CHAPTER 1: Review (See also the Precalculus notes at http://www.kkuniyuk.com) TOPIC 1: FUNCTIONS (Chapter 1: Review) 1.01 PART A: AN EXAMPLE OF A FUNCTION Consider a function f whose rule is given by f ( x)= x 2 ; f ( u)= u 2 also works. WARNING 1: f ( x) is read as f of x or f at x. It does not mean f times x. x is the input (or argument) for f, and x 2 x f x 2 is the output or function value. This function squares its input, and the result is its output. For example, f ( 3)= ( 3) 2 = 9. 3 f 9 Think of a function as a calculator button. In fact, your calculator should have a squaring button labeled x 2. f is a function, because no legal input yields more than one output. A function button on a calculator never outputs two or more values at the same time. We never get: I don t know. The answer could be 3 or 10. A function is a special type of relation. Relations that are not functions permit multiple outputs for a legal input.
(Chapter 1: Review) 1.02 PART B: POLYNOMIAL FUNCTIONS A polynomial in x can be written in the form: a n x n + a n 1 x n 1 +... + a 1 x + a 0, where n is a nonnegative integer called the degree of the polynomial, the a k coefficients are typically real numbers, and the leading coefficient a n 0. A polynomial function has a rule that can be written as: f ( x)= polynomial in x. For example: 4x 3 5 2 x2 + 1 is a 3 rd -degree polynomial in x with leading coefficient 4. The rule f ( x)= 4x 3 5 2 x2 + 1 corresponds to a polynomial function f. PART C: RATIONAL FUNCTIONS A rational expression in x can be written in the form: polynomial in x nonzero polynomial in x. Examples include: 1 x, 5x 3 1 x 2 + 7x 2, and x7 + x which equals x7 + x 1. Observe in the second example that irrational numbers such as permissible. 2 are The last example correctly suggests that all polynomials are rational expressions. A rational function has a rule that can be written as: f x = rational expression in x. PART D: ALGEBRAIC FUNCTIONS An algebraic expression in x resembles a rational expression, except that radicals and exponents that are noninteger rational numbers such as 5 7 are also permitted, even when x appears in a radicand or in a base (but not in an exponent).
Examples include: x and x 3 + 7x 5/7 3 x x + 5 + 4. (Chapter 1: Review) 1.03 All rational expressions are algebraic. Although sources such as MathWorld allow only algebraic numbers (such as rational numbers and 2 ) to be coefficients in an algebraic expression, we will typically allow all real numbers (including, for instance) in this work. An algebraic function has a rule that can be written as: f x = algebraic expression in x. A Venn diagram for expressions in x corresponding to functions is below. Each disk represents a subset of every larger disk; for example, every polynomial is a rational expression and an algebraic expression (based on the definition in this work).
(Chapter 1: Review) 1.04 PART E: DOMAIN and RANGE The domain of a function f, which we will denote by Dom( f ) (though this is not standard), is the set of all legal inputs. The range of f, which we will denote by Range( f ), is then the set of all resulting outputs. Unless otherwise specified (or in the context of a word problem ), we typically assume that the domain of a function is the set of all real input values that yield an output that is a real number. This set is the implied (or natural) domain. The implied domain of an algebraic function consists of all real numbers except those that lead to (the equivalent of): 1) a zero denominator Think: 0, or. even 2) an even root of a negative-valued radicand Think: As we study more types of functions, the list of restrictions will grow. We will also exclude real numbers that lead to: 3) logarithms of nonpositive values ( Think: log b ( 0) ), or 4) arguments of trigonometric functions that correspond to vertical asymptotes. Word problems may imply other restrictions: nonnegativity, integer values, etc. Example 1 (Domain and Range of the Squaring Function) Let f ( x)= x 2. Find the domain and the range of f. The implied domain of a polynomial function (such as this f ) is, the set of all real numbers. In interval form, is,. Its graph is the entire real number line:
(Chapter 1: Review) 1.05 WARNING 2: We use parentheses in the interval form, because ( infinity ) and ( negative infinity ) are not real numbers and are therefore excluded from the set of numbers. We will discuss infinity further in Chapter 2. If x approaches, it (generally) increases without bound. If x approaches, it (generally) decreases without bound. Note: It is debatable whether an expression like x2 + x x simplifies to x + 1, but its domain excludes 0. is a polynomial. It The resulting range of f is the set of all nonnegative real numbers (all real numbers that are greater than or equal to 0), because every such number is the square of some real number, and only those numbers are. WARNING 3: Squares of real numbers are never negative. The graph of the range is: The filled-in circle indicates that 0 is included in the range. We could also use a left bracket ( [ ) at 0; the bracket opens towards the shading. The graph helps us figure out the interval form. In interval form, the range is 0, ). The bracket next to the 0 indicates that 0 is included in the range. In set-builder form, the range is: { y y 0}, or { y : y 0}, which is read the set of all real numbers y such that y 0. Using y instead of x is more consistent with our graphing conventions in the xy-plane (since we typically associate function values in the range with y-coordinates), and it helps us avoid confusion with the domain. denotes set membership.
(Chapter 1: Review) 1.06 Example 2 (Domain of a Function) Let f ( x)= x 3, find Dom( f ), the domain of f. is real x 3 0 x 3. f x WARNING 4: We solve the weak inequality x 3 0, not the strict inequality x 3 > 0. Observe that 0 = 0, a real number. The domain of f in set-builder form is: { x x 3}, or x : x 3 { } in graphical form is: in interval form is: 3, ) Note: Range( f )= 0, Example 3 (Domain of a Function) 4 Let f ( x)= 3 x. Find Dom( f ). Solve the weak inequality: 3 x 0. Method 1 Method 2 ). Consider the graph of y = f ( x). 3 x 0 Now subtract 3 from both sides. x 3 Now multiply or divide both sides by 1. x 3 WARNING 5: We must then reverse the direction of the inequality symbol. 3 x 0 Now add x to both sides. 3 x Now switch the left side and the right side. x 3 WARNING 6: We must then reverse the direction of the inequality symbol.
(Chapter 1: Review) 1.07 The domain of f in set-builder form is: { x x 3}, or { x : x 3} in graphical form is: in interval form is: (,3 Example 4 (Domain of a Function) Let f ( x)= 1 x 3. Find Dom( f ). This is similar to Example 2, but we must avoid a zero denominator. We solve the strict inequality x 3 > 0, which gives us x > 3. The domain of f in set-builder form is: { x x > 3}, or { x : x > 3} in graphical form is: in interval form is: ( 3, ) The hollow circle on the graph indicates that 3 is excluded from the domain. We could also use a left parenthesis ( ( ) here; the parenthesis opens towards the shading. Likewise, we have a parenthesis next to the 3 in the interval form, because 3 is excluded from the domain. Types of Intervals ( 5, 7) and 3, endpoints. 5, 7 real numbers. ( 3, ) is an unbounded interval. are examples of open intervals, because they exclude their is a bounded interval, because it is trapped between two 5, 7 is a closed interval, because it includes its endpoints, and it is bounded.
(Chapter 1: Review) 1.08 Example 5 (Domain of a Function) 3 Let f ( x)= x 3. Find Dom( f ). Dom( f )=, because: The radicand, x 3, is a polynomial, and WARNING 7: The taking of odd roots (such as cube roots) does not impose any new restrictions on the domain. Remember that the cube root of a negative real number is a negative real number. Example 6 (Domain of a Function) Let g()= t t + 3. Find Dom( g). t 10 The square root operation requires: t + 3 0 t 3. We forbid zero denominators, so we also require: t 10 0 t 10. The domain of g in set-builder form is: { t t 3 and t 10}, or { t : t 3 and t 10} in graphical form is: in interval form is: 3, 10) ( 10, ) We include 3 but exclude 10. (Some instructors believe that 0 should also be indicated on the number line.) is used to separate intervals in the event that a The union symbol number or numbers need to be skipped.
(Chapter 1: Review) 1.09 PART F: GRAPHS OF FUNCTIONS The graph of y = f ( x), or the graph of f, in the standard xy-plane consists of all points [representing ordered pairs] of the form ( x, f ( x) ), where x is in the domain of f. In a sense, the graph of f = ( x, f ( x) ) x Dom( f ) We typically assume { }. x is the independent variable, because it is the input variable. y is the dependent variable, because it is the output variable. Its value (the function value) typically depends on the value of the input x. Then, it is customary to say that y is a function of x, even though y is a variable here. The form y = f x implies this. A brute force graphing method follows. Point-Plotting Method for Graphing a Function f in the xy-plane Choose several x values in Dom( f ). For each chosen x value, find f ( x), its corresponding function value. Plot the corresponding points x, f x ( ) in the xy-plane. Try to interpolate (connect the points, though often not with line segments) and extrapolate (go beyond the scope of the points) as necessary, ideally based on some apparent pattern. Ensure that the set of x-coordinates of the points on the graph is, in fact, Dom f.
(Chapter 1: Review) 1.10 Example 7 (Graph of the Squaring Function) Let f ( x)= x. Graph y = f ( x). TIP 1: As usual, we associate y-coordinates with function values. When point-plotting, observe that: Dom( f )= 0, ). For instance, if we choose x = 9, we find that f ( 9)= 9 = 3, which means that the point ( 9, f ( 9) ), or ( 9, 3), lies on the graph. is undefined (as a real number), because On the other hand, f 9 9 Dom( f ). Therefore, there is no corresponding point on the graph with x = 9. A (partial) table can help: Below, we sketch the graph of f, or y = f ( x). x Point f x 0 0 0, 0 1 1 1, 1 4 2 4, 2 9 3 9, 3 WARNING 8: Clearly indicate any endpoints on a graph, such as the origin here. The lack of a clearly indicated right endpoint on our sketch implies that the graph extends beyond the edge of our figure. We want to draw graphs in such a way that these extensions are as one would expect. WARNING 9: Sketches of graphs produced by graphing utilities might not extend as expected. The user must still understand the math involved. Point-plotting may be insufficient.
(Chapter 1: Review) 1.11 PART G: THE VERTICAL LINE TEST (VLT) The Vertical Line Test (VLT) A curve in a coordinate plane passes the Vertical Line Test (VLT) there is no vertical line that intersects the curve more than once. An equation in x and y describes y as a function of x its graph in the xy-plane passes the VLT. Then, there is no input x that yields more than one output y. Then, we can write y = f ( x), where f is a function. Example 8 (Square Root Function and the VLT; Revisiting Example 7) The equation y = x explicitly describes y as a function of x, since it is of the form y = f ( x), where f is the square root function from Example 7. Observe that the graph of y = x passes the VLT. Each vertical line in the xy-plane either misses the graph entirely, meaning that the corresponding x value is not in Dom f, or intersects the graph in exactly one point, meaning that the corresponding x value yields exactly one y value as its output.
Example 9 (An Equation that Does Not Describe a Function) (Chapter 1: Review) 1.12 Show that the equation x 2 + y 2 = 9 does not describe y as a function of x. (Method 1: VLT) The circular graph of x 2 + y 2 = 9 below fails the VLT, because there exists a vertical line that intersects the graph more than once. For example, we can take the red line ( x = 2) below: Therefore, x 2 + y 2 = 9 does not describe y as a function of x. (Method 2: Solve for y) This is also evident if we solve x 2 + y 2 = 9 for y: x 2 + y 2 = 9 y 2 = 9 x 2 y =± 9 x 2 Any input value for x in the interval ( 3, 3) yields two different y outputs. For example, x = 2 yields the outputs y = 5 and y = 5.
(Chapter 1: Review) 1.13 PART H: ESTIMATING DOMAIN, RANGE, and FUNCTION VALUES FROM A GRAPH The domain of f is the set of all x-coordinates of points on the graph of y = f x. (Think of projecting the graph onto the x-axis.) The range of f is the set of all y-coordinates of points on the graph of y = f x. (Think of projecting the graph onto the y-axis.) Domain Think: x f Range Think: y Example 10 (Estimating Domain, Range, and Function Values from a Graph) Let f ( x)= x 2 + 1. Estimate the domain and the range of f based on the graph of y = f x below. Also, estimate f 1 (). Apparently, Dom( f )=, or (, ), and Range( f )= 1, ). We will learn more about determining ranges from the graphing techniques in Chapter 4. It also appears that the point 1, 2 lies on the graph, and thus f ()= 1 2. WARNING 10: Graph analyses can be imprecise. The point 1, 2.001, for example, may be hard to identify on a graph. Not all coordinates are integers.
(Chapter 1: Review) 1.14 PART I: FUNCTIONS THAT ARE EVEN / ODD / NEITHER; SYMMETRY A function f is even f ( x)= f ( x), x Dom( f ) The graph of y = f ( x) is means for all or for every. symmetric about the y-axis. Example 11 (Proving that a Function is Even) Let f ( x)= x 2. Prove that f is an even function. Dom( f )=. x, f ( x)= ( x) 2 = x 2 = f x Q.E.D. (Latin: Quod Erat Demonstrandum) This signifies the end of a proof. It means which was to be demonstrated / proven / shown. TIP 2: Think: If we replace x with x (y) outputs. The point x, y as the input, we obtain equivalent lies on the graph if and only if ( x, y) does. Observe that the graph of y = f ( x) below is symmetric about the y-axis, meaning that the parts of the graph to the right and to the left of the y-axis are mirror images (or reflections) of each other.
(Chapter 1: Review) 1.15 The term even function may have come from the following fact: If f ( x)= x n, where n is an even integer, then f is an even function. These are the functions for:, x 4, x 2, x 0, x 2, x 4,. The graph of y = x 2 is called a parabola. The graphs of y = x 4, y = x 6, etc. are similarly bowl-shaped but are not parabolas. A function f is odd f ( x)= f ( x), x Dom( f ) The graph of y = f ( x) is symmetric about the origin. In other words, if the graph is rotated 180 about the origin, we obtain the same graph. Example 12 (Proving that a Function is Odd) Let f ( x)= x 3. Prove that f is an odd function. Dom( f )=. x, Q.E.D. f ( x)= ( x) 3 = x 3 = x 3 = f x TIP 3: Think: If we replace x with x outputs. The point x, y as the input, we obtain opposite (y) lies on the graph if and only if ( x, y) does.
(Chapter 1: Review) 1.16 The term odd function may have come from the following fact: If f ( x)= x n, where n is an odd integer, then f is an odd function. The graphs of y = x 5, y = x 7, etc. resemble the graph of y = x 3 below. WARNING 11: Zero functions are functions that only output 0 (Think: f ( x)= 0). Zero functions on domains that are symmetric about 0 on the real number line are the only functions that are both even and odd. (Can you show this?) WARNING 12: Many functions are neither even nor odd.
PART J: ARITHMETIC COMBINATIONS OF FUNCTIONS Let f and g be functions. (Chapter 1: Review) 1.17 If their domains overlap, then the overlap (intersection) Dom( f ) Dom g is the domain of the following functions with the specified rules, with one possible exception (*): x x ( x)= f x f g ( x )= f ( x ) g( x) f + g, where f + g f g, where f g fg, where fg f g, where = f ( x)+ g x = f ( x) g x g( x) (*) WARNING 13: Dom f g = { x Dom ( f ) Dom( g) g( x) 0}. Example 13 (Subtracting Functions) Let f ( x)= 4x and g( x)= x + 1. Find ( f g) ( x) and Dom( f g). x ( f g) ( x)= f ( x) g( x) = ( 4x) x + 1 x WARNING 14: Use grouping symbols when expanding g x here, since we are subtracting an expression with more than one term. = 4x x 1 x = 3x 1 x Dom( f )=. We omit only 0 from Dom( g) and also Dom( f g). Dom( f g)= \{}= 0 { x x 0}= (,0)( 0, ).
(Chapter 1: Review) 1.18 PART K: COMPOSITIONS OF FUNCTIONS We compose functions when we apply them in sequence. Let f and g be functions. The composite function f g is defined by: ( f g) x ( ) = f g x Its domain is { x x Dom( g) and g( x)dom( f )}. The domain consists of the legal inputs to g that yield outputs that are legal inputs to f. f ( ) x g g x f g x f g Think of f g as a merged function. WARNING 15: The function f g applies g first and then f. Think of pressing a g button on a calculator followed by an f button. WARNING 16: f g may or may not represent the same function as g f (in which f is applied first). Composition of functions is not commutative the way that, say, addition is. Think About It: Try to think of examples where f g and g f represent the same function. Example 14 (Composition of Functions) Let f ( u)= 1 u and g ( x )= x 1. Find ( f g) ( x) and Dom( f g). ( f g) x = f ( g( x) )= f ( x 1)= 1 x 1. In fact, Dom( f g) in set-builder form is: { x x > 1}, or { x : x > 1} in graphical form is: in interval form is: ( 1, )
(Chapter 1: Review) 1.19 Example 15 (Decomposing a Composite Function) Find component functions f and g such that ( f g) ( x)= 3x + 1. We want to decompose f g. Neither f nor g may be an identity function. For example, do not use: g( x)= x and f ( u)= 3u + 1. This would not truly be a decomposition. f does all the work! x g : x, our u g( x)= x f : f ( u)= 3u + 1 gx f g 3x + 1 f g x We need: f g( x) = 3x + 1. We can think of f and g as buttons we are designing on a calculator. We need to set up f and g so that, if x is an initial input to Dom f g, and if the g button and then the f button are pressed, then the output is 3x + 1. x g : f : u =? g( x)=? f ( u)=?? f g 3x + 1 f g x A common strategy is to let g( x), or u, be an inside expression (for example, a radicand, an exponent, a base of a power, a denominator, an argument, or something being repeated) whose replacement simplifies the overall expression. Here, we will let g( x)= 3x + 1.
(Chapter 1: Review) 1.20 We then need f to apply the square root operation. We will let f ( u)= u. The use of u is more helpful in calculus, but f ( x)= x is also acceptable. However, f ( u)= x is not acceptable. Possible Answer: Let g( x)= 3x + 1 and f ( u)= u. x g : 3x + 1, our u g( x)= 3x + 1 f : f ( u)= u gx f g 3x + 1 f g x There are infinitely many possible answers. For example, we could let g( x)= 3x and f ( u)= u + 1. x g : 3x, our u g( x)= 3x f : f ( u)= u + 1 gx f g 3x + 1 f g x These ideas will be critical to the Chain Rule of Differentiation in Section 3.6 and the u-substitution technique of integration in Section 5.2.