ALGEBRA AND TRIGONOMETRY REVIEW by Dr TEBOU, FIU. A. Fundamental identities Throughout this section, a and b denotes arbitrary real numbers.

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1 ALGEBRA AND TRIGONOMETRY REVIEW by Dr TEBOU, FIU A. Fundamental identities Trougout tis section, a and b denotes arbitrary real numbers. i) Square of a sum: (a+b) =a +ab+b ii) Square of a difference: (a-b) = iii) Difference of squares: a -b =(a-b)(a+b) iv) Sum of cubes: a 3 +b 3 =(a+b)(a -ab+b ) v) Difference of cubes: a 3 -b 3 = Exercises. 1) Use i) to complete ii) and use iv) to complete v); explain your reasoning. ) Use i)-v) to factor eac expression. a) x 4 -y 4, b) 8x 3-7, c) 4x 4-1, d) 4z -1z+9, e) 9x -8. B. Factoring. B1. Quadratic expression. Consider te quadratic expression: ax +bx+c, were a, b, c are constants tat do not depend on x, and a 0. We may ten rewrite tis expression as a(x + b a x+c a )=a(x + b b x+( a a ) ( b a ) + c), by noting tat a x + b a x=x + b x, and using i), one can a easily complete te beginning of te square to get a complete square. Now, wen we complete te square, we add an extra term, and so we sould also subtract it to keep te given expression uncanged. So te original expression may be written, (tanks to i)): ax +bx+c= a((x+ b ) -( b 4ca )). Set Δ=b -4ac, wic is called discriminant of te quadratic a 4a expression. Note tat if Δ<0, ten te expression in te parenteses is always positive and te

2 quadratic polynomial cannot be factored. If on te contrary, Δ>0, ten we can use te difference of squares formula to sow tat te quadratic polynomial as two distinct roots given by x + = b+ Δ and x - = b Δ ; tus a ax +bx+c=a(x-x + )(x-x - ). Finally, for Δ=0, te quadratic polynomial as a single double root x= - b a. B. Cubic and iger order expressions. We ave just seen tat tere is a simple rule tat may be used to determine weter a quadratic polynomial can be factored or not, and wen it can be factored, te rule, known as te quadratic formula, also elps us get te roots of te polynomial. Now for cubic and iger order polynomials, we do not ave suc a simple rule. To factor suc a polynomial, we will ave to use te trial and error metod to get one of its roots, ten use division to reduce te degree of te polynomial to be factored. To use te trial and error metod, evaluate te given polynomial at te number x=x_0 wit x_0= 0, 1, -1,, -, 3, -3, until you get te value zero. Ten x-x_0 is a factor of te given polynomial. Now divide te polynomial by x-x_0, you ten get a polynomial of a lesser degree. If te degree is 3 or more, use again te trial and error metod until you get a quadratic polynomial. Note owever tat sometimes, just using te grouping metod can save you all te work needed to carry out te trial and error metod. So, tink out of te box! Exercises. Factor eac expression: a) f(x)=6x 4 +x 3-4x -9x+10, b) g(x)=x 5-4x 3-7x +108 c) (x)=4x 3-4x -3x+3, d) k(x)= 6x 3 +17x +6x-8, e) m(x)=18x 4-9x 3-31x +x+6. a C. Rationalizing. Wen you want to rationalize te denominator of an expression involving a radical, multiply bot te numerator and denominator by te conjugate of te denominator, and expand te denominator only; if te denominator is, say, x- y, ten multiply bot numerator and denominator by x+ y, as x+ y is te conjugate of x- y. To rationalize te numerator, you proceed exactly te same way, and expand te numerator only. Exercises. 1) Rationalize te numerator and simplify eac expression: a) r(x) = 3 x+7 b) q(x) = x+ 4 x +7x+1 ) Rationalize te denominator and simplify eac expression c) p(x) = x x 1 5x+6 b) n(x) = x x 1 6 5x+16 x 5x+6 D. Functions. A function is a rule tat assigns to eac admissible input exactly one output. A simple example of a function is te assignment to eac FIU student or staff a

3 pantersoft identity number (PID). Te domain of a function is te set of all admissible inputs; for instance in te case of PID assignment, te domain of tat function consists of all te staff and students of FIU; tere are people in Miami wo ave no PID. Te range of a function is te set of all te corresponding outputs. In te case of PID assignments, te range is te set of all assigned PIDs. Te type of functions tat we will be dealing wit will be functions defined from te set of real numbers into itself; examples of functions include: polynomial functions wose domain is te wole set of real numbers, rational functions, a rational function is te ratio of two polynomial functions, if r(x)= p(x), were p and q are polynomial functions, ten te domain of r, denoted Dr, q(x) consists of all real numbers x wit q(x) 0. Anoter type of function is te absolute value function defined by x, if x < 0 x = { x, if x 0. A piecewise defined function is a function given by different formulas in different intervals; te absolute value function is an example of suc a function. Example: write te following function (x) = x + 5 3x + 7 witout te absolute value symbol. Tis amounts to writing te function in piecewise defined form; to do tis, wic intervals are we going to use? To find te intervals, solve for x eac of te arguments in te absolute value symbol: we sall solve for x: -x+5=0, getting x=5/, and 3x+7=0, getting x=-7/3. Terefore our intervals are: (, 7 3 ), [ 7, 5 ), 3 [5, ). Tus, we ave Some simple algebra yields x + 5 ( (3x + 7)), if x < 7/3 (x) = x + 5 (3x + 7), if 7 x < 5/ 3 { ( x + 5) (3x + 7), if x 5/. x + 1, if x < 7/3 (x) = 5x, if 7 x < 5/ 3 { x 1, if x 5. Exercises. 1) Find te domain of eac function: a) f(x)= x +5x+6 x x 1 b) g(x)= x 3 x + 1 ) Write eac function in a piecewise defined form: c) k(x) = 6 5x 7x + 4, d) m(x) = 4x x. Note. Te sum, difference, product and quotient of two functions is a function.

4 Composition. Te composition of a function f wit a function g is te function fog given by fog(x)=f(g(x)) wit domain Dfog={x in Dg; g(x) in Df}. Example. If f(x) = 1 x and g(x) = x 3, find fog and gof as well as Dfog and Dgof. 4x+5 fog(x)= 1 g(x)= 1 x 3 x 5/4, and x+8 4x+5 4x+5 = x+8 4x+5 ; for fog(x) to be defined, g(x) must be defined, so 0. To find were te ratio is nonnegative, we sall use a sign line /4 + Te sign line sows tat te ratio is nonnegative in te intervals (-,-4] and (-5/4, ); terefore Dfog(x)= (-,-4] (-5/4, ). To get te correct signs, pick any number in eac interval and plug it in te ratio, evaluate; te sign of te number tat you obtain is te sign of te ratio in te corresponding interval. Te case of gof is left as an exercise. Inverse functions. Let f and g be two functions suc tat g(f(x)) = x, for all x in te domain of f, and f(g(y)) = y, for every y in te domain of g, Ten f and g are called inverse functions; f is called te inverse of g, and g is called te inverse of f. Te inverse function of a function f is denoted f 1, and we ave Df =R f -1 and Rf =D f -1. Example. Let (x) = 3x+8 5x+4. Find f 1 (x), and Df. To find f 1 (x), set y = 3x+8, switc x and y 5x+4 to get x = 3y+8 5y+4. Solve te equation for y in terms of x to get f 1 (x). It follows from te equation x(5y + 4) = 3y + 8. Passing y from te rigt and side to te left, and factoring, we get y(5x + 3) = 8 4x. Consequently, it follows from te last equation, y = 8 4x 5x+3. Hence, f 1 (x) = 8 4x 5x+3 and Rf ={x; x 3/5} = (, 3 ) ( 3, ). As an exercise, do te same 5 5 for f(x) = 3x 4. Note. A function f is called one to one if f(x) f(z) wenever x z. Any one to one function f defined from its domain to its range as an inverse, or is called invertible. Te grap of an invertible function and tat of its inverse are symmetric about te line y = x. Difference quotient. Let f be a function. We define te difference quotient of f at x to be te ratio f(x+) f(x) were is a nonzero constant independent of x.

5 Example. If f(x) = x 3 + x 1, find te difference quotient of f. We ave, by using te difference of cubes formula: f(x+) f(x) = (x+)3 +(x+) 1 (x 3 +x 1) ((x+ x)((x+) +(x+)x+x )+ = ((x+)3 x 3 )+ = ((x + ) + (x + )x + x ) + 1, in te last step, was factored out in te numerator, and cancelled wit in te denominator. Remember tat in order to get f(x + ), you replace x wit x+ everywere in te expression of f(x). Exercises. If f(x) = 3x 7x + 4 and g(x) = x+8, find te difference quotients of f and g at x. 4x+5 Be sure to simplify your answers as muc as possible. Parity. A function f is called even if f(-x)=f(x) for every x in te domain of f. For instance, if f(x)=x or f(x)= cos(x), ten f is even, as (-x) =x and cos(-x)=cos(x), for every real number x. A function f is called odd if f(-x)=-f(x) for eac x in te domain of f. If f(x)=x 3 or f(x)= sin(x), ten f is odd, as (-x) 3 = - x 3 and sin(-x)= - sin(x), for every real number x. = E. Exponential and Logaritmic Functions. E1. Exponential Functions. Let a>0 be a real number. Te exponential function wit base a is te function f defined by f(x)=a x for every real number x. Te domain and range of f are given by Df =(, ), and Rf =(0, ). So exponential functions are defined everywere, and are never zero and never negative. Properties. Let a>0, x, and y, be real numbers. Ten i) a x+y =a x a y ii) a x y = a x /a y iii) a xy =(a x ) y =(a y ) x iv) a 0 =1 Note. Among all te exponential functions tere is one called natural exponential function wose base denoted e is a number between and 3; more precisely, e Any time tat te base of an exponential function is not specified, you sould understand tat te base is e. E. Logaritmic functions. Let b>0 and b 1. Te logaritmic function wit base b is te function denoted by log b and defined by y = log b x if and only if x = b y. Note. i) It follows from tis definition tat te exponential function wit base b and te logaritmic function wit base b are inverse functions. Terefore, te domain of log b is (0, ), and te range of log b is (, ). Furter, for every real number x, we ave log b (b x ) = x, and

6 for every x>0, we ave b log b(x) = x. ii) Te logaritmic function wit base e is called natural logaritm and denoted ln wile te logaritmic function wit base 10 is called common logaritm and denoted log. iii) (Cange of base formula) Let b>0 and b 1 and let a>0 and a 1. Let x>0. Ten log b (x) = log a (x) log a (b) =ln(x) ln(b). As an exercise, prove te cange of base formula. Properties. Let b>0 and b 1, and let x>0, y>0, and r be real numbers. i) log b (xy) = log b (x) + log b (y) ii) log b ( x y ) = log b(x) log b (y) iii) log b (x r ) = rlog b (x) iv) log b (1) = 0 v) log b (b) = 1. Exercises. 1) Find te domain of eac function: a) f(x) = log x ( 3x + 5), b) g(x) = log 3 (x x 1). ) Use te properties of exponential functions prove eac of te properties i) to v). 3) Expand te logaritm in terms of sums, differences and multiples of simpler logaritms 5 7x 9 a) f(x) = log b) (x) = log sin 3x sec x 5 x 3 x +x (3x 8)( cotx+5) 4) Solve eac equation for x. a) log x ( 3x + 8) =, b) ln (4x)-3ln(x ) = ln, c) e x 3e x + = 0, d) x ( ln x 3) =4. F. Algebra of infinity. It is important to always keep in mind tat and + are symbols and not numbers. Accordingly, te algebra involving tese symbols is someow different from te algebra of real numbers. We ave i) + + = + and - + ( ) = ii) + (+ ) = + and - (+ ) =, and - ( ) = + iii) For every real number a, we ave a+( ) =, a-(- )=+, and a/± =0. + if a < 0 iv) For every nonzero real number a, we ave a( ) = { if a > 0. v) Indeterminate forms: Tese are quantities wic do not ave a definite value; tey are: a) + (+ ) or, b) ±, c) 0(± ), d) ± 1±, e) 0, f) 0/0, g) 0 0. Any time tat you encounter one of tose indeterminate forms, you sould work out to get rid of tem; many examples will be discussed in class.

7 G. Trigonometric Functions. Te six fundamental trigonometric functions are stated in te cart wit teir domain and range. All angles are in radian. You ave te abbreviation of eac function into parenteses. If you do not remember, ceck teir graps in te textbook. Function Domain Range Cosine (cos) (, + ) [ 1,1] Sine (sin) (, + ) [ 1,1] Tangent (tan) {x; x π +kπ, k is an integer} (, + ) tan x= sin x cos x Cotangent (cot) cot x= cos x = 1 sin x tanx Secant (sec) sec x= 1 cos x Cosecant (csc) csc x= 1 sin x {x; x π +kπ, k is an integer} {x; x π +kπ, k is an integer} (, {x; x π +kπ, k is an integer} (, + ) 1] [1, + ) (, 1] [1, + ) G1. Important Trigonometric Identities. Pytagorean identities: a) cos x+sin x=1, for every real number x b) 1+tan x=sec x, for every x in Dtan c) 1+cot x=csc x, for every x in Dcot Complement identities cos x = sin( π x) and sin x = cos( π x), for every number x. Supplement identities sin x = sin(π x), and cos( π x) = cos x, for every number x.

8 Sum formulas cos(x + y) = cos x cos y sin x sin y sin(x + y) = sin x cos y + cos x sin y, for all numbers x and y. Exercise. Use te sum formulas to find: cos(x y) = tan(x + y) = sin(x y) = tan(x y) = Double-angle formulas cos(x) = cos x sin x = cos (x) 1 = 1 sin x, for every number x. sin(x) = sin x cos x, for every number x. Exercise. Sow tat cos 4 x sin 4 x = cos x sin x, for every number x. Product to sum formulas cos x cos y = cos(x + y) + cos (x y) cos x sin y = sin(x + y) + sin (y x) sin x sin y = cos(x y) cos (x + y) Trigonometric values to know x cos x sin x tan x sec x csc x cot x undefined undefined π/6 3/ 1/ 1/ 3 / 3 3 π/4 / / 1 1 π/3 1/ 3/ 3 / 3 1/ 3 π/ 0 1 undefined undefined 1 0 π undefined undefined G. Inverse trigonometric functions. Given tat none of te six trigonometric functions in te cart above is one-to-one on its domain, in order to be able to define an inverse for any of tose functions, we sall restrict its domain to a subdomain were it is one-to-one. Inverse cosine function. Te inverse cosine function denoted cos 1 is defined for every x in [ 1,1] by: y = cos 1 x if and only if x = cosy, y in [0, π]. Inverse sine function. Te inverse sine function denoted sin 1 is defined for every x in [ 1,1] by: y = sin 1 x if and only if x = sin y, y in [ π/, π/].

9 Inverse tangent function. Te inverse tangent function denoted tan 1 is defined for every x in (, ) by: y = tan 1 x if and only if x = tan y, y in ( π, π ). Inverse secant function. Te inverse cosine function denoted sec 1 is defined for every x in (, 1] [1, ) by: y = sec 1 x if and only if x = secy, y in [0, π ) (π/, π]. Inverse cosecant function. Te inverse cosecant function denoted csc 1 is defined for every x in (, 1] [1, ) by: y = csc 1 x if and only if x = csc y, y in [ π, 0) (0, π/]. Inverse cotangent function. Te inverse cotangent function denoted cot 1 is defined for every x in (, ) by: y = cot 1 x if and only if x = cot y, y in (0, π). Exercise. A) For wat values of x do we ave a) cos(cos 1 x) = x, b) cos 1 (cos x) = x c) sin 1 (sin x) = x, d) sin( sin 1 x) = x, e) tan(tan 1 x) = x, f) tan 1 (tan x) = x. B) Sow tat 1) sec 1 x = cos 1 (1/x), ) csc 1 x = sin 1 (1/x), 3) cot 1 x = tan 1 (1/x). Some identities involving inverse trigonometric functions i) cos 1 x + sin 1 x = π/ ii) cos( sin 1 x) = 1 x = sin( cos 1 x) iii) tan( cos 1 x) = 1 x x iv) tan( sin 1 x) = x 1 x v) sec( tan 1 x) = 1 + x vi) tan( sec 1 x) = x x x 1 vii) tan 1 x + cot 1 x = x π x Exercise. Find a) cos( tan 1 x), b) sin( tan 1 x), c) cos( sec 1 x), d) sin( sec 1 x).