, the facilitator should have participants think about how they might describe the instantaneous rate of change of a linear function by building from their work that for an arbitrary interval (c,d) ( c! d ) the average rate of change (slope of the secant line through (c,f(c)) and (d, f(d)) is always constant and equal to the slope of the linear function. Have the participants graphically determine the instantaneous rate of change as d c. Extensions For each function in exercises 1-5, have participants derive a general formula for the f (x)! f (c) function r c (x) =. The facilitator may foreshadow calculus concepts of x! c instantaneous rates of change (derivatives) and limits by asking participants what the limiting function r c (x) would be as we consider smaller and smaller intervals (c,x).the facilitator may encourage participants to rewrite the function r c (x) in terms of the change f (c +!x) " f (c) in x:!x = x " c. They would then examine r c ( c +!x) =.!x

15 15 TASK 2.2.1: AVERAGE RATES OF CHANGE TASK 2.2.1: AVERAGE RATES OF CHANGE One of the ways in which we describe functions is by whether they are increasing, decreasing, or constant on an interval in their domain. If the graph of f consistently rises on the interval then we conclude that the function f is increasing on the interval. If the graph of f consistently falls on the interval then we conclude that the function f is decreasing on the interval. If the graph of f remains constant (horizontal) then we conclude that the function f is constant on the interval. This can be expressed symbolically as: A function f is increasing on an interval if for any x 1 and x 2 in the interval, where x 2 > x 1, then f (x 2 ) > f (x 1 ). A function f is decreasing on an interval if for any x 1 and x 2 in the interval, where x 2 > x 1, then f (x 2 ) < f (x 1 ). A function f is constant on an interval if for any x 1 and x 2 in the interval, where x 2 > x 1, then f (x 2 ) = f (x 1 ). A single function can be increasing on some intervals, decreasing on others, and constant on yet others. Consider the function shown below: f is increasing on the interval (-1,3) because for any x 1 and x 2 in the interval (-1,3), where x 2 > x 1, then f (x 2 ) > f (x 1 ). f is constant on an interval (3,6) because for any x 1 and x 2 in the interval (3,6), where x 2 > x 1, then f (x 2 ) = f (x 1 ). f is decreasing on the interval (6,10) because for any x 1 and x 2 in the interval (6,10), where x 2 > x 1, then f (x 2 ) < f (x 1 ).

16 16 It is useful to not only determine whether or not a function is increasing or decreasing over an interval but also how fast the function is either increasing or decreasing. To investigate how fast a function is increasing or decreasing over an interval, we calculate the change in f(x) relative to the change in x over the interval. This yields the average rate of change over the interval. To make this notion more mathematically precise we define the average rate of change of f from c to x as follows: If c is in the domain of a function y=f(x), the average rate of change of f from c to x is defined as: Average rate of change =!y!x f (x) " f (c) =, x # c x " c This expression is also called the difference quotient of f at c. Graphically, we can see that the average rate of change of f from c to x corresponds to the slope of the secant line passing through (c, f(c)) and (x, f(x)) as in the figure below: For a given value of c, the average rate of change of f from c to x is a function of x. Thus, we can say that for a given value of c, the function r c given by: r c (x) =!y f (x) " f (c) = is a function of f and represents the average rate of change of f!x x " c from c to x for any x not equal to c. If a function is increasing on the interval (c,x) then the sign of r c is positive. Why? If a function is decreasing on the interval (c,x) then the sign of r c is negative. Why?

17 17 If a function is constant on the interval (c,x) then r c (x)=0. Why? Note that r c (x)>0 does not imply that f is increasing on (c,x). However, if we find that r c (x)>0 for any x in the interval (c,d) then we can conclude that f is increasing on the interval (c,d). The following illustrates this point: Consider the function f with graph f (8)! f (2) If we let c=2, then r 2 (8) = = 5! 4 = 1 is positive, but f is not increasing on 8! the interval (2,8). However, for any x in the interval (2,3), r 2 (x) is positive which would imply that f is increasing on the interval (2,3). Let s Try This! Complete the function tables for each of the following functions by finding f x i ( ) = f for the given values of x. You can use the graphing i calculator to make your calculations by following these steps: ( ) and Enter the x values for the given function in L1. Cursor to list L2 and move the cursor up to highlight the L2 at the top of the column. Type the function, replacing the x in the function with L1 (2 nd 1). For example, for x + 2, enter L Press ENTER. This will calculate the function values and store them in L2.

18 Move the cursor to list L3 and cursor up to highlight the L3 at the top of the column. Press 2 nd STAT (LIST), OPS, ΔList (7) Type L2 (2 nd 2), and close the parenthesis. Press ENTER. This will calculate Δy for each pair of successive function values in list L2. This is also x 2 ( ) since the x values given are such that Δx = 1. Graph each function on the graphing calculator and sketch the graph on the grid provided. 1. f (x) = 2x! 1 18 i x i f ( x i ) ( ) = f

19 19 2. f (x) =!3x + 2 i x i f ( x i ) ( ) = f

20 20 3. f (x) = 1 2 x! 2 i x i f ( x i ) ( ) = f

21 21 4. f (x) = x 2 i x i f ( x i ) ( ) = f

22 22 5. f (x) = x 3 i x i f ( x i ) ( ) = f

23 23 6. f (x) = 2 x i x i f ( x i ) ( ) = f

24 Reflect and Summarize What observations can you make about the average rate of change of linear functions compared to the average rate of change of non-linear functions? 2. Using the average rate of change of a function of f from c to x, r c (x) = f (x)! f (c), x! c and the slope-intercept form of the equation of a line, y = mx + b, prove that the slope, m, is the average rate of change of the function on any interval x 1,x 2 ( ).