Comments to the presenter.  1. All presented problems are solved.  2. Examples are suggested to be discussed, problems to be solved by students.  3. Check for the note “click”, please. If you see it, then there are steps in discussion of the concept or the solution of the problem.  4. Any comments and suggestions to improve the session would be appreciated.

Review Session Basics of Functions & Their Graphs

Pretest. 1. Domain? Range? 2. Find f(1) = f(2) = 3. Find x for which f(x) = -5 f(x) = 0 4. Identify the intervals over which the function is increasing/decreasing/constant Click to check answers.

Pretest - Solutions. 1. Domain? Range? 2. Find f(1) = -4 f(2) = Find x for which f(x) = -5 for x = 2 f(x) = 0 for x = -0.3 & Identify intervals over which the function is increasing/decreasing/constant f(x) is decreasing on, increasing on

Learning Objectives for Part I Functions, Domain & Range, Function Notation. Apply the definition of a function. Identify domain and range of a function. Identify the domain & range from a graph. Identify the domain & range from an equation. Use function notation & evaluate a function. Find the Difference Quotient. Read graphs of functions. Piecewise-Defined Functions.

Apply the definition of a function.

Definition of a Function What is a function? A correspondence (a rule that matches the elements) between two sets D and R such that to each element of the first set, D, there corresponds exactly one and only one element of the second set, R. The first set is called the domain, and the set of corresponding elements in the second set is called the range. Domain: the set of all first entries (inputs) to be used in the rule (also called domain elements or independent variables). Range: the set of all second entries (outputs) that result from applying the rule (also called range elements or dependent variables). ( x, y ) Independent Variable, input Dependent Variable, output

Functions are commonly represented in 4 ways. 1.Verbally by a sentence. 2.Numerically by a list of ordered pairs or by a table. 3.Graphically by points on a graph in a coordinate plane. 4. Algebraically by an equation in two variables. xy –4 16 – Click for each answer. Definition of function: To each element of the first set, D, there corresponds one and only one element of the second set, R. Paycheck (P) depends on hours worked (h). Use the definition of a function to determine whether each example is a function. Function Not a function Function

Not a Function. Graphical Representation. Use Vertical Line Test for Defining Functions. If any possible vertical line intersects a graph in at most one point, then the graph is a graph of a function. Function.

Identify domain and range of a function from the graph. Comment: In this presentation, please assume that the graph continues outside of the portion of the coordinate system that is shown, unless otherwise indicated.

Domain: set of first components in the relation (inputs, independent variables) Range: set of second components in the relation (outputs, dependent variables) Graphical Representation. Find the Domain & Range of the Function. Domain: Look at the x-axis. Considering all the points on the graph, identify all the x-values included on the graph. (Think about whether the graph extends further to the left and right. Include all x-values that occur on the graph.) Range: Look at the y-axis. Considering all the points on the graph, identify all the y-values included on the graph. (Think about whether the graph extends further up and down. Include all y-values that occur on the graph.) Domain: Range: How to identify the domain : (on click) How to identify the range: (on click.) Example

Function? Type of Function? Domain? Range? Yes (Vertical Line Test) Absolute Value function: Answers on clicks. Graphical Representation. Find the Domain & Range of the Function. Practice 1

Function? Name? Domain? Range? Yes (Vertical Line Test) Square Root function Graphical Representation. Find the Domain & Range of the Function. Answers on clicks. Practice 2

Function? Type of Function? Domain? Range? Yes (Vertical Line Test) Answers on clicks. Reciprocal function Graphical Representation. Find the Domain & Range of the Function. Practice 3

How to Identify the domain & range from the equation (Some Examples appear below)

If a function is specified by an equation and the domain is not indicated, we agree to assume that the domain is the set of all real inputs that produce real outputs. This set of all outputs is the range. To Identify the Correct Domain: Begin with All the Reals and exclude any values of x that would result in either of these types of problems: Type 1: A denominator that equals zero. Type 2: A negative radicand under a radical with an even index (root). Excluding the x-values that would result in these problems is referred to as “restricting the domain”. Representation by Equation. Finding the Domain & Range.

Example or [The y-value can also be any real number, since y is always 4 more than x.] Note: For any Linear function, if no restriction is specifically stated on the domain or range, both the Domain and Range will always be. Representation by Equation. Finding the Domain & Range. Function? Click for each answer. Yes. (By the definition of a function) Name? Linear function (could be written: ax +by = c, where b is not zero) Domain? Range? or (Since x may be any real number.)

This expression is defined for all real values of x except where the denominator of the fraction is zero. This means that we begin with all real numbers and then restrict the domain by excluding x = 5/3. The output values for the expression will be any real number except 0. (Why?) Representation by Equation. Finding the Domain & Range. Click for each answer. Domain: Example Range: or Because any fraction with a constant as the numerator cannot equal zero.

Consider: Is there any value of x we cannot use? For example, what is y if x = 0? Not a real number! We must Restrict the domain. Ask: For what values of x is y defined? (Note that a square root is an even root. ) Answer: Using algebra, this is written: 3x –2 ≥ 0 and solving for x, we find, y is defined only when the radicand (3x-2) is equal to or greater than zero: Example Representation by Equation. Find the Domain & Range. Domain? Domain: Range? Range: If x =, then y=0, and when x has a value greater than, the radicand is a positive number, so the square root would also be a positive number. The range is y ≥ 0. Click for each answer. This is the domain:

Conclusion. In general, the domain of a function excludes values that would cause division by zero ( Type 1 ) when the variable appears in a denominator or that would result in an even root of a negative number ( Type 2 ) when the function contains an even root [even index] of an expression that contains the variable. To find the domain, If Type 1, 1. Set any denominator equal to zero; 2. Solve the resulting equation, and 3. Exclude all these solutions from the set of all real numbers. If Type 2, 1. For any radical with even index, set the radicand greater than or equal to zero; 2. Solve the resulting inequality, and 3. Use the solution set as the domain. Representation by Equation. Find the Domain & Range. Click to continue.

Practice 4 Representation by Equation. Find the Domain & Range. Name? Domain: Range: Click for answers. Square Root function (Note: even root) What is the process to follow to identify the Domain and Range? To find the Domain: This problem has no denominator, but it does contain a radical with an even root. Set the radicand 0 and solve for x. The solution set is the domain of the function. To find the Range: 1)Determine any values that y will not equal and exclude those from the range. 2)A graph may be helpful. Set the Radicand 0 and solve:

Set the radicand -2x+3 greater than or equal to zero, and solve for x. –2x + 3 > 0 –2x > –3 – 2x/(– 2) < – 3/(– 2) x < 3/2 The domain is “ all real x < 3/2” or The range is “all real y < 0” or Practice 4 (Solutions) Representation by Equation. Find the Domain & Range. Name? Square Root function Apply the Type 2 process since the radical has an even index (root). Domain: Divide by –2 on each side and Range:For the range you may want to refer to the graph. Click once for each answer. reverse (flip) the inequality symbol!

Representation by Equation. Find the Domain & Range. Domain: Click for answers. Name? Rational function. Range: Practice 5 What is the process to follow to identify the Domain and Range? To find the Domain: This function contains no radical, but there is a denominator. Set the denominator to zero. Solve for x, and exclude any resulting values of x from. To find the Range for this function, a graph will be helpful.

Representation by Equation. Find the Domain & Range. Domain: Click once for each answer. Name?Rational function. Set the denominator equal to zero, and solve. Exclude all values of x that would result in a denominator value of 0: x 2 – x – 2 = 0 (x – 2)(x + 1) = 0 x = 2 or x = –1 As we see from the picture, the graph will eventually include or “cover” all possible values of y, so the range is "all real numbers” or. Range: The domain is "all real x ≠ –1, 2” or In interval form the domain is: Practice 5 (Solutions) Apply the Type 1 process to identify restrictions on the domain. The range requires the graph.

Representation by Equation. Find the Domain & Range. Domain: Range: Click for answers. Name?Equations like the above represent Polynomial functions. Practice 6 Polynomial functions have no denominators (so no division-by-zero problems) and no radicals (so no even-root-of-a-negative problems). For any polynomial function, the domain is "all real values of x”. For any polynomial of odd degree, the range is always. The Range of a polynomial of even degree depends on the sign of the leading coefficient: Positive Sign: ; Negative sign: Where “a” represents some real number value. Range of (A): In the graph of (A) when x is zero, y is 4, and otherwise y < 4. Range of (B): All real numbers will occur on the graph of (B) as y-values.

You may choose to sketch the graph of the function and find the range from the graph. Consider the graphs that are shown at the right. Each function has a negative leading coefficient. Degree of Equation (A): even Degree of Equation (B): odd Representation by Equation. Find the Domain & Range. Domain: Range: Click once for each answer. Name? Equations like those shown above, where y equals a sum of terms in which x has integer exponents represent Polynomial functions. For any polynomial function, the domain is "all real values of x”. Practice 6 (Solutions)

Function Notation and Evaluation.

Function Notation The notation f (x) will now be used instead of the variable y. This is read as “ f of x” and f(x) is equivalent to the y-coordinate of the function corresponding to a given x value. The equation can now be expressed as We say, “y is a function of x” to emphasize that y depends on x. We use the notation y = f (x), which is function notation. Value of the function (y or f(x)) Name of the function Independent variable Defining expression (formula) Start on click.

Find f (2) for each different function shown. a. Table. b. Set of ordered pairs. f = {(2, 6), (4, 2)} xf(x)f(x) –416 – c. Graph. d. Equation. f(x) = –x 2 f(2) = –(2) 2 Example Check each problem on click. Given a value of x, (the first coordinate or independent variable), evaluate the function f(x) using that x-value.

Function Evaluation Consider the function It tells you the “instructions” you are to follow. What does f (-3) mean? Replace x with the value -3 and evaluate the expression: The notation means: “ when x = -3, the value of f(x) is 11” So the point (-3,11) is on the graph of function f. Start on click. Example

Evaluating a function Given f(x) = x 2 + 2x – 1, find f(2), f(–3), f(k). Each location where you see an "x” in the function f(x) is just an open box, waiting for something to be put into it. f(2) = 7 since f(2) = (2) 2 +2(2) – 1 = 7 f(–3) = 2 since f(–3) = (–3) 2 +2(–3) – 1 = 2 f(k) = k 2 + 2k – 1 Functions can be defined using a letter different from x. The function below is exactly the same function as f(x) = x 2 + 2x – 1. However, below it is defined using “m”. f(m) = (m) 2 + 2(m) – 1 = m m – 1 You would still find f(2) or f(-3) using the same steps: in f(m) replace “m” with “2” the first time and the second time replace the “m” with “-3”, just as was done above. This is how to work with function notation. Click for the answers. Example

Evaluate a Function Evaluate f(x) = 3x – 7 at x = 2, a, 3-x f(2) = -1 f(2) = 3(2) – 7 = -1; f(a) = 3a – 7 f(a) = 3(a) -7 = 3a – 7; f(3 – x) = 2 – 3x f(3 – x) = 3(3 – x) – 7 = 2 – 3x. Click once for the answers. Click once for solutions. Practice 7

Difference Quotient

Find the Difference Quotient Given f(x) = 3x 2 + 2x, find When working through more involved exercises like this example, start by splitting the exercises into smaller steps, starting with the f(x + h) in the numerator:  in f( ) = 3( ) 2 + 2( ) replace the x with (x+h) to find f(x + h): f(x + h) = 3(x + h) 2 + 2(x + h) = 3x 2 + 6x h + 3 h 2 + 2x + 2h  f(x) = 3x 2 + 2x [The original function is repeated here.]  f(x + h) – f(x) = [3x 2 + 6xh + 3h 2 + 2x + 2h] – [3x 2 + 2x] = 3x 2 + 6xh + 3h 2 + 2x + 2h – 3x 2 – 2x = 6xh + 3h 2 + 2h  Example Click for steps. Note: Simplify the numerator first. Put the “h” back in the denominator later.

Find the Difference Quotient Given f(x) = 2x 2 - 2x + 1, find Practice 8 Click for the answer. Click for the solution.

Find the Difference Quotient Given f(x) = 2x 2 - 2x + 1, find Practice 8 (Solution) Click once for solution.

Reading Graphs

Increasing/Decreasing/Constant Functions Increasing: The graph goes “up” as you move from left to right. Decreasing: The graph goes “down” as you move from left to right. Constant: Graph remains horizontal as you move from left to right. For points (x 1, y 1 ) and (x 2, y 2 ) on the graph,

1. Is it a function? 2. Domain: Range: Use the graph to determine the following. 3. Find f(1) = f(2) = f(4) = f(5) = 4. Find x for which f(x) = 2 for x = 2, 3 f(x) = 1 for x =0 or x from 5. Intervals on which f is increasing, decreasing, or constant. Example Yes (Vertical Line Test) f is increasing (up) on (2, 3), constant (horizontal) on (1, 2) undefined decreasing (down) on Check each problem on click.

Piecewise-Defined Function.

Evaluate Piecewise-Defined Function Evaluate f(0), f(2), f(3) for the function Notice that this function is defined by different rules I and II for different parts of its domain. A piecewise-defined function is a function whose definition involves more than one formula (has two or more “pieces”). f(0)=? Since 0 < 2, use f(0) = 2 –2(0) = 2, f(2)=? Since 2 ≥ 2, use f(2) = (2) – 2 = 0, f(3)= ? Since 3 ≥ 2, use f(3) = (3) – 2 = 1. Click for each evaluation step. (I) (II) (I) Example Graphing such functions involves graphing each rule over the appropriate portion of the domain on the same coordinate plane.

Since the function has two pieces (rules), we use two T-charts. The break between the two “pieces" of this function (the point at which the function changes rules) is at x = 2, so that is where T-charts will break. The function is graphed in two pieces: the first piece includes only x-values before x = 2, and the second piece has domain from x = 2 to positive infinity. Why is the last point in T-chart I in parentheses? Actually, x = 2 is not a domain value for T-chart I, but it is helpful to know the y-value that the piece "approaches" close to this endpoint of its domain (as x gets closer to 2). Graph (I) (II) (I)(II) xy Example xy (2) Notice that the point (2,-2) is not included in the graph of part I (shown with an open circle), but the point (2,0) is included in graph of part II (shown with a closed circle) (-2) Click to complete each T-Chart (table) below Click for each part of the graph. Click for the answer.

Graph Practice 9 (I) (II) Click for the graph. (I) (II)

xy (1)(-1) xy Why would you need two T-charts for the graph? Why is the point (1,-1) in T-chart I in parentheses? Describe what happens to the graph of f(x) at x = 1? The endpoint (1,-1) on the graph of piece I is an open circle. (Why?). The endpoint (1,2) of piece II is a closed circle. (Why?). Graph Practice 9 (Solution) (I) (II) (I) (II) Click for the graph. Answer the questions. There are two pieces. It’s not on the graph. The rule for the function changes. (This graph has a break or jump at x=1.) This point is not on the graph. 1 is not in the domain for this piece of the graph. The point (1, 2) is on this piece of the graph. 1 is included in the domain for this piece of the graph

Graph Transformations.

The basic graph transformations are:  Vertical and Horizontal translations (shifts),  Vertical and Horizontal stretches and shrinks.  Reflection in the x-axis and y-axis, Learning Objectives for Part II Graph Transformations.

Vertical shifts : Move the graph up (+k) or down (-k); Impact only the “y” values; No changes are made to the “x” values. Horizontal shifts: Move the graph left (+k) or right (-k) (HINT: go the opposite direction); Impact only the “x” values; No changes are made to the “y” values. Vertical and Horizontal Translations of Function f(x) g(x) = f(x) ± k (k > 0) and g(x) = f(x ± k) (k > 0) Start on click.

Combining a Vertical & Horizontal shifts List the transformations to the basic function. 1. Horizontal shift right 2 units, 2. Vertical shift down 4 units. What is the basic function? Graph: Click for each answer. Absolute Value function Click for the graph. Practice 10 (with solution)

Vertical Stretches & Shrinks Stretch graph vertically (k > 1), or Compress graph vertically (0< k < 1). Impact only the “y” values; No changes are made to the “x” values. Horizontal Stretches & Shrinks Stretch graph horizontally (0 1). Impact only the “x” values; No changes are made to the “y” values. Vertical and Horizontal Stretches and Compressions of Function f(x) g(x) = kf(x) (k >0) or g(x) = f(kx) (k >0) Start on click.

Vertical Stretching (by factor of 3) Transformation of the Graph: Vertical Compression (by factor of 1/4) Graph: Transformation of the Graph: Click once for solution. Basic function: Graph: Click once for solution. Practice 11 (with solution)

Reflection in the (around the) x-axis -f(x) reflects f(x) around the x-axis. Impacts only the sign of the “y” values, (e.g., positive y values become negative); No changes are made to the “x” values. Reflection in the (around the) y-axis f(-x) reflects f(x) around the y-axis. Impacts only the sign of the “x” values, (e.g., negative x values are changed to positive); No changes are made to the “y” values. Reflections of Function f(x) around y -axis and x -axis g(x) = -f(x) or g(x) = f(-x) Click once to start. g(x) = -f(x) g(x) = f(-x)

Summary of Graph Transformations Reflection in y-axis: y = f (-x) Reflect the graph of y = f (x) in the y-axis. Horizontal Stretch or Shrink: y = f (ax) (a >0) a > 1: Stretch graph of y = f (x) horizontally by multiplying each x-value by a. 0< a <1: Shrink graph of y = f (x) horizontally by multiplying each x-value by a. Horizontal Shift: y = f (x ± c) (c >0) + c Shift graph of y = f (x) left c units. - c Shift graph of y = f (x) right c units. Vertical Stretch or Shrink: y = af (x) (a >0) a > 1: Stretch graph of y = f (x) vertically by multiplying each ordinate value by a. 0< a <1: Shrink graph of y = f (x) vertically by multiplying each ordinate value by a. Reflection in x-axis: y = -f (x) Reflect the graph of y = f (x) in the x-axis. Vertical Shift: y = f (x) ± d (d >0) + d Shift graph of y = f (x) up d units. - d Shift graph of y = f (x) down d units.

Multiple Transformations This would be the order of the steps to follow if all these types of transformations are needed: Reflect in the y-axis Horizontal Stretch or Shrink Horizontal Shift Vertical Stretch or Shrink Reflect in the x-axis Vertical Shift 1.Horizontal Shift 2.Vertical Stretch or Shrink 3.Reflection in the x-axis 4.Vertical Shift However, the examples included in this presentation require only these steps:

Basic function: List of transformations: 1.Horizontal shift: 4 units left; 2.Vertical stretch: by factor 2; 3.Reflection: around x-axis; 4.Vertical shift: 2 units down. Domain h(x): Range h(x): Example Show sequence of transformations of f(x) to get State the domain and range of h(x). Click for each step.

Graph using key points: Basic function: Click for each step. Key points: List of transformations of key points: 1.Horizontal shift 4 units left (subtract 4 from each x): (-4,0), (-3, 1), (0,2); 2.Vertical Stretch by factor 2 (2 times all the y’s): (-4,0), (-3,2), (0,4); 3.Reflection in the x-axis (change signs of all y’s): (-4,0), (-3,-2), (0,-4); 4.Vertical shift: 2 units down (subtract 2 from each y): (-4,-2),(-3,-4),(0,-6). Example (0,0), (1,1), (4,2)

Basic function: y= x 3 Key points: (0,0), (1,1), (-1,-1) List of transformations of key points: Sequence of transformations Practice 10 Graph: 1.Horizontal shift 3 units right (add 3 to each x): (3,0), (4, 1), (2,-1); 2.Vertical Stretch by factor 4: (4 times all the y’s): (3,0), (4,4), (2,-4); 3.Reflection in the (around) x-axis: (change signs of all y’s): (3,0), (4,-4), (2,4); 4.Vertical shift: 1 unit down: (add 1 to each y): (3,-1), (4, -5), (2,3).

Basic function: Click for each step. Key points: List of transformations of key points: 1.Horizontal shift 2 units right (add 2 to each x): (2,0), (3, 1), (1,1); 2.Vertical Stretch by factor 3: (3 times all the y’s): (2,0), (3,3), (1,3); 3.Reflection in the (around) x-axis: (change signs of all y’s): (2,0), (3,-3), (1,-3); 4.Vertical shift: 1 unit up: (add 1 to each y): (2,1), (3, -2), (1,-2). Sequence of transformations Practice 10 (Solution) Graph: (0,0), (1,1), (-1,1)

Learning Objectives for Part III Function Operations and Composition Arithmetic Operations on Functions and Domain. Composition of Functions and Domain. Inverse Functions.

Arithmetic Operations on Functions and Domain.

Given two functions  and g, and for all x common to both domains, the sum, difference, product & quotient of  and g are defined as follows. 1.Sum: 2.Difference: 3.Product: 4.Quotient: The domain of an arithmetic combination of functions f and g consists of all real numbers that are common to the domains of f and g. For the quotient f(x) / g(x), there’s further restriction that g(x) ≠ 0. Arithmetic Operations on Functions

Solutions Using Operations on Functions, Evaluating New Functions. Practice 11 Click for each answer.

The Domains For functions  and g, the domains of  + g,  – g, and  g include all real numbers in the intersection (common part) of the domains of  and g, while the domain of  / g includes those real numbers in the intersection of the domains of  and g for which g(x) ≠ 0. The two rules: (1)Never allow a negative radicand under a radical sign with an even index (root). (2)No denominator of a fraction can be allowed to equal zero.

Example Finding the Domains of Quotients of Functions Remember that any restrictions on the domains of f or g must be considered to determine the domain when performing arithmetic operations on functions. The domain of f is (why?), and the domain of g is (why?) Click for steps. So, the domain of (f/g) is, and the domain of (g/f) is. In f/g x=4 would give a 0 denominator as would x=0 in g/f. Domain of : Since no square root can have a negative radicand, and no denominator can be zero, this domain can be found by solving the inequalities and and finding the intersection of the two solution sets. The domain of is [0,4). The intersection (common part) of these domains is (refer to a number line). For a quotient, we also have the restriction that the denominator cannot be 0. Why are the two domains different?

Practice 12 (with answers) Using Operations on Functions, Finding Domains. Click once for the solutions to (A). (B) The domains of  + g,  – g,  g: The domain  /g: Click again for the answers to (B). See next slide for the solutions to (B).

1) Find the domains of  and g. The domain of  is the set of all real numbers: (– ,  ); The domain of g is the set of all real numbers that make 2x – 1 nonnegative: 2) The domain each of these functions is the intersection of the domains of  and g: Finding Domains. The domain of  / g : The domain for each of  + g,  – g,  g is: Click for steps. Practice 12 (solutions for B) Domain of Numerator: Domain of Denominator: But no denominator can be equal to zero so, which means so we know The intersection is.

Composition of Functions and its Domain.

If  and g are functions, then the composite function, or composition of g and , is defined by The domain of is the set of all real numbers x in the domain of  such that  (x) is in the domain of g. Composition of Functions and its Domain.

Composition of Functions Example Click for steps.

Problem 5 Example Click for answers (answer color matches problem color).

Example Composition of Functions and Domain. (B) Domain? Would it be correct to say that the domain for this composition function is the set of all real numbers??? Click for steps. Begin with the domain of f, which is all real numbers except 3. Exclude any values of x for which f(x) is not in the domain of function g and any other values of x for which would be undefined. Since 3 is the only x-value that needs to be excluded, the domain of the function is all reals except 3.

Composition of Functions and Domain. Example (cont.) 1). The domain of  is all real numbers except 3. (why?) 2). The domain of g is all real numbers except 0. (why?) Will f(x)= 0 for any x-value? 3).The function values of f(x) are used as “x” in g. Will the denominator of g equal 0? 4). For the domain of, begin with the domain of the first function applied, function f. Exclude values of x where f(x) is not in the domain of function g and any other values of x for which would be undefined. For this example, only the number 3 must be excluded from the set of all real numbers. Input for g(x) The domain of is or Click for steps.

Composition of Functions and Domain. Practice 13 Click for answers. Begin with the domain of function g, which is all real values of x except zero. Exclude any values of x for which g(x) is not in the domain of function f and any other values of x for which would be undefined. Exclude both 0 and 1/3 to arrive at the domain (B) of the composite function. (Note: Continue for additional information concerning deriving this composite function.)

Composition of Functions. To simplify the complex fraction multiply both the numerator and denominator by x. Or change the denominator to fraction form and then multiply by its reciprocal. Practice 13 (Solution for A) Click for steps.

Composition of Functions - Domain. (B) Domain: 1). The domain of  is all real numbers except 3. (why?) 2). The domain of g is all real numbers except 0. (why?) 4). g(x) = 3 for some x. However, 3 cannot be used in function f; Then solve g(x) = 3 and exclude the solution from the domain of Input for f(x) 5).Therefore the domain of is the set of all real numbers except 0 and : Click for steps. Practice 13 (Solution for B)

Inverse Functions.

Inverse Functions For a function  to have an inverse,  must be a one-to-one function. A function is said to be one-to-one if each range value corresponds to exactly one domain value. By the definition of inverse function, the domain of  is the range of  -1, and the range of  is the domain of  -1 Note: Any function that is either increasing or decreasing on its entire domain has an inverse function.

Decide if the Given Functions are Inverses. Example 1) Apply the Horizontal Line Test to graphs to see if the functions are one-to-one. 2) Check to see if both statements are true: a) (  ◦ g)(x) = x and b) (g ◦  )(x) = x. 1) Both functions are one-to-one. (Why? - give your reason) 2) a) (  ◦ g)(x) = f(g(x)) Prove on your own. Check the proof. b) (g ◦  )(x) = Conclusion: Functions f and g are inverse functions. (g = f -1 ) Steps Click for steps.

1. If  is one-to-one, then  -1 exists. ( For graphs use the Horizontal Line Test) 2. The domain of  is the range of  -1, and the range of  is the domain of  If the point (a, b) lies on the graph of , then (b, a) lies on the graph of  -1, so the graphs of  and  -1 are reflections of each other across the line y = x. 4. To find the equation for  -1, 1) replace  (x) with y; 2) interchange x and y; 3) solve for y; 4) replace y with  -1. This is your candidate for the inverse function. 5) prove that (  º  -1 )(x)=x and (  -1 º  )(x)=x Important Facts About Inverses Click once) 6) Celebrate!

FINDING THE INVERSE OF A FUNCTION WITH A RESTRICTED DOMAIN Example First, notice that the domain of  is restricted to the interval [– 5,  ). Function  is one-to-one because it is increasing on its entire domain and, thus, has an inverse function. Apply steps. 1) replace  ( x ) with y ; 2) interchange x and y; 3) solve for y; 4) replace y with  -1 Click for each step. Continued on next slide.

Example (cont.) However, we cannot define  -1 as x 2 – 5. FINDING THE INVERSE OF A FUNCTION WITH A RESTRICTED DOMAIN As a check, graphs of  and  -1 are helpful. 5) Prove that (  º  -1 )(x)=x and (  -1 º  )(x)=x on your own. [If the result is “x” both times, it proves that the two functions are inverses functions.] The domain of  is [– 5,  ), and range is [0,  ). The domain of  -1 must match the range of , so we define the domain of  -1 as shown in red here: The domain of  -1, [0,  ), is the range of . The line y = x is included on the graphs to show that the graphs are mirror images with respect to this line. Click to continue.

Finding the Inverse of a Function Practice 14 Given decide whether the equation defines a one-to-one function. If so, find the equation of the inverse. Click for the answer. The given equation does not define a one-to-one function. The given function does not have an inverse on its current domain. However, if the domain were restricted so that the function would then be a one-to-one function, the function with restricted domain would then have an inverse. If the domain were restricted to the interval, then the function would have an inverse on this restricted domain. The equation of the inverse function would be

Post-test.

Find the Domain & Range. Problem 1

Find the Domain & Range. Problem 2

Graph the Function Using Sequence of Transformations Problem 3

Find the Inverse of a Function. Problem 4

Find the Domain & Range. Domain: Range: Click for each answer. Problem 1(Solutions) Set the radicand to be greater than or equal to zero and solve the resulting inequality: Multiply both sides by 2: Add 8 to both sides: Since y=0 when x=8 and the radicand is non-negative, for all x-values greater than or equal to 8, the range is y ≥ 0.

The range: all real numbers except 0. (No function with a constant numerator can be equal to zero.) Find the Domain & Range. Domain: Range: Click for each solution. The domain is all the values that x can equal. The domain for this function includes all real numbers except 1. Problem 2 (Solutions)

Graph the Function Using Sequence of Transformations Key-points: (-1,-1), (0,0), (1,1) 1 st transformation would be Horizontal shift: 2 units left (subtract 2 from each x). The points are now: (-3,-1), (-2,0), (-1,1) 2 nd transformation would be Vertical stretch: 3 times all the y’s. The points are now: (-3,-3), (-2,0), (-1,3) 3 rd transformation would be Vertical shift: 1 unit down (subtract 1 from all y’s). The points are now: (-3,-4), (-2,-1), (-1,2) Basic function: Problem 3 (Solution)

Find the Inverse of a Function. First, notice that the domain of  is restricted to the interval [0,  ). Function  is one-to-one because it is increasing on its entire domain and, thus, has an inverse function. The steps: 1) replace  (x) with y; 2) interchange x and y; 3) solve for y; 4) replace y with  -1 Click for each step. 5) prove that (  º  -1 )(x) = x and (  -1 º  )(x) = x. Problem 4 (Solution ) Continued on next slide.

Solution (cont.) FINDING THE INVERSE OF A FUNCTION WITH A RESTRICTED DOMAIN As a check, graphs of  and  -1 are helpful.

Be Happy! Do Algebra! Developed by Irina Safina. North Harris LSCS