Precalculus Review Differential Calculus Definitions, Rules an Teorems Sara Brewer, Alabama Scool of Mat an Science Functions, Domain an Range f: X Y a function f from X to Y assigns to eac x X a unique y Y te omain of f is te set X - te set of all real numbers for wic te function is efine Y is te image of x uner f, enote f(x) te range of f is te subset of Y consisting of all images of numbers in X x is te inepenent variable, y is te epenent variable fis one-to-one if to eac y-value in te range tere correspons exactly one x-value in te omain f is onto if its range consists of all of Y Implicit v. Explicit Form 3x + 4y = 8 implicitly efines y in terms of x y = 3 x + 2 explicit form 4 Graps of Functions A function must pass te vertical line test A one-to-one function must also pass te orizontal line test Given te grap of a basic function y = f(x), te grap of te transforme function, y = af(bx + c) + = af b x + c +, can be foun using te following rules: b a - vertical stretc (mult. y-values by a) b - orizontal stretc (ivie x-values by b) c orizontal sift (sift left if c > 0, rigt if c < 0) b b b - vertical sift (sift up if > 0, own if < 0) You soul know te basic graps of: y = x, y = x 2, y = x 3 3, y = x, y = x, y = x, y = 1, x y = sin x, y = cos x, y = tan x, y = csc x, y = sec x, y = cot x For furter iscussion of graping, see Precal an Trig Guies to Graping Elementary Functions Algebraic functions (polynomial, raical, rational) - functions tat can be expresse as a finite number of sums, ifferences, multiples, quotients, & raicals involving x n Functions tat are not algebraic are trancenental (eg. trigonometric, exponential, an logaritmic functions)
Polynomial Functions f(x) = a n x n + a n 1 x n 1 +... +a 2 x 2 + a 1 x + a 0 a n = leaing coefficient a n x n = lea term n = egree of polynomial (largest exponent) a 0 = constant term (term wit no x) Lea term test for en beavior: even functions beave like y = x 2, o functions beave like y = x 3 O an Even Functions y = f(x) is even if f( x) = f(x) y = f(x) is o if f( x) = f(x) Rational Functions f(x) = p(x), q(x) p, q polynomials, q(x) 0 Zeros of a function Solutions to te equation f(x) = 0; Wen written in te form (x, 0) are calle te x-intercepts of te function Y-intercepts Foun by evaluating f(0) (plugging 0 in for x) Vertical Asymptotes Foun by setting te enominator equal to 0 an solving for x. Note tat any factors foun in bot numerator an enominator will result in oles rater tan vertical asymptotes. Te grap will approac but never cross a vertical asymptote. Horizontal/Oblique Asymptotes If egree of te numerator is smaller tan te egree of enominator, tere will be a orizontal asymptote at y = 0. If te egree of te numerator is te same as te egree of te enominator, tere will be a orizontal asymptote at y = c, were c is te constant ratio of te leaing coefficients. If te egree of te numerator is one iger tan te egree of te enominator, tere will be an oblique asymptote (one egree iger - linear, 2 egrees iger, quaratic, etc), foun by long ivision.
Limits Informal Definition of a Limit If f(x) becomes arbitrarily close to a single number L as x approaces c from eiter sie, te limit of f(x), as x approaces c, is L. Note: te existence or nonexistence of f(x) at x = c as no bearing on te existence of te limit of f(x) as x approaces c. Formal Definition of a Limit Let f be a function efine on an open interval containing c (except possibly at c) an let L be a real number. Te statement lim x c f(x) = L means tat for eac ε > 0, tere exists a δ > 0 suc tat if 0 < x c < δ, ten f(x) L < ε. Translation: If given any arbitrarily small positive number ε, tere exists anoter small positive number δ suc tat f(x) is ε-close to L wenever x is δ-close to c, ten te limit of f as x approaces c exists an is equal to L. Note: if te limit of a function exists, ten it is unique. Example proof of a limit using te ε δ efinition Problem: Prove tat lim x 3 2x + 5 = 1 Proof: We want to sow tat for eac ε > 0 tere exists a δ > 0 suc tat (2x + 5) ( 1) < ε wenever x ( 3) < δ. Since we can rewrite 2x + 5 + 1 = 2x + 6 = 2 x + 3, if we take δ = ε 2, ten wenever x ( 3) < δ, we ave (2x + 5) ( 1) = 2 x + 3 < 2 ε = ε. Since we ave 2 foun a δ tat works for any ε, we ave prove tat te limit of te function is 1. Evaluating Limits Analytically Basic Limits Let b, c R, n > 0 an integer, f, g functions, lim x c f(x) = L, lim x c g(x) = K 1. Constant lim x c b = b 2. Ientity lim x c x = c 3. Polynomial lim x c x n = c n 4. Scalar Multiple lim x c [bf(x)] = bl 5. Sum or Difference lim x c [f(x) ± g(x)] = L ± K 6. Prouct lim x c [f(x)g(x)] = LK 7. Quotient lim x c f(x) g(x) = L K, K 0 8. Power lim x c [f(x)] n = L n Note: If substitution yiels 0, an ineterminate form, te expression must be rewritten in orer to 0 evaluate te limit.
Examples: 1. lim x 1 2x 2 x 3 x+1 2. lim x 0 2+x 2 x = lim x 1 (x+1)(2x 3) x+1 = lim x 1 (2x 3) = 5 2+x 2 = lim x 0 2+x+ 2 = lim x 2+x+ 2 x 0 (x+) 3. lim 2 x 2 x 0 = lim 2 +2x+ 2 x 2 2x+ 0 = lim 2 0 2+x 2 x 2+x+ 2 = lim x 0 = lim 0 (2x+) 1 = 1 2+x+ 2 2 2 = lim 0 (2x + ) = 2x Squeeze Teorem If g(x) f(x) (x), tat is, a function is boune above an below by two oter functions, an lim x c g(x) = lim x c (x) = L, tat is, te two upper an lower functions ave te same limit, ten lim x c f(x) = L, tat is, te limit of te function tat is squeeze between te two functions wit equal limits must ave te same limit. Example: Prove tat lim x 0 x 2 sin x = 0 using te Squeeze Teorem. Solution: 1 sin x 1. Multiplying bot sies of eac inequality by x 2 yiels x 2 x 2 sin x x 2. Now, applying limits, we ave lim x 0 x 2 lim x 0 x 2 sin x lim x 0 x 2 0 lim x 0 x 2 sin x 0 Since our limit is boune above an below by 0, by te Squeeze teorem, we ave tat lim x 0 x 2 sin x = 0. Two important limits tat can be proven using te Squeeze Teorem tat we will use to fin many oter limits are: sin x lim x 0 x Examples: sin 2x 1. lim x 0 = lim sin 2x sin 3x x 0 3x 2 = 1 1 2 = 2 2x sin 3x 3 3 3 sin(x+) sin x 2. lim 0 lim 0 ( sin x) 1 cos + lim 1 cos x = 1 an lim = 0 x 0 x sin x cos +cos x sin sin x = lim 0 (cos x) sin 0 = lim 0 sin x(cos 1)+cos x sin = ( sin x) 0 + cos x 1 = cos x = Continuity If lim x c f(x) = f(c), we say tat te function f is continuous at c. Discontinuities occur wen eiter 1. f(x) is unefine, 2. lim x c f(x) oes not exist, or 3. lim x c f(x) f(c) If a iscontinuity can be remove by inserting a single point, it is calle removable. Oterwise, it is nonremovable (e.g. verticals asymptotes an jump iscontinuities)
One-Sie Limits lim f(x) = L x c + lim f(x) = L x c Examples: 1. lim x 2 + x 2 x 2 = 1 limit from te rigt limit from te left 2. lim x 1 + f(x), were f(x) = x 2 1, x 2 Note tat = x 2 1, x < 2 x, x 1 Since we want te limit from te rigt, we look at te 1 x, x > 1 piece were te function is efine > 1. Plugging 1 into 1 x gives us tat te limit is 0. Continuity at a point A function f is continuous at c if te following 3 conitions are met: 1. f(c) is efine 2. Limit of f(x) exists wen x approaces c 3. Limit of f(x) wen x approaces c is equal to f(c) Continuity on an open interval A function is continuous on an open interval if it is continuous at eac point in te interval. A function tat is continuous on te entire real line (, )is everywere continuous. Continuity on a close interval A function f is continuous on te close interval [a, b] if it is continuous on te open interval l (a, b) an lim x a + f(x) = f(a) an lim x b f(x) = f(b). Examples: 1. Discuss te continuity of te function f(x) = 1 on te close interval [ 1,2]. We know tat tis x 2 4 function as vertical asymptotes at x = 2 an x = 2, as a orizontal asymptote at x = 0, an as no zeros. Te only iscontinuities are at 2 an 2, an bot are non-removable. So wile we can t say tat te function is continuous on [ 1,2] (because we woul nee te limit from te left to exist at x = 2), we can say tat te function is continuous on [ 1,2). 2. Discuss te continuity of te function f(x) = x 1 x 1. Factoring yiels (x) =. Since te x 2 +x 2 (x 1)(x+2) x 1 factors cancel, tis tells us tat tere is a ole in te grap at x = 1, an tat te function beaves like f(x) = 1 everywere except at x = 1. Tis latter function as a vertical asymptote at x+2 x = 2. Tus, our original function as a removable iscontinuity at x = 1 an a non-removable iscontinuity at x = 2. Continuity of a Composite Function If g is continuous at c an f is continuous at g(c), ten (f g)(x) = f g(x) is continuous at c. Intermeiate Value Teorem If f is continuous on te close interval [a, b] an k is any number between f(a) an f(b), ten tere is at least one number c in [a, b] suc tat f(c) = k.
Infinite Limits means te function increases or ecreases witout boun; i.e. te grap of te function approaces a vertical asymptote Fining Vertical Asymptotes x-values at wic a function is unefine result in eiter oles in te grap or vertical asymptotes. Holes result wen a function can be rewritten so tat te factor wic yiels te iscontinuity cancels. Factors tat can t cancel yiel vertical asymptotes. Example: as vertical asymptotes at an as a vertical asymptote at an a ole at Rules involving infinite limits Let an 1. [ ] 2. [ ] { 3. Te Derivative Te slope of te tangent line to te grap of at te point is given by Te erivative of f at x is given by An alternate form of te erivative tat we often use to sow tat certain continuous functions are not ifferentiable at all points (eiter aving sarp points or vertical tangent lines) is given by Basic Differentiation Rules 1. Te erivative of a constant function is zero, i.e., for [ ] 2. Power Rule for [ ] Special case: [ ] 3. Constant Multiple Rule [ ] 4. Sum & Difference Rules [ ]
Derivatives of Trig Functions 1. [sin x] = cos x x 2. [cos x] = sin x x 3. x [tan x] = sec2 x 4. x [cot x] = csc2 x 5. [sec x] = sec x tan x x 6. [csc x] = csc x cot x x Rates of Cange Te average rate of cange of a function f wit respect to a variable t is given by f t = f(t 2 ) f(t 1) t 2 t 1 Te instantaneous rate of cange of a function f at an instance of te variable t is given by f (t) Te rate of cange of position is velocity. Te rate of cange of velocity is acceleration. Te position of a free-falling object uner te influence of gravity is given by s(t) = 1 2 gt2 + v 0 t + s o, were s o = initial eigt of te object, v 0 = initial velocity of te object, g = 32 ft 2 or 9.8 m/s2 Prouct Rule x [f(x)g(x)] = f(x)g (x) + f (x)g(x) Quotient Rule x f(x) g(x) = g(x)f (x) f(x)g (x) g 2 (x) low ee ig less ig ee low, raw te line an square below Cain Rule x [f(g(x))] = f g(x) g (x) Wen one or more functions are compose, you take te erivative of te outermost function first, keeping its inner function, an ten multiply tat entire ting by te erivative of te inner function. If more tan two functions are compose, you just repeat tis process for all inner functions, e.g. x [f(g((x)))] = f g (x) g (x) (x) x [f(g((k(x))))] = f g k(x) g k(x) k(x) k (x) s
Example: f(x) = sin 3 sec 2 (5x 4 7x) + cos x First we want to rewrite any roots as fractional powers, an any f n (x) as [f(x)] n. f(x) = sin (3 [sec(5x 4 7x)] 2 + cos x) 1 2 Let's ientify our sequence of neste functions to etermine wat orer we will take erivatives. sin a, a = b 1 2, b = 3c 2 + cos x, c = sec, = 5x 4 7x Te erivatives of tese functions are, respectively, (cos a)a, a = 1 2 b 1 2 b, b = (6c)c sin x, c = (sec tan ), = 20x 3 7 So, using te cain rule, te erivative of our original function is f (x) = cos 3 sec 2 (5x 4 7x) + cos x 1 2 (3 [sec(5x4 7x)] 2 + cos x) 1 2 ([6sec(5x 4 7x)] (sec(5x 4 7x) tan(5x 4 7x)) (20x 3 7) sin x) Implicit Differentiation Wen taking te erivative of a function wit respect to x, te only variable wose erivative is 1 is x, since all oter variables are treate as functions of x (except variables staning for constants, like π). So, wen we ave an equation in x an y written implicitly (i.e. not solve for y in terms of x) an we want to fin y, we take te erivative of bot sies, but every erivative involving y must treat y as a function of x, meaning it as to be multiplie by te erivative of te "insie" function, y. Example: x 2 + y 3 = 3xy Taking te erivative of bot sies wit respect to x, we ave 2x + 3y 2 y = 3xy + 3y (we use te power rule twice on te LHS, an te prouct rule on te RHS) Ten we put all our y -terms on one sie, factor out te y, an ivie. 3y 2 y 3xy = 3y 2x y (3y 2 3x) = 3y 2x y 3y 2x = 3y 2 3x If we want to fin y in terms of x an y, we just take te erivative of y, again remembering to treat y as a function of x. We'll en up wit an expression in x, y, an y, so we substitute te expression we foun for y in terms of just x an y. Using te above implicit ifferentiation for y, we ave y = (3y2 3x)(3y 2) (3y 2x)(6yy 3) (3y 2 3x) 2 (3y 2 3y 2x 3y 2x 3x) 3 3y = 2 2 (3y 2x) 6y 3x 3y 2 3 3x (3y 2 3x) 2
Relate Rates Relate rate problems are basically just applie implicit ifferentiation problems, typically ealing wit 2 or more variables tat are eac functions of an aitional variable. For example, a problem ealing wit te volume of a cone, were volume, raius, an eigt are all functions of time. We take te erivative of bot sies of our equation implicitly, remembering tat eac variable is a function of time. Basic equations/formulas to know: Volume of a Spere V = 4 3 πr3 Surface Area of a Spere A = 4πr 2 Volume of a cone V = 1 3 πr2 Volume of a rigt prism V = (area of base) (perpenicular eigt) Example problem: A conical tank is 10 feet across at te top an 10 feet eep. If it is being fille wit water at a rate of 5 cubic feet per minute, fin te rate of cange of te ept of te water wen it is 3 feet eep. Steps to solve: 1. Ientify knowns/unknowns r = 5 wen = 10 r = 5 = 1 r = 10 2 2 v = 5 t ft3 /min =? wen = 3ft t 2. Ientify equation V = 1 3 πr2 Note tat tis equation involves variables V, r, an, all of wic are functions of t. Wen we take te erivative, we will en up wit v, t t r v, an. We know te value of, an we're t t trying to solve for r, but we on't know anyting about, so we want to rewrite te equation t t so tat it oesn't inclue r at all. To o tis, we use te ratio of raius to eigt of te tank. In oter problems, tis step may inclue rearranging te Pytagorean teorem, or using trig functions - watever information you know about te given sape/area/volume. 3. Rewrite equation V = 1 3 πr2 = 1 π 3 2 2 = 1 3 1 π3 4 4. Take te erivative of bot sies wit respect to t (or watever te inepenent variable is). V = 1 t 4 π2 t 5. Rearrange to solve for te unknown. = 4 V t t π 2 6. Plug in known values. Note tat we are not plugging in any known values until te VERY en of te problem. Oterwise, our erivative woul ave been inaccurate. t = 4(5) π3 2 = 20 9π ft/min
Relative Extrema, Increasing & Decreasing, an te First Derivative Test Recall tat te erivative of a function at a point is te slope of te tangent line at tat point. Wen te erivative is zero, te slope is zero, wen te erivative is positive, te slope is positive, an wen te erivative is negative, te slope is negative. f (x) f (x) = 0 f (x) > 0 f (x) < 0 f(x) f is eiter constant or as a relative maximum or minumum f is increasing f is ecreasing Absolute Extrema on a Close Interval To fin absolute maxima an minima on a close interval (te igest an lowest y-values a function acieves on te interval), first take te erivative of te function, set it equal to zero, an solve for x. Tese values are calle critical values or critical numbers. Ten plug any of tese x-values wic lie witin te given close interval, an te x-values of te enpoints of te interval into te original function in orer to fin te y-values of tese points. Te largest y-value is your absolute maximum an te smallest y-value is your absolute minimum. Open Intervals on wic a Function is Increasing or Decreasing If we want to iscuss te beavior of a function on its entire omain, we take all of te critical numbers we foun by setting te erivative equal to zero, along wit any oter x-values for wic te erivative is unefine, an use tese numbers to split te number line into intervals. For example, say we ave a function wit te following erivative: (x 2)(x + 5)(x 8) f (x) = (x + 1) 2 Te critical numbers are 5, 1, 2, an 8. So, we split up te real number line into te following intervals, an ten coose a number witin eac interval to plug into te erivative to see if it is positive or negative in tat interval (if it is positive/negative for one value in tat interval, it will be for all values). Note tat te enominator is always positive, so we can concentrate on te numerator to etermine if f' is positive/negative for eac value. (, 5) ( 5, 1) ( 1,2) (2,8) (8, ) f ( 6) < 0 f ( 2) > 0 f (0) > 0 f (3) < 0 f (9) > 0 Tus, te function f is increasing on ( 5, 1) ( 1,2) (8, ) an ecreasing on (, 5) (2,8).
Concavity an te Secon Derivative Te secon erivative is te rate of cange of te slope of a function. Wen slope increases, a function is concave up. Wen slope ecreases, te function is concave own. Wen a function canges concavity, we get wat are calle inflection points. To etermine possible inflection points, take te secon erivative of te function, set it equal to zero an solve for x. Tose values are te x-coorinates of your inflection points. Ten, ivie up te real number line into intervals base on tose numbers, an plug a number from eac interval into te secon erivative. If positive, te function is concave up on tat interval. If negative, te function is concave own on te interval. f (x) f (x) = 0 f (x) > 0 f (x) < 0 f(x) f as inflection point f is concave up f is concave own Te secon erivative can also be use to etermine weter a particular relative extremum, foun using te first erivative, is in fact a maximum or a minimum, witout aving to etermine weter te function is increasing or ecreasing on eiter sie of te extremum. If te function is concave own at an extremum, it is a maximum; if te function is concave up at an extremum, it is a minimum. Optimization Problems Wen trying to maximize or minimize a function, you're basically using te first erivative test to fin relative extrema. Wen you set te first erivative equal to zero an solve for x, if tere is more tan one solution, it will usually be obvious tat only one of tem makes sense as te answer. For example, a negative value for area, raius, etc. certainly oesn't make sense, so you coose te positive value. Tese problems will typically involve formulas wit multiple variables, an so require multiple equations to solve. Before taking any erivatives, figure out every equation you can tat relate te variables, ten rewrite te formula you are trying to maximize or minimize by substituting for one or more of te variables so tat you en up wit a function of a single variable.
Rolle s Teorem: Let f be continuous on te close interval [a, b] an ifferentiable on te open interval (a, b). If f(a) = f(b), ten tere is at least one number c in (a, b) suc tat f (c) = 0. Basically, Rolle s Teorem states tat if a continuous, ifferentiable function acieves te same y-value at bot enpoints of a given interval, ten tere must be some value in tat interval tere te function as a orizontal tangent line. Mean Value Teorem: If f is continuous on te close interval [a, b] an ifferentiable on te open interval (a, b), ten tere exists a number c in (a, b) suc tat f (c) = f(b) f(a). b a Te Mean Value Teorem is really a generalize version of Rolle s Teorem. It states tat if a function is continuous an ifferentiable on a given interval, ten tere must be some value in tat interval were te slope of te tangent line to te grap of te function is te same as te slope of te secant line troug te enpoints of te interval. Limits at Infinity Te line y = L is a orizontal asymptote of te grap of f if lim x f(x) = L or lim x f(x) = L. c If r is a positive rational number an c is any real number, ten lim x = 0. If x r xr is efine wen c x < 0, ten lim x xr = 0. Guielines for fining limits at infinity of rational functions: If te egree of te numerator is less tan te egree of te enominator, ten te limit is 0. If te egree of te numerator is equal to te egree of te enominator, ten te limit is te ratio of leaing coefficients. If te egree of te numerator is larger tan te egree of te enominator, ten te limit is ±. (te above 3 rules follow te rules for orizontal/oblique asymptotes we learne in Precal) If te numerator or enominator as a raical, remember tat te nt root of x to te n is efine as: n x n x, if n is o = x, if n is even an tat absolute value of x is efine as: x, if x 0 x = x, if x < 0 Ineterminate Forms an l Hopital s Rule 0 0,, 0, 1, 0 0, an are calle ineterminate forms. l Hopital s Rule: Let f an g be functions tat are ifferentiable on an open interval (a, b) containing c, except possibly at c itself. Assume tat g (x) 0 for all x in (a,b), except possibly at c itself. If te limit of f(x)/g(x) as x approaces c prouces an ineterminate form 0/0,, ( ), or ( ), ten lim x c f(x) g(x) = lim x c f (x) g (x)
DERIVATIVE RULES Power Rule: x [xn ] = nx n 1 Constant Multiple Rule: x [cf(x)] = c x [f(x)] Sum & Difference: x [f(x) ± g(x)] = f (x) ± g (x) Prouct Rule: x [f(x)g(x)] = f (x)g(x) + f(x)g (x) Quotient Rule: x f(x) g(x) = f (x)g(x) f(x)g (x) g 2 (x) Cain Rule: x [f(g(x))] = f g(x) g (x) Trig Functions: [sin x] = cos x x [cos x] = sin x x x [tan x] = sec2 x x [cot x] = csc2 x [sec x] = sec x tan x x [csc x] = csc x cot x x Exponential an Logaritmic Functions: x [au ] = a u u ln a x [log a u] = u u ln a Inverse Functions: 1 x [f 1 (x)] = f (f 1 (x)) Inverse Trig Functions: u [arcsin u] = x 1 u 2 [arctan u] = u x 1 + u 2 u [arcsec u] = x u u 2 1 [arccos u] = u x 1 u 2 [arccot u] = u x 1 + u 2 u [arccsc u] = x u u 2 1