General Form: y = a n x n + a n 1 x n a 2 x 2 + a 1 x + a 0

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1 Families of Functions Prepared by: Sa diyya Hendrickson Name: Date: Definition: function A function f is a rule that relates two sets by assigning to some element (e.g. x) in a set A exactly one element (e.g. y = f(x)), in a set B. The set A is called the domain and B is called the range. Functions are usually represented in one of the following ways: 1. Verbally (description of the rule in words) 2. Algebraically (via some formula) 3. Graphically (using the Coordinate Plane) 4. Numerically (via a table of values or set of points) Family Examples General Form: y = a n x n + a n 1 x n a 2 x 2 + a 1 x + a 0 Polynomials Absolute Value y = g(x) Specific Examples f(x) = 2 (horizontal line), g(t) = 3t 2 5t 2 (quadratic) h(v) = v 3 4v 2 + 2v 8 (cubic) f(x) = x (most basic) or h(x) = 3x 1 nth root y = n! g(x) f(x) = x (most basic) or h(x) = 3 x + 1 (cube root) Fractional Functions y = g(x) h(x) 1. Rational Functions: f(x) = g(x) h(x) 2. f(x) = x x Here, g(x) = x and h(x) = x where g and h are polynomials Exponential, logarithmic and trigonometric functions COMING SOON! Making Math Possible 1 of 16 c Sa diyya Hendrickson

2 Domains Guest List, including Dietary (Domain) Restrictions I. NO Domain Restrictions Family Restrictions Details Polynomials Absolute Value nth root, for n ODD NO NO NO General Form: f(x) = a n x n + a n 1 x n a 2 x 2 + a 1 x + a 0 Polynomials are defined for all values of x! f(x) = g(x), dom(f) = R where g(x) is a polynomial For any real number returned by g(x), we can always determine the distance of this number from zero! Thererefore, f(x) = n! g(x), dom(f) = R n ODD and g(x) is a polynomial As long as n is odd, we can always determine the nth root, for any real number returned by g(x). dom(f) = R II. Domain Restrictions Family Restrictions Details nth root, for n EVEN Fractional Functions YES YES f(x) = n! g(x), n EVEN Recall that whenever n is even, whatever is under the radical must be greater than or equal to zero. This requires us to solve the following inequality: g(x) 0 dom(f) = {x R g(x) 0} f(x) = g(x) h(x) = g(x) 1 h(x) Here, we must satisfy the following requirements: 1. g(x) must be defined 2. h(x) must be defined AND h(x) = 0 In other words, we must satisfy the restrictions of both functions at the same time. So, we take the intersection of the domain of g(x) and the domain of 1 h(x) dom(f) = dom(g) dom(1/h) MORE FUNCTIONS ARE COMING SOON! When they arrive, expand your guest list! Making Math Possible 2 of 16 c Sa diyya Hendrickson

3 Domains Exercise 1: Determine the domain for each of the following functions a) f(x) = x 7 4x 3 + 2x 7 b) g(t) = t+1 t 4 +2 c) h(k) = 4 k 12k 2 +4k 5 Making Math Possible 3 of 16 c Sa diyya Hendrickson

4 Evaluating Functions First let s recall a few important facts about functions: 1 f(x) represents a rule f that is being applied to an element x (in the domain of the function). Please don t mistaken this notation for multiplication. 2 If f is defined with the variable x, then the parentheses are like the jaws of the function and x is the placeholder for the food that f can eat. 3 In the rule, x helps us easily identify where to put the food! Each time you are given a function f algebraically, it will be defined in terms of some variable (e.g. x or t). It s a good idea to make a mental note of that variable. Afterwards, if you are given an expression of the form f( 1), f(a+3), f(h), f(x+5), etc... basically f at anything other than just the variable that it was defined by, we re being asked to evaluate! In other words, wherever we see the variable in the original rule, we must replace it with the current food in the parentheses! Exercise 2: If f(x) = x2 4 2x 1, evaluate f(1). Use the definition of f(x) to create a blueprint for the rule! You can achieve this by rewriting the rule with just the parentheses! Blueprint: f( ) = ( )2 4! 2( ) 1 So, f(1) means that f is eating the value x = 1. To find the resulting value, we can use the blueprint and simply put the number 1 in each pair of brackets! f(1) = (1)2 4! 2(1) 1 = = 3 1 = 3 1 = 3 Making Math Possible 4 of 16 c Sa diyya Hendrickson

5 Evaluating Functions A piecewise function is a function that is defined in pieces/cases, based on specific intervals of x. g(x) if x... (Case 1) f(x) = h(x) if x... (Case 2) Exercise 3: Consider the following piecewise function: 1 if x 3 (Case 1) f(x) = 1 x 2 if 3 < x 2 (Case 2) 3 x if x > 2 (Case 3) a) Evaluate f( 1000) b) Evaluate 2f(0) f(27) Making Math Possible 5 of 16 c Sa diyya Hendrickson

6 Evaluating Functions Exercise 4: Let f(x) = 2x f(a + h) f(a). Evaluate and simplify, h = 0. x 1 h Note: this expression is very important in calculus and is known as the difference quotient! Solution: Notice that the given expression involves f(a + h) and f(a), meaning that we need to first evaluate f at x = a + h and x = a. First, consider the blueprint: Blueprint: f( ) = 2( ) ( ) 1 Now, breaking the problem into smaller/simpler parts, we have: f(a + h) = 2(a + h) (a + h) 1 f(a) = 2(a) (a) 1 The numerator of the expression is given by: f(a + h) f(a) = 2(a + h) a + h 1 2a a 1 LCD = 2(a + h)(a 1) 2a(a + h 1) (a + h 1)(a 1) DP (2a + 2h)(a 1) 2a 2 2ah + 2a = (a + h 1)(a 1) DP 2a 2 2a + 2ah 2h 2a 2 2ah + 2a = (a + h 1)(a 1) collect = 2h (a + h 1)(a 1) The entire expression becomes: f(a + h) f(a) h = = = 2h (a + h 1)(a 1) h 1 2h (a + h 1)(a 1) 1 h 2 (a + h 1)(a 1) Making Math Possible 6 of 16 c Sa diyya Hendrickson

7 Combining Functions We know that we can add, subtract, multiply and divide numbers. Because functions provide us with formulas for numbers in disguise, it makes sense that we can use these same operations on functions! Consider the functions f(x) and g(x). The notations below are used to express operations on functions: a Addition/ Subtraction: (f ± g)(x) = f(x) ± g(x) b Multiplication: (fg)(x) = (f g)(x) = f(x) g(x)! " c Division: For g(x) = 0, f g (x) = f(x) g(x) Exercise 5: Perform the indicated operations to determine the resulting function, h(x). Then, identify the family of functions that h(x) belongs to. a) For f(x) = x and g(x) = 1, h(x) = (f g)(x). x b) For f(x) = x, h(x) = (f f)(x). c) For f(x) = x and g(x) = 1 x + 3, h(x) = # f g $ (x). Making Math Possible 7 of 16 c Sa diyya Hendrickson

8 Composing Functions A very interesting and useful operation on functions is known as composition, which occurs when one function eats another. The notation is given below: (f g)(x) = f (g(x)) Formally, this reads, f composed of g at x. Informally, we can see this as f eating g at x, since g is in the jaws of f. Be careful not to confuse this notation with multiplication! Note ORDER MATTERS! Exercise 6: Consider the functions f(x) = x and g(x) = x 2. Determine (f g)(x) and (g f)(x). Exercise 7: Who ate who? Determine two functions f(x) and g(x) such that h(x) = (f g)(x) a) h(x) = 4! x 2 5 b) h(x) = ( 7 x + 3) 2 c) h(x) = 1 x + 2 Making Math Possible 8 of 16 c Sa diyya Hendrickson

9 Composing Functions Exercise 8: Let f(x) = x 2 6 and g(x) = 2x + 5. Evaluate the following:! " 11 a) (f g)(2) (g f)(2) b) (g f g) 2 Making Math Possible 9 of 16 c Sa diyya Hendrickson

10 Graphing Functions We ll consider poster child functions to be the most basic representatives for popular types of functions. Making Math Possible 10 of 16 c Sa diyya Hendrickson

11 Graphing Quadratics We will explore the components necessary for sketching a parabola, the graph of a quadratic function, by answering the following questions: Consider the following quadratic function: y = 2x x 12. a) Does the parabola that represents this function open up or down? How do you know? b) Does this quadratic have a maximum or a minimum value? c) What helps us to quickly determine if this quadratic has any x -intercepts/roots/zeros? d) Give two methods to help you find the x-intercepts (if they exist)? e) What is the y -intercept? f) Give three methods for finding the vertex of a quadratic. Making Math Possible 11 of 16 c Sa diyya Hendrickson

12 Graphing Quadratics g) Roughly sketch the graph of the given quadratic function, showing the coordinates of the vertex, the y -intercept and the x -intercepts, if any. Making Math Possible 12 of 16 c Sa diyya Hendrickson

13 Graphing Functions The graph of a function f is a visual representation of the range of f when it has eaten all of the values in its domain. The strategies below will give us an organized approach to constructing the graph. Exercise: Sketch the piecewise function: 2 if x 0 f(x) = 4 x 2 if 0 < x 2 x 2 if x > 2 S1 Identify the type of function in each piece. This helps us to have some idea of what the graphs in each piece should look like. 2 if x 0 (horizontal line) f(x) = 4 x 2 if 0 < x 2 (quadratic) x 2 if x > 2 (line) Notice that in the middle piece we have a quadratic function. Therefore, the graph should be a parabola. S2 Separate the graph into the appropriate number of pieces, using the endpoints in the given intervals. Then, label each piece with its given function. This function produces a graph separated into three parts, where the endpoints occur at x = 0 and x = 2 Making Math Possible 13 of 16 c Sa diyya Hendrickson

14 Graphing Functions S3 Find the endpoints for each piece. This helps us know where one graph ends and another begins! 1 For y = 2 (ending at x = 0) When x = 0, y = 2 (since y is constant) endpoint: (0, 2) 2 For y = 4 x 2 (beginning at x = 0 and ending at x = 2) When x = 0, y = 4 (0) 2 = 4 (by evaluation!) endpoint: (0, 4) When x = 2, y = 4 (2) 2 = 0 (by evaluation!) endpoint: (2, 0) 3 For y = x 2 (beginning at x = 2) When x = 2, y = (2) 2 = 0 (by evaluation!) endpoint: (2, 0) S4 Plot the endpoints. Notes 1. Be sure to shade in the endpoints where f can actually take/ eat the x (i.e. when the interval involves equality). 2. If two graphs share that same endpoint with one requiring shading while the other doesn t, always shade the point. If you don t, you ll be claiming that x never takes on that value, which is not true! Notice that based on the given intervals: (0,2) should be shaded since x 0, as well as (2, 0) since 0 < x 2 Making Math Possible 14 of 16 c Sa diyya Hendrickson

15 Graphing Functions S5 Do the rough work necessary and sketch the graphs in each piece. Notes 1. Linear functions don t require much rough work. This is because you only need two points to draw a line, and the endpoints found in S3 already give you one! To find a second point, simply use the slope or evaluate the equation of the line at any other x value in the interval. 2. For nonlinear functions, little to no rough work is required if it s a poster-child function (e.g. y = x 2, y = x 3, y = x, etc.). For more complicated functions, use the techniques you ve been given, including strategies for transformations. For y = 2, there is nothing to do! Simply draw a horizontal line extending to the left of the point (0, 2). For y = 4 x 2, we can look at this in a couple of ways! Option 1 Option 2 A quadratic that can be easily expressed in vertex form: a(x h) 2 + k A transformed quadratic with the base function y = x 2. (See the Transformations Handout for strategies for this approach!) We will proceed with Option 1. 1 Notice that: y = 4 x 2 = (x) = (x 0) vertex : (h, k) = (0, 4), which is also the y intercept! 2 a = 1 < 0 parabola opens down (i.e. vertex is a maximum) 3 For x intercepts, solve: y = 0 y = 0 4 x 2 = 0 Note: 4 x 2 is a Difference of Squares! 2 2 x 2 = 0 (2 + x)(2 x) = x = 0 or 2 x = 0 by ZPP x = 2 or x = 2 x intercepts are: ( 2, 0) and (2, 0) Note: If the quadratic has irrational roots or no roots at all, just select an x value to the left or right of the vertex, evaluate y = f(x) at that value, then sketch the graph so that it s symmetric about the x coordinate of the vertex. Because we are interested in the interval 0 < x 2, we only care about the x intercept (2, 0). The sketch is as follows: Making Math Possible 15 of 16 c Sa diyya Hendrickson

16 Graphing Functions For y = x 2 (which is linear), we only need two points! The endpoint (2, 0) gives us one. To find the second point, consider the following approaches: 1. Use the slope! i.e. From the point (2, 0) rise 1 unit and run 1 unit to the right (since the slope is 1 = 1 1 = rise/fall run right/left ). This gives the point (3, 1). 2. Evaluate y = x 2 at an x value to the right of 2. For example, when x = 3, y = (3) 2 = 1. So, a second point is (3, 1). Combining our results, we have the following final graph: Making Math Possible 16 of 16 c Sa diyya Hendrickson