Calculus Refresher, version c , Paul Garrett, garrett@math.umn.edu garrett/

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1 Calculus Refresher, version c , Paul Garrett, garrett@math.umn.eu garrett/

2 Contents () Introuction (2) Inequalities (3) Domain of functions (4) Lines (an other items in Analytic Geometry) (5) Elementary limits (6) Limits with cancellation (7) Limits at infinity (8) Limits of exponential functions at infinity (9) The iea of the erivative of a function (0) Derivatives of polynomials () More general power functions (2) Quotient rule (3) Prouct Rule (4) Chain rule (5) Tangent an Normal Lines (6) Critical points, monotone increase an ecrease (7) Minimization an Maximization (8) Local minima an maxima (First Derivative Test) (9) An algebra trick (20) Linear approximations: approximation by ifferentials (2) Implicit ifferentiation (22) Relate rates (23) Intermeiate Value Theorem, location of roots (24) Newton s metho (25) Derivatives of transcenental functions (26) L Hospital s rule (27) Exponential growth an ecay: a ifferential equation (28) The secon an higher erivatives (29) Inflection points, concavity upwar an ownwar (30) Another ifferential equation: projectile motion (3) Graphing rational functions, asymptotes (32) Basic integration formulas (33) The simplest substitutions (34) Substitutions (35) Area an efinite integrals (36) Lengths of Curves (37) Numerical integration 2

3 (38) Averages an Weighte Averages (39) Centers of Mass (Centrois) (40) Volumes by Cross Sections (4) Solis of Revolution (42) Surfaces of Revolution (43) Integration by parts (44) Partial Fractions (45) Trigonometric Integrals (46) Trigonometric Substitutions (47) Historical an theoretical comments: Mean Value Theorem (48) Taylor polynomials: formulas (49) Classic examples of Taylor polynomials (50) Computational tricks regaring Taylor polynomials (5) Prototypes: More serious questions about Taylor polynomials (52) Determining Tolerance/Error (53) How large an interval with given tolerance? (54) Achieving esire tolerance on esire interval (55) Integrating Taylor polynomials: first example (56) Integrating the error term: example 3

4 . Introuction The usual trouble that people have with calculus (not counting general math phobias) is with algebra, not to mention arithmetic an other more elementary things. Calculus itself just involves two new processes, ifferentiation an integration, an applications of these new things to solution of problems that woul have been impossible otherwise. Some things which were very important when calculators an computers in t exist are not so important now. Some things are just as important. Some things are more important. Some things are important but with a ifferent emphasis. At the same time, the essential ieas of much of calculus can be very well illustrate without using calculators at all! (Some not, too). Likewise, many essential ieas of calculus can be very well illustrate without getting embroile in awful algebra or arithmetic, not to mention trigonometry. At the same time, stuy of calculus makes clear how important it is to be able to o the necessary algebra an arithmetic, whether by calculator or by han. 2. Inequalities It is worth reviewing some elementary but important points: First, a person must remember that the only way for a prouct of numbers to be zero is that one or more of the iniviual numbers be zero. As silly as this may seem, it is inispensable. Next, there is the collection of slogans: positive times positive is positive negative times negative is positive negative times positive is negative positive times negative is negative Or, more cutely: the prouct of two numbers of the same sign is positive, while the prouct of two numbers of opposite signs is negative. Extening this just a little: for a prouct of real numbers to be positive, the number of negative ones must be even. If the number of negative ones is o then the prouct is negative. An, of course, if there are any zeros, then the prouct is zero. Solving inequalities: This can be very har in greatest generality, but there are some kins of problems that are very o-able. One important class contains problems like Solve: 5(x )(x + 4)(x 2)(x + 3) < 0 That is, we are asking where a polynomial is negative (or we coul ask where it s positive, too). One important point is that the polynomial is alreay factore: to solve this problem we nee to have the polynomial factore, an if it isn t alreay factore this can be a lot of aitional work. There are many ways to format the solution to such a problem, an we just choose one, which oes have the merit of being more efficient than many. 4

5 We put the roots of the polynomial P (x) = 5(x )(x + 4)(x 2)(x + 3) = 5 (x ) (x ( 4)) (x 2) (x ( 3)) in orer: in this case, the roots are, 4, 2, 3, which we put in orer (from left to right)... < 4 < 3 < < 2 <... The roots of the polynomial P break the numberline into the intervals (, 4), ( 4, 3), ( 3, ), (, 2), (2, + ) On each of these intervals the polynomial is either positive all the time, or negative all the time, since if it were positive at one point an negative at another then it woul have to be zero at some intermeiate point! For input x to the right (larger than) all the roots, all the factors x + 4, x + 3, x, x 2 are positive, an the number 5 in front also happens to be positive. Therefore, on the interval (2, + ) the polynomial P (x) is positive. Next, moving across the root 2 to the interval (, 2), we see that the factor x 2 changes sign from positive to negative, while all the other factors x, x + 3, an x + 4 o not change sign. (After all, if they woul have one so, then they woul have ha to be 0 at some intermeiate point, but they weren t, since we know where they are zero...). Of course the 5 in front stays the same sign. Therefore, since the function was positive on (2, + ) an just one factor change sign in crossing over the point 2, the function is negative on (, 2). Similarly, moving across the root to the interval ( 3, ), we see that the factor x changes sign from positive to negative, while all the other factors x 2, x + 3, an x + 4 o not change sign. (After all, if they woul have one so, then they woul have ha to be 0 at some intermeiate point). The 5 in front stays the same sign. Therefore, since the function was negative on (, 2) an just one factor change sign in crossing over the point, the function is positive on ( 3, ). Similarly, moving across the root 3 to the interval ( 4, 3), we see that the factor x + 3 = x ( 3) changes sign from positive to negative, while all the other factors x 2, x, an x + 4 o not change sign. (If they woul have one so, then they woul have ha to be 0 at some intermeiate point). The 5 in front stays the same sign. Therefore, since the function was positive on ( 3, ) an just one factor change sign in crossing over the point 3, the function is negative on ( 4, 3). Last, moving across the root 4 to the interval (, 4), we see that the factor x + 4 = x ( 4) changes sign from positive to negative, while all the other factors x 2, x, an x + 3 o not change sign. (If they woul have one so, then they woul have ha to be 0 at some intermeiate point). The 5 in front stays the same sign. Therefore, since the function was negative on ( 4, 3) an just one factor change sign in crossing over the point 4, the function is positive on (, 4). In summary, we have In particular, P (x) < 0 on the union of the intervals (, 2) an ( 4, 3). That s it. P (x) = 5(x )(x + 4)(x 2)(x + 3) > 0 on (2, + ) P (x) = 5(x )(x + 4)(x 2)(x + 3) < 0 on (, 2) P (x) = 5(x )(x + 4)(x 2)(x + 3) > 0 on ( 3, ) P (x) = 5(x )(x + 4)(x 2)(x + 3) < 0 on ( 4, 3) P (x) = 5(x )(x + 4)(x 2)(x + 3) > 0 on (, 4) (, 2) ( 4, 3) 5

6 As another example, let s see on which intervals P (x) = 3( + x 2 )(x 2 4)(x 2 2x + ) is positive an an on which it s negative. We have to factor it a bit more: recall that we have nice facts so that we get x 2 a 2 = (x a) (x + a) = (x a) (x ( a)) x 2 2ax + a 2 = (x a) (x a) P (x) = 3( + x 2 )(x 2)(x + 2)(x )(x ) It is important to note that the equation x 2 + = 0 has no real roots, since the square of any real number is non-negative. Thus, we can t factor any further than this over the real numbers. That is, the roots of P, in orer, are 2 << (twice!) < 2 These numbers break the real line up into the intervals (, 2), ( 2, ), (, 2), (2, + ) For x larger than all the roots (meaning x > 2) all the factors x + 2, x, x, x 2 are positive, while the factor of 3 in front is negative. Thus, on the interval (2, + ) P (x) is negative. Next, moving across the root 2 to the interval (, 2), we see that the factor x 2 changes sign from positive to negative, while all the other factors + x 2, (x ) 2, an x + 2 o not change sign. (After all, if they woul have one so, then they woul have be 0 at some intermeiate point, but they aren t). The 3 in front stays the same sign. Therefore, since the function was negative on (2, + ) an just one factor change sign in crossing over the point 2, the function is positive on (, 2). A new feature in this example is that the root occurs twice in the factorization, so that crossing over the root from the interval (, 2) to the interval ( 2, ) really means crossing over two roots. That is, two changes of sign means no changes of sign, in effect. An the other factors ( + x 2 ), x + 2, x 2 o not change sign, an the 3 oes not change sign, so since P (x) was positive on (, 2) it is still positive on ( 2, ). (The rest of this example is the same as the first example). Again, the point is that each time a root of the polynomial is crosse over, the polynomial changes sign. So if two are crosse at once (if there is a ouble root) then there is really no change in sign. If three roots are crosse at once, then the effect is to change sign. Generally, if an even number of roots are crosse-over, then there is no change in sign, while if an o number of roots are crosse-over then there is a change in sign. 6

7 #2. Fin the intervals on which f(x) = x(x )(x + ) is positive, an the intervals on which it is negative. #2.2 Fin the intervals on which f(x) = (3x 2)(x )(x + ) is positive, an the intervals on which it is negative. #2.3 Fin the intervals on which f(x) = (3x 2)(3 x)(x + ) is positive, an the intervals on which it is negative. 3. Domain of functions A function f is a proceure or process which converts input to output in some way. A traitional mathematics name for the input is argument, but this certainly is confusing when compare with orinary English usage. The collection of all legal reasonable or sensible inputs is calle the omain of the function. The collection of all possible outputs is the range. (Contrary to the impression some books might give, it can be very ifficult to figure out all possible outputs!) The question What s the omain of this function? is usually not what it appears to be. For one thing, if we are being formal, then a function hasn t even been escribe if its omain hasn t been escribe! What is really meant, usually, is something far less mysterious. The question usually really is What numbers can be use as inputs to this function without anything ba happening?. For our purposes, something ba happening just refers to one of trying to take the square root of a negative number trying to take a logarithm of a negative number trying to ivie by zero trying to fin arc-cosine or arc-sine of a number bigger than or less than Of course, iviing by zero is the worst of these, but as long as we insist that everything be real numbers (rather than complex numbers) we can t o the other things either. For example, what is the omain of the function f(x) = x 2? Well, what coul go wrong here? No ivision is inicate at all, so there is no risk of iviing by 0. But we are taking a square root, so we must insist that x 2 0 to avoi having complex numbers come up. That is, a preliminary escription of the omain of this function is that it is the set of real numbers x so that x 2 0. But we can be clearer than this: we know how to solve such inequalities. Often it s simplest to see what to exclue rather than inclue: here we want to exclue from the omain any numbers x so that x 2 < 0 from the omain. We recognize that we can factor x 2 = (x )(x + ) = (x ) (x ( )) 7

8 This is negative exactly on the interval (, ), so this is the interval we must prohibit in orer to have just the omain of the function. That is, the omain is the union of two intervals: (, ] [, + ) #3.4 Fin the omain of the function f(x) = x 2 x 2 + x 2 That is, fin the largest subset of the real line on which this formula can be evaluate meaningfully. #3.5 Fin the omain of the function f(x) = x 2 x2 + x 2 #3.6 Fin the omain of the function f(x) = x(x )(x + ) 4. Lines (an other items in Analytic Geometry) Let s review some basic analytic geometry: this is escription of geometric objects by numbers an by algebra. The first thing is that we have to pick a special point, the origin, from which we ll measure everything else. Then, implicitly, we nee to choose a unit of measure for istances, but this is inee usually only implicit, so we on t worry about it. The secon step is that points are escribe by orere pairs of numbers: the first of the two numbers tells how far to the right horizontally the point is from the origin (an negative means go left instea of right), an the secon of the two numbers tells how far up from the origin the point is (an negative means go own instea of up). The first number is the horizontal coorinate an the secon is the vertical coorinate. The ol-fashione names abscissa an orinate also are use sometimes. Often the horizontal coorinate is calle the x-coorinate, an often the vertical coorinate is calle the y- coorinate, but the letters x, y can be use for many other purposes as well, so on t rely on this labelling! The next iea is that an equation can escribe a curve. It is important to be a little careful with use of language here: for example, a correct assertion is The set of points (x, y) so that x 2 + y 2 = is a circle. It is not strictly correct to say that x 2 + y 2 = is a circle, mostly because an equation is not a circle, even though it may escribe a circle. An conceivably the x, y might be being use for something other than horizontal an vertical coorinates. Still, very often the language is shortene so that the phrase The set of points (x, y) so that is omitte. Just be careful. The simplest curves are lines. The main things to remember are: Slope of a line is rise over run, meaning vertical change ivie by horizontal change (moving from left to right in the usual coorinate system). The equation of a line passing through a point (x o, y o ) an having slope m can be written (in so-calle point-slope form) y = m(x x o ) + y o or y y o = m(x x o ) 8

9 The equation of the line passing through two points (x, y ), (x 2, y 2 ) can be written (in so-calle twopoint form) as y = y y 2 x x 2 (x x ) + y...unless x = x 2, in which case the two points are aligne vertically, an the line can t be written that way. Instea, the escription of a vertical line through a point with horizontal coorinate x is just x = x Of course, the two-point form can be erive from the point-slope form, since the slope m of a line through two points (x, y ), (x 2, y 2 ) is that possibly irritating expression which occurs above: m = y y 2 x x 2 An now is maybe a goo time to point out that there is nothing sacre about the horizontal coorinate being calle x an the vertical coorinate y. Very often these o happen to be the names, but it can be otherwise, so just pay attention. #4.7 Write the equation for the line passing through the two points (, 2) an (3, 8). #4.8 Write the equation for the line passing through the two points (, 2) an (3, 8). #4.9 Write the equation for the line passing through the point (, 2) with slope 3. #4.0 Write the equation for the line passing through the point (, 5) with slope. 5. Elementary limits The iea of limit is intene to be merely a slight extension of our intuition. The so-calle ε, δ-efinition was invente after people ha been oing calculus for hunres of years, in response to certain relatively pathological technical ifficulties. For quite a while, we will be entirely concerne with situations in which we can either irectly see the value of a limit by plugging the limit value in, or where we transform the expression into one where we can just plug in. So long as we are ealing with functions no more complicate than polynomials, most limits are easy to unerstan: for example, lim x 3 4x2 + 3x 7 = 4 (3) (3) 7 = 38 4x 2 + 3x 7 lim x 3 2 x 2 = 4 (3)2 + 3 (3) 7 2 (3) 2 = 38 7 The point is that we just substitute the 3 in an nothing ba happene. This is the way people evaluate easy limits for hunres of years, an shoul always be the first thing a person oes, just to see what happens. #5. Fin lim x 5 2x 2 3x

10 #5.2 Fin lim x 2 x+ x #5.3 Fin lim x x Limits with cancellation But sometimes things blow up when the limit number is substitute: x 2 9 lim x 3 x 3 = 0 0????? Ick. This is not goo. However, in this example, as in many examples, oing a bit of simplifying algebra first gets ri of the factors in the numerator an enominator which cause them to vanish: x 2 9 lim x 3 x 3 = lim (x 3)(x + 3) (x + 3) (3 + 3) = lim = = 6 x 3 x 3 x 3 Here at the very en we i just plug in, after all. The lesson here is that some of those arn algebra tricks ( ientities ) are helpful, after all. If you have a ba limit, always look for some cancellation of factors in the numerator an enominator. In fact, for hunres of years people only evaluate limits in this style! After all, human beings can t really execute infinite limiting processes, an so on. #6.4 Fin lim x 2 x 2 x 2 4 #6.5 Fin lim x 3 x 2 9 x 3 #6.6 Fin lim x 3 x 2 x 3 Next, let s consier 7. Limits at infinity 2x + 3 lim x 5 x The hazar here is that is not a number that we can o arithmetic with in the normal way. Don t even try it. So we can t really just plug in to the expression to see what we get. On the other han, what we really mean anyway is not that x becomes infinite in some mystical sense, but rather that it just gets larger an larger. In this context, the crucial observation is that, as x gets larger an larger, /x gets smaller an smaller (going to 0). Thus, just base on what we want this all to mean, lim x x = 0 an so on. lim x lim x x 2 = 0 x 3 = 0 0

11 This is the essential iea for evaluating simple kins of limits as x : rearrange the whole thing so that everything is expresse in terms of /x instea of x, an then realize that lim x is the same as lim x 0 So, in the example above, ivie numerator an enominator both by the largest power of x appearing anywhere: 2x + 3 lim x 5 x = lim x x 5 x = lim 2 + 3y y 0 5y = = 2 The point is that we calle /x by a new name, y, an rewrote the original limit as x as a limit as y 0. Since 0 is a genuine number that we can o arithmetic with, this brought us back to orinary everyay arithmetic. Of course, it was necessary to rewrite the thing we were taking the limit of in terms of /x (rename y ). Notice that this is an example of a situation where we use the letter y for something other than the name or value of the vertical coorinate. #7.7 Fin lim x x+ x #7.8 Fin lim x x 2 +3 x+. #7.9 Fin lim x x x 2 +x+. #7.20 Fin lim x x 2 5x 2 +x+. 8. Limits of exponential functions at infinity It is important to appreciate the behavior of exponential functions as the input to them becomes a large positive number, or a large negative number. This behavior is ifferent from the behavior of polynomials or rational functions, which behave similarly for large inputs regarless of whether the input is large positive or large negative. By contrast, for exponential functions, the behavior is raically ifferent for large positive or large negative. As a reminer an an explanation, let s remember that exponential notation starte out simply as an abbreviation: for positive integer n, 2 n = (n factors) 0 n = (n factors) ( ) n ( ) ( ) ( ) ( ) =... (n factors) From this iea it s not har to unerstan the funamental properties of exponents (they re not laws at all): a m+n = a a a... a (m + n factors) } {{ } m+n = (a a a... a) (a a a... a) = a m a n } {{ } } {{ } m n

12 an also = (a a a... a) } {{ } m a mn = (a a a... a) = } {{ } mn... (a a a... a) } {{ } m } {{ } n = (a m ) n at least for positive integers m, n. Even though we can only easily see that these properties are true when the exponents are positive integers, the extene notation is guarantee (by its meaning, not by law) to follow the same rules. Use of other numbers in the exponent is something that came later, an is also just an abbreviation, which happily was arrange to match the more intuitive simpler version. For example, a = a an (as consequences) a n = a n ( ) = (a n ) = a n (whether n is positive or not). Just to check one example of consistency with the properties above, notice that a = a = a ( ) ( ) = a = /a = a This is not suppose to be surprising, but rather reassuring that we won t reach false conclusions by such manipulations. Also, fractional exponents fit into this scheme. For example a /2 = a a /3 = [3]a a /4 = [4]a a /5 = [5]a This is consistent with earlier notation: the funamental property of the n th root of a number is that its n th power is the original number. We can check: a = a = (a /n ) n = a Again, this is not suppose to be a surprise, but rather a consistency check. Then for arbitrary rational exponents m/n we can maintain the same properties: first, the efinition is just a m/n = ( [n]a) m One hazar is that, if we want to have only real numbers (as oppose to complex numbers) come up, then we shoul not try to take square roots, 4 th roots, 6 th roots, or any even orer root of negative numbers. For general real exponents x we likewise shoul not try to unerstan a x except for a > 0 or we ll have to use complex numbers (which wouln t be so terrible). But the value of a x can only be efine as a limit: let r, r 2,... be a sequence of rational numbers approaching x, an efine a x = lim i a ri We woul have to check that this efinition oes not accientally epen upon the sequence approaching x (it oesn t), an that the same properties still work (they o). 2

13 The number e is not something that woul come up in really elementary mathematics, because its reason for existence is not really elementary. Anyway, it s approximately e = but if this ever really mattere you have a calculator at your sie, hopefully. With the efinitions in min it is easier to make sense of questions about limits of exponential functions. The two companion issues are to evaluate lim x + ax lim x ax Since we are allowing the exponent x to be real, we better eman that a be a positive real number (if we want to avoi complex numbers, anyway). Then lim x + ax = + if a > if a = 0 if 0 < a < 0 if a > lim x ax = if a = + if 0 < a < To remember which is which, it is sufficient to use 2 for a > an 2 for 0 < a <, an just let x run through positive integers as it goes to +. Likewise, it is sufficient to use 2 for a > an 2 for 0 < a <, an just let x run through negative integers as it goes to. 9. The iea of the erivative of a function First we can tell what the iea of a erivative is. But the issue of computing erivatives is another thing entirely: a person can unerstan the iea without being able to effectively compute, an vice-versa. Suppose that f is a function of interest for some reason. We can give f some sort of geometric life by thinking about the set of points (x, y) so that f(x) = y We woul say that this escribes a curve in the (x, y)-plane. (An sometimes we think of x as moving from left to right, imparting further intuitive or physical content to the story). For some particular number x o, let y o be the value f(x o ) obtaine as output by plugging x o into f as input. Then the point (x o, y o ) is a point on our curve. The tangent line to the curve at the point (x o, y o ) is a line passing through (x o, y o ) an flat against the curve. (As oppose to crossing it at some efinite angle). The iea of the erivative f (x o ) is that it is the slope of the tangent line at x o to the curve. But this isn t the way to compute these things Derivatives of polynomials There are just four simple facts which suffice to take the erivative of any polynomial, an actually of somewhat more general things. 3

14 First, there is the rule for taking the erivative of a power function which takes the nth power of its input. That is, these functions are functions of the form f(x) = x n. The formula is x xn = n x n That is, the exponent comes own to become a coefficient in front of the thing, an the exponent is ecrease by. The secon rule, which is really a special case of this power-function rule, is that erivatives of constants are zero: x c = 0 for any constant c. The thir thing, which reflects the innocuous role of constants in calculus, is that for any functions f of x x c f = c The fourth is that for any two functions f, g of x, the erivative of the sum is the sum of the erivatives: x f (f + g) = x x f + x g Putting these four things together, we can write general formulas like x (axm + bx n + cx p ) = a mx m + b nx n + c px p an so on, with more summans than just the three, if so esire. An in any case here are some examples with numbers instea of letters: x 5x3 = 5 3x 3 = 5x 2 x (3x7 + 5x 3 ) = 3 7x x 2 0 = 2x 6 + 5x 2 x (2 3x2 2x 3 ) = 0 3 2x 2 3x 2 = 6x 6x 2 x ( x4 + 2x 5 + ) = 4x x = 4x 3 + 0x 4 Even if you o catch on to this iea right away, it is wise to practice the technique so that not only can you o it in principle, but also in practice. #0.2 Fin x (3x7 + 5x 3 ) #0.22 Fin x (x2 + 5x 3 + 2) #0.23 Fin x ( x4 + 2x 5 + ) #0.24 Fin x ( 3x2 x 3 ) 4

15 . More general power functions It s important to remember some of the other possibilities for the exponential notation x n. For example x 2 = x x = x x 2 = x an so on. The goo news is that the rule given just above for taking the erivative of powers of x still is correct here, even for exponents which are negative or fractions or even real numbers: Thus, in particular, x xr = r x r x = x x x 2 = 2 x 2 x x = x x = x 2 = x 2 When combine with the sum rule an so on from above, we have the obvious possibilities: x (3x2 7 x + 5 x 2 = x (3x2 7x 2 + 5x 2 ) = 6x 7 2 x 2 0x 3 The possibility of expressing square roots, cube roots, inverses, etc., in terms of exponents is a very important iea in algebra, an can t be overlooke. #.25 Fin x (3x7 + 5 x ) #.26 Fin x ( 2 x x + 3) #.27 Fin x (7 5 x 3 + 5x 7 ) 2. Quotient rule The quotient rule is one of the more irritating an goofy things in elementary calculus, but it just couln t have been any other way. The general principle is x ( ) f = f g g f g g 2 The main hazar is remembering that the numerator is as it is, rather than accientally reversing the roles of f an g, an then being off by ±, which coul be fatal in real life. ( ) = x x 2 x (x 2) x (x 2) 0 (x 2) (x 2) 2 = (x 2) 2 = (x 2) 2 5

16 x ( ) x = (x ) (x 2) (x )(x 2) (x 2) (x ) x 2 (x 2) 2 = (x 2) 2 x = ( 5x 3 ) + x 2 x 7 (x 2) (x ) (x 2) 2 = (x 2) 2 = (5x3 + x) (2 x 7 ) (5x 3 + x) (2 x 7 ) (2 x 7 ) 2 = (5x2 + ) (2 x 7 ) (5x 3 + x) ( 7x 6 ) (2 x 7 ) 2 an there s harly any point in simplifying the last expression, unless someone gives you a goo reason. In general, it s not so easy to see how much may or may not be gaine in simplifying, an we won t make ourselves crazy over it. #2.28 Fin x ( x x 2 ) #2.29 Fin x ( x 2 ) #2.30 Fin x ( x x 2 5 ) #2.3 Fin x ( x3 2+ x ) 3. Prouct Rule Not only will the prouct rule be of use in general an later on, but it s alreay helpful in perhaps unexpecte ways in ealing with polynomials. Anyway, the general rule is x (fg) = f g + fg While this is certainly not as awful as the quotient rule just above, it is not as simple as the rule for sums, which was the goo-souning slogan that the erivative of the sum is the sum of the erivatives. It is not true that the erivative of the prouct is the prouct of the erivatives. Too ba. Still, it s not as ba as the quotient rule. One way that the prouct rule can be useful is in postponing or eliminating a lot of algebra. For example, to evaluate ( (x 3 + x 2 + x + )(x 4 + x 3 + 2x + ) ) x we coul multiply out an then take the erivative term-by-term as we i with several polynomials above. This woul be at least milly irritating because we have to o a bit of algebra. Rather, just apply the prouct rule without feeling compelle first to o any algebra: ( (x 3 + x 2 + x + )(x 4 + x 3 + 2x + ) ) x = (x 3 + x 2 + x + ) (x 4 + x 3 + 2x + ) + (x 3 + x 2 + x + )(x 4 + x 3 + 2x + ) = (3x 2 + 2x + )(x 4 + x 3 + 2x + ) + (x 3 + x 2 + x + )(4x 3 + 3x 2 + 2) 6

17 Now if we were somehow still oblige to multiply out, then we still have to o some algebra. But we can take the erivative without multiplying out, if we want to, by using the prouct rule. For that matter, once we see that there is a choice about oing algebra either before or after we take the erivative, it might be possible to make a choice which minimizes our computational labor. This coul matter. #3.32 Fin x (x3 )(x 6 + x 3 + )) #3.33 Fin x (x2 + x + )(x 4 x 2 + ). #3.34 Fin x (x3 + x 2 + x + )(x 4 + x 2 + )) #3.35 Fin x (x3 + x 2 + x + )(2x + x)) 4. Chain rule The chain rule is subtler than the previous rules, so if it seems trickier to you, then you re right. OK. But it is absolutely inispensable in general an later, an alreay is very helpful in ealing with polynomials. The general assertion may be a little har to fathom because it is of a ifferent nature than the previous ones. For one thing, now we will be talking about a composite function instea of just aing or multiplying functions in a more orinary way. So, for two functions f an g, x ((f(g(x))) = f (g(x)) g (x) There is also the stanar notation (f g)(x) = f(g(x)) for this composite function, but using this notation oesn t accomplish so very much. A problem in successful use of the chain rule is that often it requires a little thought to recognize that some formula is (or can be looke at as) a composite function. An the very nature of the chain rule picks on weaknesses in our unerstaning of the notation. For example, the function F (x) = ( + x 2 ) 00 is really obtaine by first using x as input to the function which squares an as to its input. Then the result of that is use as input to the function which takes the 00th power. It is necessary to think about it this way or we ll make a mistake. The erivative is evaluate as x ( + x2 ) 00 = 00( + x 2 ) 99 2x To see that this is a special case of the general formula, we nee to see what correspons to the f an g in the general formula. Specifically, let f(input) = (input) 00 g(input) = + (input) 2 The reason for writing input an not x for the moment is to avoi a certain kin of mistake. But we can compute that f (input) = 00(input) 99 g (input) = 2(input) 7

18 The hazar here is that the input to f is not x, but rather is g(x). So the general formula gives x ( + x2 ) 00 = f (g(x)) g (x) = 00g(x) 99 2x = 00( + x 2 ) 99 2x More examples: 3x + 2 = x x (3x + 2)/2 = 2 (3x + 2) /2 3 It is very important to recognize situations like for any constants a, b, n. An, of course, this inclues x (3x5 x + 4) = (3x 5 x + 4) 0 (5x 4 ) x (ax + b)n = n(ax + b) n a ax + b = x 2 (ax + b) /2 a x ax + b = (ax + a b) 2 a = (ax + b) 2 Of course, this iea can be combine with polynomials, quotients, an proucts to give enormous an excruciating things where we nee to use the chain rule, the quotient rule, the prouct rule, etc., an possibly several times each. But this is not har, merely teious, since the only things we really o come in small steps. For example: x by the quotient rule, which is then ( ) + x + 2 ( + 7x) 33 = ( + x + 2) ( + 7x) 33 ( + x + 2) (( + 7x) 33 ) (( + 7x) 33 ) 2 ( 2 (x + 2) /2 ) ( + 7x) 33 ( + x + 2) (( + 7x) 33 ) (( + 7x) 33 ) 2 because our observations just above (chain rule!) tell us that x + 2 = x 2 (x + 2) /2 (x + 2) = (x + 2) /2 2 Then we use the chain rule again to take the erivative of that big power of + 7x, so the whole thing becomes ( 2 (x + 2) /2 ) ( + 7x) 33 ( + x + 2) (33( + 7x) 32 7) (( + 7x) 33 ) 2 Although we coul simplify a bit here, let s not. The point about having to o several things in a row to take a erivative is pretty clear without oing algebra just now. #4.36 Fin x (( x2 ) 00 ) #4.37 Fin x x 3 8

19 #4.38 Fin x (x2 x 2 3) #4.39 Fin x ( x 2 + x + ) #4.40 Fin x ( 3 x 3 + x 2 + x + ) #4.4 Fin x ((x3 + x + ) 0 ) 5. Tangent an Normal Lines One funamental interpretation of the erivative of a function is that it is the slope of the tangent line to the graph of the function. (Still, it is important to realize that this is not the efinition of the thing, an that there are other possible an important interpretations as well). The precise statement of this funamental iea is as follows. Let f be a function. For each fixe value x o of the input to f, the value f (x o ) of the erivative f of f evaluate at x o is the slope of the tangent line to the graph of f at the particular point (x o, f(x o )) on the graph. Recall the point-slope form of a line with slope m through a point (x o, y o ): y y o = m(x x o ) In the present context, the slope is f (x o ) an the point is (x o, f(x o )), so the equation of the tangent line to the graph of f at (x o, f(x o )) is y f(x o ) = f (x o )(x x o ) The normal line to a curve at a particular point is the line through that point an perpenicular to the tangent. A person might remember from analytic geometry that the slope of any line perpenicular to a line with slope m is the negative reciprocal /m. Thus, just changing this aspect of the equation for the tangent line, we can say generally that the equation of the normal line to the graph of f at (x o, f(x o )) is y f(x o ) = f (x o ) (x x o) The main conceptual hazar is to mistakenly name the fixe point x, as well as naming the variable coorinate on the tangent line x. This causes a person to write own some equation which, whatever it may be, is not the equation of a line at all. Another popular boo-boo is to forget the subtraction f(x o ) on the left han sie. Don t o it. So, as the simplest example: let s write the equation for the tangent line to the curve y = x 2 at the point where x = 3. The erivative of the function is y = 2x, which has value 2 3 = 6 when x = 3. An the value of the function is 3 3 = 9 when x = 3. Thus, the tangent line at that point is The normal line at the point where x = 3 is y 9 = 6(x 3) y 9 = (x 3) 6 So the question of fining the tangent an normal lines at various points of the graph of a function is just a combination of the two processes: computing the erivative at the point in question, an invoking the point-slope form of the equation for a straight line. 9

20 #5.42 Write the equation for both the tangent line an normal line to the curve y = 3x 2 x + at the point where x =. #5.43 Write the equation for both the tangent line an normal line to the curve y = (x )/(x + ) at the point where x = Critical points, monotone increase an ecrease A function is calle increasing if it increases as the input x moves from left to right, an is calle ecreasing if it ecreases as x moves from left to right. Of course, a function can be increasing in some places an ecreasing in others: that s the complication. We can notice that a function is increasing if the slope of its tangent is positive, an ecreasing if the slope of its tangent is negative. Continuing with the iea that the slope of the tangent is the erivative: a function is increasing where its erivative is positive, an is ecreasing where its erivative is negative. This is a great principle, because we on t have to graph the function or otherwise list lots of values to figure out where it s increasing an ecreasing. If anything, it shoul be a big help in graphing to know in avance where the graph goes up an where it goes own. An the points where the tangent line is horizontal, that is, where the erivative is zero, are critical points. The points where the graph has a peak or a trough will certainly lie among the critical points, although there are other possibilities for critical points, as well. Further, for the kin of functions we ll eal with here, there is a fairly systematic way to get all this information: to fin the intervals of increase an ecrease of a function f: Compute the erivative f of f, an solve the equation f (x) = 0 for x to fin all the critical points, which we list in orer as x < x 2 <... < x n. (If there are points of iscontinuity or non-ifferentiability, these points shoul be ae to the list! But points of iscontinuity or non-ifferentiability are not calle critical points.) We nee some auxiliary points: To the left of the leftmost critical point x pick any convenient point t o, between each pair of consecutive critical points x i, x i+ choose any convenient point t i, an to the right of the rightmost critical point x n choose a convenient point t n. Evaluate the erivative f at all the auxiliary points t i. Conclusion: if f (t i+ ) > 0, then f is increasing on (x i, x i+ ), while if f (t i+ ) < 0, then f is ecreasing on that interval. Conclusion: on the outsie interval (, x o ), the function f is increasing if f (t o ) > 0 an is ecreasing if f (t o ) < 0. Similarly, on (x n, ), the function f is increasing if f (t n ) > 0 an is ecreasing if f (t n ) < 0. It is certainly true that there are many possible shortcuts to this proceure, especially for polynomials of low egree or other rather special functions. However, if you are able to quickly compute values of (erivatives of!) functions on your calculator, you may as well use this proceure as any other. Exactly which auxiliary points we choose oes not matter, as long as they fall in the correct intervals, since we just nee a single sample on each interval to fin out whether f is positive or negative there. Usually we pick integers or some other kin of number to make computation of the erivative there as easy as possible. 20

21 It s important to realize that even if a question oes not irectly ask for critical points, an maybe oes not ask about intervals either, still it is implicit that we have to fin the critical points an see whether the functions is increasing or ecreasing on the intervals between critical points. Examples: Fin the critical points an intervals on which f(x) = x 2 + 2x + 9 is increasing an ecreasing: Compute f (x) = 2x + 2. Solve 2x + 2 = 0 to fin only one critical point. To the left of let s use the auxiliary point t o = 2 an to the right use t = 0. Then f ( 2) = 2 < 0, so f is ecreasing on the interval (, ). An f (0) = 2 > 0, so f is increasing on the interval (, ). Fin the critical points an intervals on which f(x) = x 3 2x + 3 is increasing, ecreasing. Compute f (x) = 3x 2 2. Solve 3x 2 2 = 0: this simplifies to x 2 4 = 0, so the critical points are ±2. To the left of 2 choose auxiliary point t o = 3, between 2 an = 2 choose auxiliary point t = 0, an to the right of +2 choose t 2 = 3. Plugging in the auxiliary points to the erivative, we fin that f ( 3) = 27 2 > 0, so f is increasing on (, 2). Since f (0) = 2 < 0, f is ecreasing on ( 2, +2), an since f (3) = 27 2 > 0, f is increasing on (2, ). Notice too that we on t really nee to know the exact value of the erivative at the auxiliary points: all we care about is whether the erivative is positive or negative. The point is that sometimes some teious computation can be avoie by stopping as soon as it becomes clear whether the erivative is positive or negative. #6.44 Fin the critical points an intervals on which f(x) = x 2 + 2x + 9 is increasing, ecreasing. #6.45 Fin the critical points an intervals on which f(x) = 3x 2 6x + 7 is increasing, ecreasing. #6.46 Fin the critical points an intervals on which f(x) = x 3 2x + 3 is increasing, ecreasing. 7. Minimization an Maximization The funamental iea which makes calculus useful in unerstaning problems of maximizing an minimizing things is that at a peak of the graph of a function, or at the bottom of a trough, the tangent is horizontal. That is, the erivative f (x o ) is 0 at points x o at which f(x o ) is a maximum or a minimum. Well, a little sharpening of this is necessary: sometimes for either natural or artificial reasons the variable x is restricte to some interval [a, b]. In that case, we can say that the maximum an minimum values of f on the interval [a, b] occur among the list of critical points an enpoints of the interval. An, if there are points where f is not ifferentiable, or is iscontinuous, then these have to be ae in, too. But let s stick with the basic iea, an just ignore some of these complications. Let s escribe a systematic proceure to fin the minimum an maximum values of a function f on an interval [a, b]. Solve f (x) = 0 to fin the list of critical points of f. Exclue any critical points not insie the interval [a, b]. A to the list the enpoints a, b of the interval (an any points of iscontinuity or non-ifferentiability!) At each point on the list, evaluate the function f: the biggest number that occurs is the maximum, an the littlest number that occurs is the minimum. Fin the minima an maxima of the function f(x) = x 4 8x on the interval [, 3]. First, take the erivative an set it equal to zero to solve for critical points: this is 4x 3 6x = 0 2

22 or, more simply, iviing by 4, it is x 3 4x = 0. Luckily, we can see how to factor this: it is x(x 2)(x + 2) So the critical points are 2, 0, +2. Since the interval oes not inclue 2, we rop it from our list. An we a to the list the enpoints, 3. So the list of numbers to consier as potential spots for minima an maxima are, 0, 2, 3. Plugging these numbers into the function, we get (in that orer) 2, 5,, 4. Therefore, the maximum is 4, which occurs at x = 3, an the minimum is, which occurs at x = 2. Notice that in the previous example the maximum i not occur at a critical point, but by coincience i occur at an enpoint. You have 200 feet of fencing with which you wish to enclose the largest possible rectangular garen. What is the largest garen you can have? Let x be the length of the garen, an y the with. Then the area is simply xy. Since the perimeter is 200, we know that 2x + 2y = 200, which we can solve to express y as a function of x: we fin that y = 00 x. Now we can rewrite the area as a function of x alone, which sets us up to execute our proceure: area = xy = x(00 x) The erivative of this function with respect to x is 00 2x. Setting this equal to 0 gives the equation 00 2x = 0 to solve for critical points: we fin just one, namely x = 50. Now what about enpoints? What is the interval? In this example we must look at physical consierations to figure out what interval x is restricte to. Certainly a with must be a positive number, so x > 0 an y > 0. Since y = 00 x, the inequality on y gives another inequality on x, namely that x < 00. So x is in [0, 00]. When we plug the values 0, 50, 00 into the function x(00 x), we get 0, 2500, 0, in that orer. Thus, the corresponing value of y is = 50, an the maximal possible area is = #7.47 Olivia has 200 feet of fencing with which she wishes to enclose the largest possible rectangular garen. What is the largest garen she can have? #7.48 Fin the minima an maxima of the function f(x) = 3x 4 4x on the interval [ 2, 3]. #7.49 The cost per hour of fuel to run a locomotive is v 2 /25 ollars, where v is spee, an other costs are $00 per hour regarless of spee. What is the spee that minimizes cost per mile? #7.50 The prouct of two numbers x, y is 6. We know x an y. What is the greatest possible sum of the two numbers? #7.5 Fin both the minimum an the maximum of the function f(x) = x 3 + 3x + on the interval [ 2, 2]. 8. Local minima an maxima (First Derivative Test) A function f has a local maximum or relative maximum at a point x o if the values f(x) of f for x near x o are all less than f(x o ). Thus, the graph of f near x o has a peak at x o. A function f has a local minimum or relative minimum at a point x o if the values f(x) of f for x near x o are all greater 22

23 than f(x o ). Thus, the graph of f near x o has a trough at x o. (To make the istinction clear, sometimes the plain maximum an minimum are calle absolute maximum an minimum.) Yes, in both these efinitions we are tolerating ambiguity about what near woul mean, although the peak/trough requirement on the graph coul be translate into a less ambiguous efinition. But in any case we ll be able to execute the proceure given below to fin local maxima an minima without worrying over a formal efinition. This proceure is just a variant of things we ve alreay one to analyze the intervals of increase an ecrease of a function, or to fin absolute maxima an minima. This proceure starts out the same way as oes the analysis of intervals of increase/ecrease, an also the proceure for fining ( absolute ) maxima an minima of functions. To fin the local maxima an minima of a function f on an interval [a, b]: Solve f (x) = 0 to fin critical points of f. Drop from the list any critical points that aren t in the interval [a, b]. A to the list the enpoints (an any points of iscontinuity or non-ifferentiability): we have an orere list of special points in the interval: a = x o < x <... < x n = b Between each pair x i < x i+ of points in the list, choose an auxiliary point t i+. Evaluate the erivative f at all the auxiliary points. For each critical point x i, we have the auxiliary points to each sie of it: t i < x i < t i+. There are four cases best remembere by rawing a picture!: if f (t i ) > 0 an f (t i+ ) < 0 (so f is increasing to the left of x i an ecreasing to the right of x i, then f has a local maximum at x o. if f (t i ) < 0 an f (t i+ ) > 0 (so f is ecreasing to the left of x i an increasing to the right of x i, then f has a local minimum at x o. if f (t i ) < 0 an f (t i+ ) < 0 (so f is ecreasing to the left of x i an also ecreasing to the right of x i, then f has neither a local maximum nor a local minimum at x o. if f (t i ) > 0 an f (t i+ ) > 0 (so f is increasing to the left of x i an also increasing to the right of x i, then f has neither a local maximum nor a local minimum at x o. The enpoints require separate treatment: There is the auxiliary point t o just to the right of the left enpoint a, an the auxiliary point t n just to the left of the right enpoint b: At the left enpoint a, if f (t o ) < 0 (so f is ecreasing to the right of a) then a is a local maximum. At the left enpoint a, if f (t o ) > (so f is increasing to the right of a) then a is a local minimum. At the right enpoint b, if f (t n ) < 0 (so f is ecreasing as b is approache from the left) then b is a local minimum. At the right enpoint b, if f (t n ) > (so f is increasing as b is approache from the left) then b is a local maximum. The possibly bewilering list of possibilities really shouln t be bewilering after you get use to them. We are alreay acquainte with evaluation of f at auxiliary points between critical points in orer to see whether the function is increasing or ecreasing, an now we re just applying that information to see whether the graph peaks, troughs, or oes neither aroun each critical point an enpoints. That is, the geometric meaning of the erivative s being positive or negative is easily translate into conclusions about local maxima or minima. Fin all the local (=relative) minima an maxima of the function f(x) = 2x 3 9x 2 + on the interval [ 2, 2]: To fin critical points, solve f (x) = 0: this is 6x 2 8x = 0 or x(x 3) = 0, so there are two critical points, 0 an 3. Since 3 is not in the interval we care about, we rop it from our list. Aing the enpoints to the list, we have 2 < 0 < 2 23

24 as our orere list of special points. Let s use auxiliary points,. At the erivative is f ( ) = 24 > 0, so the function is increasing there. At + the erivative is f () = 2 < 0, so the function is ecreasing. Thus, since it is increasing to the left an ecreasing to the right of 0, it must be that 0 is a local maximum. Since f is increasing to the right of the left enpoint 2, that left enpoint must give a local minimum. Since it is ecreasing to the left of the right enpoint +2, the right enpoint must be a local minimum. Notice that although the processes of fining absolute maxima an minima an local maxima an minima have a lot in common, they have essential ifferences. In particular, the only relations between them are that critical points an enpoints (an points of iscontinuity, etc.) play a big role in both, an that the absolute maximum is certainly a local maximum, an likewise the absolute minimum is certainly a local minimum. For example, just plugging critical points into the function oes not reliably inicate which points are local maxima an minima. An, on the other han, knowing which of the critical points are local maxima an minima generally is only a small step towar figuring out which are absolute: values still have to be plugge into the funciton! So on t confuse the two proceures! (By the way: while it s fairly easy to make up story-problems where the issue is to fin the maximum or minimum value of some function on some interval, it s harer to think of a simple application of local maxima or minima). #8.52 Fin all the local (=relative) minima an maxima of the function f(x) = (x + ) 3 3(x + ) on the interval [ 2, ]. #8.53 Fin the local (=relative) minima an maxima on the interval [ 3, 2] of the function f(x) = (x + ) 3 3(x + ). #8.54 Fin the local (relative) minima an maxima of the function f(x) = 2x + x 3 on the interval [ 3, 3]. #8.55 Fin the local (relative) minima an maxima of the function f(x) = 3x 4 8x 3 + 6x on the interval [ 3, 3]. 9. An algebra trick The algebra trick here goes back at least 350 years. This is worth looking at if only as an aitional review of algebra, but is actually of consierable value in a variety of han computations as well. The algebraic ientity we use here starts with a prouct of factors each of which may occur with a fractional or negative exponent. For example, with 3 such factors: f(x) = (x a) k (x b) l (x c) m The erivative can be compute by using the prouct rule twice: f (x) = = k(x a) k (x b) l (x c) m + (x a) k l(x b) l (x c) m + (x a) k (x b) l m(x c) m Now all three summans here have a common factor of (x a) k (x b) l (x c) m 24