A BRIEF REVIEW OF ALGEBRA AND TRIGONOMETR Some Key Concepts:. The slope and the equation of a straight line. Functions and functional notation. The average rate of change of a function and the DIFFERENCE- QUOTIENT OF A FUNCTION. Domains of some common types of functions 5. Graphs of functions and piecewise functions 6. Epanding, factoring, and simplifying epressions 7. Solving algebraic equations 8. Basic trigonometry 9. Solving trigonometric equations
. The slope and the equation of a straight line: (a) Slope of a straight line y Q,y Dy=y -y y P,y D= - X Slope of the line PQ m rise run change of y change of y y y Four different kinds of lines and their slopes: Negative Positive Zero Undefined X (b) Equation of a straight line The equation of a straight line can be written in any one of the following three ways: (i) The Point-Slope Form The equation of a straight line that passes through the point, ) and having slope m is given by ( y
y y m( ) (ii) The Slope-Intercept Form The equation of a line with slope m and having the y-intercept ( 0, b ) is given by y m b (iii) The Standard Form In this form, you move all and y terms of the equation of the given line, to one side, and the constant term to the other side. It is normally written as Eample: A By C. Find the equation of the line that passes through the points (, ) and (, 9). Provide the eact answer in the standard form with integral coefficients. Make sure to draw your own sketch (not included here). Let P, y ) be (, ) and Q, y ) be (, 9). Then the slope m of the line PQ is given by ( ( m ( y y) /( ) (9 ) /( ) 7 / The equation of this line PQ can be found either by using the point-slope form or by the slope-intercept form. For eample, using the point-slope form y y m ), its equation is given by, ( 7 y ( ) Multiply both sides by to get: ( y ) 7( ) y 6 7 7 In the standard form, the equation is: 7 y (c) Parallel and perpendicular lines: If two non-vertical lines are parallel, then they have equal slopes. In other words, if m and m are the slopes of the parallel lines, then m m. On the other hand, if two-non vertical lines are perpendicular, then the slope of one line is equal to the negative reciprocal of the slope of the other. In other words, if m and m are the slopes of the perpendicular lines, then m / m.
Eample:. Find the equation of the line that is perpendicular to the line y and which passes through the point (,). Provide the eact answer in the slope-intercept form. Make sure to make a brief sketch coordinate aes are not necessary (not included). First, find the slope of the given line y. To do this, arrange this equation in the form y m b (i.e. solve for y) and look for the coefficient of : y; y ; y Therefore, the slope of the given line is /. So, the slope of a line perpendicular to the original line is the negative reciprocal of /, i.e. /( / ) or /. Now, to find the equation of the required perpendicular line, use the point-slope form with m / and, y ) (, ) : ( y ( ) To put in the slope-intercept form, solve the above equation for y: y. Functions and the functional notation: Quite often, a function can be given by an equation, such as y. The same function can be re-written as f ( ), which is called the functional notation. Thus, f () is just another name for y. Thus, as a specific eample, f () means the y-value of the function when is equal to. So, this can be found by substituting each -variable in the epression by. Therefore, we have f () () () 8 9 f ( ) Note that f ( ) just means that, when, the y-value of the function f is equal to.
Eamples:. Given f ( ) Sin( ), find the eact value of f ( / ). f Sin 6 f 6. Given f ( ), find if f ( ). DO NOT PLUG IN in the given equation, as that would be a major mistake! Rather set f ( ), and solve the resulting equation for, as follows: 0 0 The above is a quadratic equation. BE VER FAMILIAR with the methods of solving quadratic equations and other equations. For eample, one can use the Factoring Method or the Quadratic Formula ( b b ac) /(a). Let us use the Factoring Method, since the right-hand side of the last equation can be found to be factorable: ( 0)( ) 0 Now use the zero-product property: 0 0 or 0 0 / or Thus, there are two -values for which f ( ).. VER IMPORTANT EXAMPLE: Given f ( ), find f ( h). Epand and simplify your answer. This problem illustrates one of the subtle points of functional notation: DO NOT simply add h to the right-hand side of the equation to obtain h as the answer for f ( h). In fact, h is equal to f ( ) h, which is quite different from the desired f ( h). In order to find f ( h), replace all the -terms on the right hand-side of the given equation by h : f ( h) ( h) ( h) 5
Now use rules of algebra such as the Foil Method or special squaring formulas to epand the right-hand side: f ( h) ( h h 6h h ) h h f ( h) 6h h h Just to summarize, given a function f (), the two quantities f ( h) and f ( ) h are totally different! For eample, for the function f ( ), we can find these two quantities as follows: Given : f ( ) f ( h) ( h) But f ( ) h h In general, f ( h) f ( ) h h. Average rate of change of a function and the difference-quotient: (a) The average rate of change of a function f y=f Q Dy=f -f f P D= - In the above diagram, when the -value is changed from to, the y-value of the function f () changes from y f ) to y f ). Then, the change of, or the increment of, is given by ( (, and is denoted by. Similarly, the change of y, or the increment of y, is given by y y f ) f ( ), and is denoted by y. ( 6 X
We define the average rate of change of f () from to as the ratio y / important to observe that, in the above diagram. It is very y / is equal to the slope of the secant line PQ. A secant line of a curve is a line joining any two points of the curve. Thus we have: The average rate of change of y y y f ( ) from f ( ) f ( ) to is eqaul to any of slope of the sec ant line PQ Eamples:. Find the average rate of change of the function f ( ) Cos () from 0 to /. Solution First make sure to draw your own diagram to see the corresponding secant line (not included). The graphical view point is very essential for a successful study of especially calculus. Let 0 and /. Then the average rate of change of f () from to is given by: f ( ) f ( ) f ( / ) f (0) ( / ) 0 Cos ( / ) Cos (0) ( / ) ( / ) () ( / ) ( / ) ( / ) 9 (b) The difference-quotient of a function In part (a), we just defined the average rate of change of a function f () from to. The difference quotient of a function is just a special case of the average rate of change in other words, the difference-quotient is just the average rate of change of f () from to h. In order to find a formula for the difference-quotient, let rate of change of formula f ( ) f ( )] / ( ) as follows: [ Difference Quotient and h, and rewrite the previous average f ( h) f ( ) f ( h) ( h) h f ( ) Thus, one can look at the difference-quotient of a function in three different ways. As the algebraic meaning, or the definition, the difference-quotient of a function is [ f ( h) f ( )] / h. As the geometric meaning of the difference-quotient, it is equal to the slope of the secant line PQ where P (, f ( ) ) and Q( h, f ( h) ) are two points on the graph of y f (). As the physical meaning of the difference-quotient, it is equal to the average rate of change of the function f () from to h. All these ideas are illustrated in the writings below: 7
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Here are some worked eamples on difference-quotients of several types of functions: 9
0
. Domains of some common types of functions Given a function f (), its domain is the set of all values of, for which f () is defined. As a very simple eample, the domain of the function f ( ) is the entire real line, or the interval (, ), since the epression is defined for any real value. The function f ( ) is just one eample of a polynomial. In fact, the domain of any polynomial is the entire real line. However, on the other hand, the domain of the function f ( ) /( ) is not the entire real line. Note that the epression /( ) is not defined at, since / 0 is not defined. However, for any other -value, the epression /( ) is always defined. In other words, the domain of the function f ( ) /( ) is all real numbers ecept. In the interval notation, we can write this domain as (, ) (, ). Here are two very useful principles: For / T to be defined, T must be a nonzero quantity, i.e. T 0 For T to be defined as a real number, T must be a nonnegative quantity, i.e. T 0 Eamples:. Find the domain of ( f ). Epress the eact answer in the interval notation. 6 The numerator is always defined. By the previous remarks, the domain of f () is all -values such that the denominator 6 is nonzero. In order to find when this does happen, ask the opposite question, i.e. set 6 equal to zero, and solve for. 6 0 6 8 However, DO NOT be confused in saying that domain is equal to. The domain is in fact, all real values ecept. Thus, in the interval notation, the domain of f () is equal to (, ) (, ) (, )
. Find the domain of f ( ) 7. Epress the eact answer in the interval notation. Recall that For T to be defined as a real number, T must be a nonnegative quantity, i.e. T 0. Therefore, the domain of f () is all -values such that 7 0. We can solve this linear inequality as follows: 7 0 7 Now divide both sides of the inequality by. However, note that when you multiply or divide both sides of an equality by an negative quantity, the inequality sign must be reversed!! 7 Therefore, the domain of f () is all real values less than or equal to 7/. In the interval notation, the answer is (, 7 / ]. 5. Graphs of functions and piecewise functions (a) Some basic graphs our study of calculus will become more enjoyable, if you can memorize the shapes of some important graphs: 7 (a) y (b) - - - - X y (c) 5 0 5 0 5 - - X y 0 0 - - - -0-0 X (d) y (e) y (f) y - - - 0 X.5 0.5 0 X.5 0.5 X
(g) y (h) 8 6 y (i) 8 6 y X - - - - - - --- X --- 0 X (j) y e (k) ln() y (l) y 7 6 5-6 8 X.0 0.5 -.0-0.5 0.5.0 X -0.5 - - X -.0 Now here is the main advantage: Once you memorize the above shapes, you can build the graphs of more complicated functions using transformation techniques. Eamples:. Draw a quick graph of f ( ). Note that the parent function for the given equation is y, and we already know the shape of its graph by part (d) above. Now start with this graph (d), translate it units to the left, and then units down, to obtain the required graph, The verte of the final graph is (, ). 5
-6 - - - X - - (b) Piecewise functions and their graphs One of the biggest concepts of calculus is the notion of the limit of a function. In order to understand this concept successfully, it is better to be familiar with the idea of piecewise functions. These functions normally have two or more pieces in its defining equation. Here is an eample: Eamples:. Consider the function given by f ( ) 8 if if if (a) Find f () (b) Find f () (c) Find f () (d) Draw a clear graph of f () (a) Note that satisfies the inequality, Therefore, to find f (), plug in in the epression, the first piece of the piecewise function. Thus, f ( ). (b) Here satisfies the inequality. Therefore, to find f () the given function. Thus, f () ()., we use the second piece of (c) This part is VER INTERSTING! In this case. does not satisfy either inequality or. Therefore, f () is undefined, or simply does not eist. What this means is that, the graph of f () will have a hole corresponding to. See the graph in part (d) below. (d) First the graph the three functions y, y, and y 8 by completing disregarding the conditions given by the inequalities. This is where knowing the basic graphs will really pay dividends! 6
y y y 8 5 5 0 0 5 - - - - 5 0 5 0 5 0 5 - - - - Now we are ready to cut each of the above graphs using appropriate vertical lines: As shown above, cut the first graph using the vertical line, and take the part to the left of this vertical line. The second graph must be cut using two vertical lines, and, and take the part of the graph between these two vertical lines. The third graph is cut by the vertical line, and take the part to the right of this vertical line. Finally, we must glue together the remaining parts of these three graphs. However, one must take care of the end points of the pieces which are glued together. The final assembled graph is given below note that a solid dot means the corresponding point is included, while a small circle means the point is ecluded. 8 6 0 8 6-6 -5 - - - - 5 6 7 8 9 0 X 7
6. Epanding, factoring, and simplifying epressions Here are some crucial algebra formulas: (A) Epansion Formulas:.... ( a b) a ab b ( a b) a ab b ( a b) a a b ab b ( a b) a a b ab b (B) Factoring Formulas:. a b does not have any real factors. However, the comple factors are ( a bi)( a bi). a b ( a b)( a b). a b ( a b)( a ab b ). a b ( a b)( a ab b ) Eamples:. Factor: 6 5 ( 6 9) 6 5 (8 7) [() () ]. Factor: 6y y ( y )( y ) ( y)( y)( y ) 6y. Factor: y y y Use factoring by grouping method. y y y ( y) y ( y) ( y)( 8 y ) ( y)( y)( y) ( y) ( y)
. VER IMPORTANT (LEARN THIS!!) Simplify the following algebraic epression, and leave the answer in the factored form: ( ).( ) () ( )..( ) ( 8( ( ( ( ).( ) () ( ) ( ) )( ) )( ) )( ) ( [ ( [ ( ) ( ) ] 8 )( ) 8) )..( ) ] 5. VER IMPORTANT (LEARN THIS!!) 9 Simplify the following epression, and leave the answer in the factored form: 9 9 9 ( )( ) 6. EXTREMEL IMPORTANT (LEARN THIS - OU WILL NEED THIS SEVERAL TIMES!!) Simplify the following epression completely: /. ( ).() ( ) / /. ( ).() ( / ( ) / ( ) / ( ) / ( ) ( ) / ( ) ( ) ) / 9
7. EXTREMEL IMPORTANT (LEARN THIS - OU WILL NEED THIS SEVERAL TIMES!!) Simplify the following epression completely: (9 ) /.. (9 ) (9 ) / ( ) (9 ) / (9 (9.. (9 ) (9 ) / (9 ) (9 ) (9 ) ) ) / [8 / (9 ) / / (9 ) (9 ) (9 ).(9 ) / / / ] (8 ) / (9 ) ( ).(9 ) / REMINDER: Now try several problems from the transparency on Some Factoring Problems 7. Solving algebraic equations Solving various types of equations is an essential skill for anyone studying calculus. Here are some eamples. Eamples:. Solve 0 The above is not a quadratic equation. However, with a simple substitution, it can be transformed into a quadratic equation. Let u. Then the given equation becomes: u u 0 The right-hand side of the above equation does not factor. So use the quadratic formula with a, b, and c. 0
u b b ac a () () ()() 8 Therefore, the four solutions are,,, and. Solve ( ) ( ) ( ) ( ) 0 DO NOT epand the right-hand side of the given equation instead factor the right-hand side ( ) ( ) ( ) ( ) 0 ( ) ( ) (6 ) 0 ( ) ( ) ( ) 0 Use the zero product property to get: 0 or 0 or 0 /, /, / REMINDER: Now try several problems from the transparency on Some Equation Solving Problems
8. Basic trigonometry Make every attempt to be familiar with the following facts from trigonometry: (a) Trig functions of special acute angles wheel (b) Trig functions of all special angles in color 0 0 5 60 90 Sin 0 Cos Tan 0 0 undefined (c) Signs of trigonometric functions in various quadrants
(d) Basic Trigonometric Identities (e) The Double Angle Formulas: Eamples:. Find the eact value of the epression Cos Tan Cos ( ) 6 Cos Tan Cos 6 ( ) ( ) ( ) 8. SOMETHING VER USEFUL TO KNOW! Find all the values for which Sine, Cosine, and Tangent functions are zero. 9 Just use the information gained by above (a) and (b). (a) Sin is zero for any angle that falls on the -ais. In other words, this happens for...,,,0,,,,.... All these values are integer multiples of. Thus we can conclude the following: Sin 0 if and only if n where n is any integer
(b) Cos 0 is zero for any angle that falls on the y-ais. In other words, this happens for..., /, /, /, /, 5 /,.... All these values are odd integer multiples of /. Thus we can conclude the following: Cos 0 if and only if n where n is any odd integer. (c) Just like Sin, Tan is zero for any angle that falls on the -ais. This we can write: Tan 0 if and only if n where n is any integer 9. Solving trigonometric equations Solving trigonometric equations quite often uses the facts (a) and (b) of the previous section 8. Eamples: Try your BEST to understand the first two eamples. The others are built from these two.. Solve 0 where 0. First, isolate the trigonometric function by solving for : Now draw the quadrant diagram, and use the ASTC principle to find out which quadrant the angle belongs to. We know that is a negative number. So the angle must belong to either quadrant II or quadrant III. We will locate these two quadrants by circling them as given in the diagram below: II III I IV The net step is to find the reference angle. It is the acute angle between the X-ais and the terminal side of the angle. So, by dropping the negative side of the right-hand-side of the equation /, we ask that the cosine of what acute angle is equal to /. It is 60 or / radians. Thus, the reference angle is equal to /. We will now draw this reference angle in the circled quadrants:
II III / / / / I IV All you have to do now is to read off the angles: The angles are always read from the positive X-ais. This is where we use the given domain 0 for the first time. So here no negative angles are allowed. Thus, we will turn counterclockwise up to two complete turns, because radians represents two complete turns see the red turnings in the given diagram. The first answer is /. The second answer is /. These are the only two answers up to one complete turn of radians. However, recall that we have to go up to two complete turns. They are not shown in the diagram, but the final two angles are 8/ and 0/. Thus, there are solutions to the given equation. They are /, /, 8/, 0/.. Solve Cos 0 where 0. Cos Cos Cos We can treat the above like two separate equations, Cos / and Cos / and solve them. For both of these equations, the reference angle is 0 or / 6 radians. (i) To solve Cos /, observe that Cos is negative only in quadrants II and III. Draw the reference angle of / 6 radians in these two quadrants as follows: Never attempt Without a diagram! Therefore, the two solutions for this part are 5/6 and 7/6. 5
(ii) On the other hand, to solve Cos /, observe that Cos is positive in quadrants I and IV. Draw the reference angle of / 6 radians in these two quadrants as follows: Therefore, the two solutions for this part are /6 and /6. The final solutions are obtained by combining the answers for the above parts (i) and (ii). Therefore, the solutions are / 6, 5 / 6, 7 / 6, and / 6. Solve Sin ( ) Sin 0 where Use the Double Angle Formula Sin( ) Sin Cos to convert the entire right-hand side of the equation to the single angle, as follows: Sin Cos Sin 0 Sin ( Cos ) 0 Sin 0 or Cos 0 Sin 0 or Cos Now solve like two different problems. Make sure to supply your own diagrams (omitted here). For any, Cos is always between - and, so the equation Cos doesn t have any solutions. Thus, only solutions for the original equation can come from Sin 0. However, in eample of section 8 we learned that Sin 0 only happens when the angle lies along the -ais. Therefore, all the solutions strictly between and are, 0,,, and. 6
. Solve Cos ( ) Sin( ) where 0. Use the Double Angle Formula Cos( ) Sin to convert the entire right-hand side of the equation to the single angle. Also, in the process, the entire equation will be in terms of one type of trigonometric function, i.e. in terms of Sin () only. ( Sin ( )) Sin( ) Sin ( ) Sin( ) 0 ( Sin )( Sin ) 0 Sin ( ) / or Sin ( ) Now solve like two different problems. Make sure to supply your own diagrams (omitted here). First, the equation Sin ( ) / does not have any solutions, since for any angle, Sin () always lies between - and. On the other hand, the only solution for Sin ( ) is / in the given domain 0. Therefore, the only solution to the given equation is /. 7