Chapter 2: Analysis of Graphs of Functions 2.1 Graphs of Basic Functions and Relations; Symmetry 2.2 Vertical and Horizontal Shifts of Graphs 2.3 Stretching, Shrinking, and Reflecting Graphs 2.4 Absolute Value Functions 2.5 Piecewise-Defined Functions 2.6 Operations and Composition

Equations and Inequalities Involving Absolute Value Example Solve |2x + 1| = 7. (Graphically and Analytically)

Equations and Inequalities Involving Absolute Value Analytic Solution For |2x + 1| to equal 7, 2x + 1 must be 7 units from 0 on the number line. This can only happen when 2x + 1 = 7 or 2x + 1 = −7. 2x + 1 = 7 or 2x + 1 = −7 2x = 6 or 2x = −8 x = 3 or x = −4 The solution set is {−4, 3}.

Solving Absolute Value Inequalities Graphically and Analytically Solve the inequalities (a) |2x + 1| > 7 and (b) |2x + 1| < 7.

Solving Absolute Value Inequalities Graphically Solve the previous equations graphically by letting y1 = |2x + 1| and y2 = 7 and find all points of intersection. The graph of y1 = |2x + 1| lies below the graph of y2 = 7 for x-values between –4 and 3, supporting the solution set (–4, 3). The graph of y1 = |2x + 1| lies above the graph of y2 = 7 for x-values greater than 3 or less than −4, confirming the analytic result.

Solving Absolute Value Inequalities Analytically Solve the inequalities (a) |2x + 1| > 7 and (b) |2x + 1| < 7. The expression 2x + 1 must represent a number that is more than 7 units from 0 on either side of the number line. 2x + 1 > 7 or 2x + 1 < −7 2x > 6 or 2x < −8 x > 3 or x < −4 The solution set is the interval (−∞, −4) U (3, ∞). The expression 2x + 1 must represent a number that is less than 7 units from 0 on the number line. −7 < 2x + 1 < 7 −8 < 2x < 6 Subtract 1 from each part. −4 < x < 3 Divide each part by 2. The solution set is the interval (−4, 3).

Solving Special Cases of Absolute Value Equations and Inequalities Solve Analytically and Graphically a) |3x + 5| = −5 b) |3x + 5| < −5 c) |3x + 5| > −5

Solving Special Cases of Absolute Value Equations and Inequalities Solve Analytically a) |3x + 5| = −5 b) |3x + 5| < −5 c) |3x + 5| > −5 Because the absolute value of an expression is never negative, the equation has no solution. The solution set is Ø. Using similar reasoning as in part (a), the absolute value of an expression will never be less than −5. The solution set is Ø. Because absolute value will always be greater than or equal to 0, the absolute value of an expression will always be greater than −5. The solution set is (−∞, ∞). Graphical Solution The graphical solution is seen from the graphing of y1 = |3x + 5| and y2 = −5.

Solving |ax + b| = |cx + d| Graphically and Analytically Solve |x + 6| = |2x − 3|

Solving |ax + b| = |cx + d| Graphically Solve |x + 6| = |2x − 3| graphically. Let y1 = |x + 6| and y2 = |2x − 3|. The equation y1 = y2 is equivalent to y1 − y2 = 0, so graph y3 = |x + 6| − |2x − 3| and find the x-intercepts. From the graph below, we see that they are −1 and 9, supporting the analytic solution.

Solving |ax + b| = |cx + d| Analytically To solve the equation |ax + b| = |cx + d| analytically, solve the compound equation ax + b = cx + d or ax + b = −(cx + d). Example Solve |x + 6| = |2x − 3| analytically. The equation is satisfied if x + 6 = 2x −3 or x + 6 = −(2x − 3) 9 = x or x + 6 = −2x + 3 3x = −3 x = −1 The solution set is {−1, 9}.

Piecewise-Defined Functions Example Find each function value given the piecewise-defined function

Piecewise-Defined Functions Example Find each function value given the piecewise-defined function

Graph of a Piecewise-Defined Function

Graph of a Piecewise-Defined Function

FYI: Graphing a Piecewise-Defined Function with a Graphing Calculator Use the test feature Returns 1 if true, 0 if false when plotting the value of x In general, it is best to graph piecewise-defined functions in dot mode, especially when the graph exhibits discontinuities. Otherwise, the calculator may attempt to connect portions of the graph that are actually separate from one another.

Graphing a Piecewise-Defined Function

Graphing a Piecewise-Defined Function

The Graph of the Greatest Integer Function If using a graphing calculator, put the calculator in dot mode.

The Greatest Integer (Step) Function Solution (a) −5 (b) 2 (c) −7 Using the Graphing Calculator The command “int” is used by many graphing calculators for the greatest integer function.

Application of a Piecewise-Defined Function Downtown Parking charges a $5 base fee for parking through 1 hour, and $1 for each additional hour or fraction thereof. The maximum fee for 24 hours is $15. Sketch a graph of the function that describes this pricing scheme. Solution Sample of ordered pairs (hours, price): (0.25, 5), (0.75, 5), (1, 5), (1.5, 6), (1.75, 6). During the 1st hour: price = $5 During the 2nd hour: price = $6 During the 3rd hour: price = $7 During the 11th hour: price = $15 It remains at $15 for the rest of the 24- hour period. Graph on the interval (0, 24].

Using a Piecewise-Defined Function to Analyze Data (1 of 2) Due to acid rain, the percentage of lakes in Scandinavia that had lost their population of brown trout increased dramatically between 1940 and 1975. Based on a sample of 2850 lakes, this percentage can be approximated by the piecewise-defined function f.

Using a Piecewise-Defined Function to Analyze Data (2 of 2) (a) Use the first rule with x = 1950.