Unit # - Families of Functions, Taylor Polynomials, l Hopital s Rule Some problems an solutions selecte or aapte from Hughes-Hallett Calculus. Critical Points. Consier the function f) = 54 +. b) a) Fin all intervals where the function is increasing. b) Fin all intervals where the function is ecreasing. c) Finally, fin all critical points in the graph of f). The function is increasing an ecreasing where f ) > 0 an f ) < 0; critical points are where f ) = 0. For this function, f ) = 54 = 9) = ) + ). f ) = ) 4) 4 ) 4) = 4 4) f ) = 4 4) 4) 4 ) 4)) 4) ) cancel one 4) everywhere, = 4 4) 4) 4 )4) 4) Epan: = 45 4 9) 4 5 48 ) 4) a) f is increasing on, ] [, ). b) f is ecreasing on [, ]. c) Critical points are = an =. Simplify: = 8 + 9 4) Factor: f ) = 8 + ) 4). Consier the function f) = 4 efine on the interval [ 9, 9]. a) At what coorinates oes the function f) have vertical asymptotes? b) On what intervals) is f) concave up? c) At what coorinate is there an inflection point for this function? The + factor is always positive. At = 0, the sign of the numerator can change, an there are asymptotes at = an =. Therefore, we look at the intervals 9, ),, 0), 0, ), an, 9). The 9 s ue to the eplicit omain given.) On the intervals, 0] an, 9], the function is concave up because f > 0) there; ite at 9 by the omain [ 9, 9] given in the question). c) At = 0 there is a point of inflection, because the function changes from concave up to concave own. = an = cannot be points of inflection because they aren t points on the graph, but vertical asymptotes. a) The enominator of f) is 4 = )+). There is no cancellation with the numerator when the enominator is zero, so = an are vertical asymptotes..
5. Consier the function f) = 8 ) /. For this function there are two important intervals:, A) an A, ) where A is a critical value. a) Fin A. 4. For the function f given above, etermine whether the following conitions are true or false. a) True. b) True. c) True. ) True. e) True. a) f ) = 0. b) f ) oes not eist. c) f ) < 0 if < <. ) f ) > 0 if < or <. e) f ) < 0 if. For the function f given above, etermine whether the following conitions are true. a) True. a) f ) < 0 if 0 < <. b) f ) > 0 if >. c) f ) < 0 if 0 <. ) f ) > 0 if < < 4. e) f ) < 0 if > 4. b) False. c) True. ) True. e) True. f) There are two inflection points on f). Ientify their approimate coorinates. f) =, 4. b) For each of the following intervals, tell whether f) is increasing or ecreasing:, A), an A, ) c) For each of the following intervals, tell whether f) is concave up or concave own:, A), an A, ). f ) = ) / a) A = is the only critical point of this function. f ) is unefine there, but f) is efine. b) Since f ) is negative for <, f) is ecreasing on, ). Similarly, f) is increasing on, ). c) Analyzing concavity requires f ). a) f ) = ) ) 4/ The enominator, with the even power 4/), will always be positive. Therefore, f ) is negative everywhere, ecept where the erivative isn t efine =. Therefore, f) is concave own on, ) an, ).. Consier the function f) = e 9. f) has two inflection points, at = C an = D, with C < D. a) Fin C an D. Determine whether f) is concave up or concave own on the following intervals. b), C) c) C, D) ) D, ) f) = e 9 so f ) = e 9 + 9 e 9 an f ) = e 9 + 8e 9 ) + 8e 9 + 8 e 9 ) = e 9 + + 8 ) Setting f ) = 0 to fin transitions between concave up an concave own, we use the quaratic formula an fin C = 0.79 an 9
7. D = + 0.05. 9 Net, we will verify that these points actually reflect a change in concavity by fining the concavity on the intervals aroun them. b) The earlier roots inicate that we can factor f ): f ) = e 9 C) D) Base on the sign of f, the function will be concave up on, C). Check sample values of f ) on the interval.) c) Concave own on C, D). ) Concave up on D, ). b) f 0) = 0, an the first or secon erivative test will inicate that this is a local maimum. c) f ) = e + e. The points of inflection, where f ) = 0 an f changes sign, will be at = ± / ±0.408 9. Consier the function f) = /5 0). This function has two critical points = A an = B. a) Fin A an B. b) For each of the following intervals, tell whether f) is increasing or ecreasing:, A], [A, B], an [B, ). c) Classify the critical points A an B as relative min, ma or neither. a) f ) = 5 4/5 0) + /5 Ientify the graphs A blue), Bre) an C green) as the graphs of a function f) an its erivatives f ) an f ). Hint: Remember that f ) is itself a function, an we can fin the erivative of the function f ), which is calle the secon erivative of the function f) an enote by f ). f) is given by the graph A. f ) is given by the graph C. Note that A is always increasing, so C is always positive. f ) is given by the graph B. Note the negative slopes in C, which lea to negative values in B f ) is unefine at = 0, but f0) is efine, so = 0 is a critical point. To fin where f ) = 0, we set Mult through by 4/5 : 0 = 5 4/5 0) + /5 0 = 0) + 5 0 = 5 = 5 5 = = 5.7 8. Let f) = e. a) Fin the coorinates) where f) has a local minimum if any). b) Fin the coorinates) where f) has a local maimum if any). c) Fin the coorinates) where f) has a point of inflection. a) f ) = e, so has a single critical point at = 0. At that point, the function is at a local maimum use either the first or secon erivative tests). Therefore, there are no local minima of f). So the two critical points are at A = 0 an B = 5/. b) Base on test points in f ), we can fin that f) is ecreasing on, 0], an on [0, 5/]. increasing on [5/, ) c) Since the function is ecreasing on both sies of = 0, = 0 is neither a local min nor local ma. At = 5/, the transition is from ecreasing to increasing, so the critical point is a local minimum.
0. Consier the function f) = 8 5)e. a) Fin the critical points of f), the intervals on which it is increasing an ecreasing, an then classify the critical points as local maima or minima. b) Fin where f is concave up, concave own, an has inflection points. c) Fin any horizontal an vertical asymptotes of f. ) Sketch a graph of the function f, using the information from the previous parts, an by aing the an y intercepts.. Consier the function f) = 4 4 +. a) Fin the critical points of f), the intervals on which it is increasing an ecreasing, an then classify the critical points as local maima or minima. b) Fin where f is concave up, concave own, an has inflection points. c) Fin any horizontal an vertical asymptotes of f. ) Sketch a graph of the function f, using the information from the previous parts, an by aing the an y intercepts. a) The only critical point is = 0.. The function is increasing on the interval, 0.], an ecreasing on [0., ). Since the function increases up to = 0., an ecreases afterwars, = 0. is a local maimum of the function first erivative test). b) The function is concave up on, 0.4], an concave own on [ 0.4, ). There is an inflection point where it switches between these two intervals, at = 0.4. c) There is a horizontal asymptote at y = 0, because f) = 0. The function is continuous for all values, so there are no vertical asymptotes. a) f ) =. The function has no critical + ) points; the point at = where the erivative is unefine is not a point on the graph because f ) is also unefine, so can t be a critical point. The function is increasing on the intervals, ), ). Note that we can t pool this all into one interval because there is a jump in the function across the vertical asymptote at =. There are no local etrema of the function, because there are no critical points. b) f ) = 4. The function is concave up on + ), ), an concave own on, ). The transition occurs at the vertical asymptote, so there are no inflection points. c) There is a horizontal asymptote at y = 4, because 4 4 + = 4 an 4 4 + = 4 also. There is a vertical asymptote at =. ) 4
. The graph of the function f is Fill in the function analysis table below by inicating whether the values on each interval are positive +), negative -), or change sign both). 0 < < < < < < 4 4 < < 0 f f f 0 < < < < < < 4 4 < < 0 f + + both - f + - - - f - - + -. Consier the graph of the function f) = + ) ) a) What is the y-intercept of the graph? b) What are the -intercepts of the graph, if any? c) At what values of oes the graph have a vertical asymptote, if any? ) At what values of y oes the graph have a horizontal asymptote, if any? e) For what values of is f ecreasing? f) For what values of is f concave up? a) y intercept at = 0, y = / ) =. b) Setting y = 0, we fin =. c) There is a vertical asymptote at =, because f) is unefine. ) The graph has no horizontal asymptotes, because both f) an f) are unefine go off to infinity or negative infinity). e) f ) = 0 at = 5, so the intervals of interest are, ) up to the vertical asymptote),, 5) between the asymptote an the critical point), an 5, ). By checking the sign of f ), we fin that the function is only ecreasing, or f ) < 0, on, 5]. f) f ) = 0 at =, so the intervals of interest for concavity are, ) up to the possible inflection point),, ) between the inflection point an the asymptote), an, ). By stuying the sign of f ) on each of those intervals, we fin that function is concave up on, ), ). 4. Suppose that f) = 8 ln), > 0. a) Fin all critical values of f. b) Fin the intervals on which f) is increasing, an on which it is ecreasing. c) Ientify all the critical points of f) as either local min, local ma, or neither. ) Fin the intervals on which f) is concave up, an on which it is concave own. e) Fin all inflection points of f. f) Fin the y location of the critical points), then use all of the preceing information to sketch a graph of f). a) f ) = 8 / is efine for all > 0 omain of f)). The only point where f ) = 0 is = 0.75. b) The function is increasing on [0.75, ), an is ecreasing on 0, 0.75]. c) The only critical point, at = 0.75, is a local minimum by the first erivative test). ) f ) = /. This is always positive on the omain > 0, so the graph is concave up on 0, ). There are no intervals where it is concave own. e) An inflection point where the concavity on the graph changes from positive to negative, or vice versa. Since this function is concave up everywhere, there are no inflection points. f) At = 0.75, y = 80.75) ln0.75) 7.7. 5
. Let f) = e. 5. Consier the function a) Compute f ). f) = e 4 + e b) Fin the intervals on which f) is increasing, an on which it is ecreasing. c) Ientify the location of any local maima or minima. ) Compute f ). e) Fin the intervals on which f) is concave up, an on which it is concave own. f) Ientify the location of any inflection points of f). f ) = 4e 4 + e ) Since all the terms in f ) are positive, f) is always increasing. Since f ) > 0 everywhere, there are no critical points, so the function has no local maima or minima. f ) = 4e 4 e ) 4+e )). Note that this will equal zero when 4 e = 0, or when = ln4). f ) will be positive when < ln4), so the function will be concave up on, ln4)). f ) will be negative when > ln4), so the function will be concave own on ln4), ). There is a single point of inflection at = ln4). a) Fin the intervals) on which f is increasing. b) Fin the intervals) on which f is ecreasing. c) Fin any local maima of f. Provie the an y coorinate of the points. ) Fin any local minima of f. Provie the an y coorinate of the points. e) Fin the intervals) on which f is concave upwar. f) Fin the intervals) on which f is concave ownwar. f ) = e e = e ). This is only efine for 0, as is the original function. To analyze the signs of f, it is helpful to multiply through by /, or think of it as factoring out / : f ) = e ) a) e an are always positive. Therefore, the erivative is positive when the factor is positive, which is for the interval 0, 0.5). This means f) is increasing on the interval [0, 0.5]. b) When is negative, f ) is negative, so the function is ecreasing on [0.5, ) c) The critical point is at, y) = 0.5, 0.5e 0.5 ) 0.5, 0.858). Since the function increases before this value an ecreases afterwars, this is a local maimum. ) The function only has one critical point, an it is a local maimum; there are no local minima. e) To fin f ), it is easier to start with a ifferent form of f ), where the roots are written out as powers: f ) = e ) = e ) = e / /) Differentiating, f ) = e / /) ) + e / / = e ) = e / / )
Values where f ) = 0 or unefine will be when = 0 or = 0.5 ±. Only the + option gives a positive value, or = 0.5 +.07. Checking the sign of f ) on each interval, the concavity will be positive on.07, ), so the function is concave up on this interval. f) On the interval 0,.07), the function will be concave own. 7. The function f) = 9 + has one local minimum an one local maimum. Fin their, y) locations. 8. The following shows graphs of three functions, A in black), B in blue), an C in green). If these are the graphs of three functions f, f, an f, ientify which is which. f ) = 9 Setting f = 0, 0 = 9 = 9 = ± 9 ±0.474 These are the two critical points of f). Looking at the sign of f ) = 9 / on ifferent intervals, we fin that The local maimum is at 0.474, 8.485). The local minimum is at 0.474, 8.485). To etermine which graphe function is the the erivative of another, we use the fact that the erivative is zero where the original function has a maimum or minimum. We note that C has zeros where B has a maimum or minimum, so it coul be the erivative, so we can have C = B. In aition, B has zeros where A has a maimum or minimum, so that it is possible that B = A. To confirm this, we compare intervals where B is increasing an ecreasing an verify that these are where C is positive an negative, respectively, an similarly compare A an B. This gives us that f = A, f = B, an f = C. Families of Functions 9. Consier the function f) = a + sin4), where a is a positive constant. For what range of a values is this function always increasing? For the function to be always increasing, we note that we nee the erivative to be larger than zero or at worst equal to zero. If the erivative equals zero at separate points, but is always positive otherwise, then the function is always increasing; think of y = as an eample.) The erivative will be f ) = a + 4 cos4). To guarantee that f ) will be positive, we nee to know how low it can go. Base on our knowlege of the cos graph, an recognizing that f ) is just a shifte an scale version of it, we fin that the smallest value of f ) = a + 4 cos4) will be a 4, so a 4 will be require for f ) to be greater than or equal to zero everywhere. 0. A function of the form y = be a) has a local maimum at, 4). What are the values of a an b? We have two pieces of information: y = 4 when =, or y) = 4, an y = 0 when =, or y ) = 0. The y) = 4 fact gives us the relation 4 = be a) which isn t immeiately helpful can t solve for a or b). The y ) = 0 fact gives us y = be a) ) a) Subbing in = an y = 0 0 = be a) ) a) 7
Out of all these factors, b can t equal zero or our function woul become just y = 0, an have no critical points at all). The eponential can t equal zero either, so we must have a equal 0, or a =. If we have selecte a =, then we can go back to our first equation, using =, y = 4 an a =, an solve for b: so b = 4. y = be a) 4 = be ) = be 0 = b. For what values of the numbers a an b oes the function f) = ae b have the maimum value f) =? f ) = ae b + abe b ) = ae b + b ) For there to be a critical point at all, f ) = 0, we must have b being negative otherwise all the factors are always positive). For there to be a ma at =, we nee a critical point at =, so f ) = 0 0 = ae b) + b )) 0 = + 8b b = /8 Since a scales the function vertically, we can select it so that f) = : f) = = a)e /8 = ae / e / = a So the values require for f) = to be a critical point of f) are a = e / an b = /8. By looking at f ) with those a an b values, we can confirm by the first erivative test that this point is a local maimum, as esire.. Fin the formula for a function of the form y = A sinb) + C with a maimum at.5, 7), a minimum at 4.5, ), an no critical points between these two points. Since the maimum is y = 7 an the minimum is y =, the amplitue is A = 7 + )/ = 4. Between the maimum an the minimum, the -value changes by. There is half a perio between a maimum an the net minimum, so the perio is. Thus π B =, so. The mi-line is y = C = 7 )/ =, so the B = π function is y = 4 sin π ) +.. Fin a formula for a curve of the form y = e a) /b for b > 0 with a local maimum at = 5 an points of inflection at = 7 an =. The maimum of y = e a) /b occurs at = a. This is because the eponent a) /b is zero when = a an negative for all other -values. The same result can be obtaine by taking erivatives.) Thus we know that a = 5. Points of inflection occur where y/ changes sign, that is, where y/ = 0. Differentiating gives y y = b e +5) /b + = b e +5) /b + 5) = e +5) /b, so b 4 + 5) b e +5) /b + b + 5) ). Since e +5) /b is never zero, y/ = 0 where + b + 5) = 0. We know y/ = 0 at =, so substituting = gives + b + 5) = 0. Solving for b gives b = 8. Since a = 5, the function is y = e +5) /8. You can check that at = 5, we have y = 8 e 0 + 0) < 0 so the point = 5 oes inee give a maimum. 4. Let f) = e k, for k > 0. a) f) has a local minimum. Fin the an y coorinates of this minimum. b) Fin the value of k for which this y- coorinate is largest. c) How o you know that this value of k maimizes the y-coorinate? Fin y/k to use the secon-erivative test. Note: this is not the traitional use of the secon erivative test! The erivative you get shoul be negative for all positive values of k, an confirm that you agree that this means that your value of k maimizes the y-coorinate of the minimum. Graphing the function suggests that each graph ecreases to a local minimum an then increases sharply. The local minimum appears to move to the right as k increases. It appears to move up until k = /, an then to move back own. 8
f ) = e k = 0 when = 0 = lnk/). Since f ) < 0 for < 0 an f ) > 0 for > 0, f is ecreasing to the left of = 0 an increasing to the right, so f reaches a local minimum at = lnk/). The minimum value of f is f lnk/) ) = e lnk/a)/ klnk/a)/) = k k ln k ). Since we want to maimize this epression, we can imagine a function gk) = k ln k ). To maimize this function we simply take its erivative an fin the critical points. Differentiating, we obtain g k) = lnk ) + ) = lnk ). Thus g k) = 0 when k =, g k) > 0 for k <, an g k) < 0 for k >. Thus k = is a local maimum for gk). That is, the largest global minimum for f occurs when k =. We can also verify that this is the maimum by noting that g k) = /k < 0, so that the function is always concave own. Thus, because this is the only critical point, we are guarantee that it is a global minimum. 5. Consier f) = a e b ) for a > 0 an b > 0. a) Fin f ). b) Base on your epression for f ), is f) increasing or ecreasing for all values of? c) Fin f ). ) Base on your epression for f ), is f) concave up or concave own for all? a) Noting that f) = a ae b, we fin its erivative using the chain rule: f ) = abe b. b) Eponentials are always positive, so that, given that a > 0 an b > 0, we must have f ) > 0 an we know that f) must be increasing. c) f ) = ab e b < 0, so the function must be concave own.. Fin constants a an b in the function f) = ae b such that f 7 ) = an the function has a local maimum at = 7. Using the prouct rule on the function f) = ae b, we have f ) = ae b + abe b = ae b + b). We want f 7 ) =, an since this is to be a maimum, we require f 7 ) = 0. These conitions give an f/7) = a 7 eb/7 = f /7) = ae b/7 + b/7) = 0. Since ae b/7 is non-zero, we can ivie both sies of the secon equation by ae b/7 to obtain + b 7 = 0. This implies b = 7. Plugging b = 7 into the first equation gives us a 7 )e =, or a = 7e. How o we know we have a maimum at = 7 an not a minimum? Since f ) = ae b + b) = )e 7 7), an 7e)e 7 is always positive, it follows that f ) > 0 when < 7 an f ) < 0 when > 7. Since f is positive to the left of = 7 an negative to the right of = 7, f 7 ) is a local maimum. 7. For some positive constant C, a patient s temperature change, T, ue to a ose, D, of a rug is given by T = C ) D D. We have an so a) What osage maimizes the temperature change? b) The sensitivity of the boy to the rug is efine as T/D. What osage maimizes sensitivity? T D) = C D ) D = CD D, T D = CD D = DC D). Since, by this formula, T/D is zero when D = 0 or D = C, negative when D > C, an positive when D < C, we have by the first erivative test) that the temperature change is maimize when D = C. The sensitivity is T/D = CD D ; its erivative is T/D = C D, which is zero if D = C/, negative if D > C/, an positive if D < C/. Thus by the first erivative test the sensitivity is maimize at D = C/. 9
Taylor Polynomials For reference, the general formula for the Taylor polynomial centere at = a is: P ) = fa) + f a)! a) + f a)! a) +... + f n a) a) n n! where n! means n factorial, or n n ) n ).... E.g. 4! = 4 = 4. 8. Suppose g is a function which has continuous erivatives, an that g5) =, g 5) =, g 5) =, g 5) = 5. We have a) What is the Taylor polynomial of egree for g near 5? b) What is the Taylor polynomial of egree for g near 5? Note: you will nee to look up this formula, but it follows the same pattern as the lower-egree Taylor polynomials.) c) Use the two polynomials that you foun in parts a) an b) to approimate g5.). g) = g5)+g 5) 5)+ g 5)! Substituting gives 5) + g 5) 5) +! g) = 5) +! 5) + 5! 5) + a) The egree Taylor polynomial, P ), is obtaine by truncating after the 5) term: P ) = + ) 5) +! 5). b) The egree Taylor polynomial, P ), is obtaine by truncating after the 5) term: P ) = + ) 5)+! 5) + 5) 5)! c) Substitute = 5. into the Taylor polynomial of egree : P 5.) = + )5. 5)+! 5. 5) =.95. From the Taylor polynomial of egree, we obtain P 5.) = + )5. 5) + 5. 5)! +! 5)5. 5) =.958. 9. Fin the Taylor polynomial of egree n = 4 for near the point a = π 4 for the function cos4). Note: you will nee to look up the general formula for the higher-orer terms, but they follow the same pattern as the lower-orer terms. Let f) = cos4). Then f π 4 ) =, an f ) = 4 sin4) f π 4 ) = 0 f ) = cos4) f π 4 ) = f ) = 4 sin4) f π 4 ) = 0 f ) = 5 cos4) f π 4 ) = 5 So, P 4 ) = + π ) 5 + π ) 4. 4 4 4 0. a) Fin the fifth orer Taylor polynomial for sin) near 0. b) Use this to fin the it 0 sin) a) The fifth orer Taylor polynomial is b) The it is P 5 ) = + 0 5, sin) + 0 0 0 5 =.. Fin the Taylor polynomial of egree aroun the point = 4 of f) = 5 +. Let f) = 5 + = 5 + ) /. Then an f ) = 5 +, f ) = 45 + ), / f ) = 85 + ) 5/. The Taylor polynomial of egree three about = 4 is thus P ) = 5 4 + + 4) 5 4 + )! ) + 4) 45 4) / + )! ) + 4). 85 4) 5/ 0
. Calculate the Taylor polynomials P ) an P ) centere at = for f) = ln + ). Recall the general formula for the Taylor polynomial centere at = a: P ) = fa)+ f a)! a)+ f a)! f) = ln + ) f ) = + f ) = + ) f ) = + ) a) +...+ f n a) a) n n! So, f) = ln + ) an P ) = A+B )+C ) where A = f) = ln) = 0.947 B = f ) = ) = 0.5 C = f )! = 8) = 0.5 Thus, or, P ) = ln) + ) ) ) 8) P ) = 0.947 + 0.5 ) 0.5 ) Similarly, P ) = D + E ) + F ) + G ) where D = f) = ln) = 0.947 E = f ) = ) = 0.5 F = f )! = 8) = 0.5 G = f )! = 4) = 0.047 Thus, P ) = ln) + ) ) 8) ) + ) 4) or, P ) = P ) + 0.047 ). Calculate the Taylor polynomials P ) an P ) centere at = π for f) = sin). f) = sin) an P ) = A + B π ) + C π ) where A = f π ) = = 0.5 B = f π ) = 0.8 C = f π )! = 4 = 0.5 Thus, or P ) = + π ) 4 π ) P ) = 0.5 + 0.8 π ) 0.5 π ) Similarly, P ) = D+E π )+F π ) +G π ) where D = f π ) = = 0.5 E = f π ) = = 0.8054078449 F = f π )! = 4 = 0.5 G = f π )! = = 0.447579740 Thus, P ) = + π ) 4 π ) + π ) or P ) = 0.5+0.8 π ) 0.5 π ) 0.44 π ) 4. Calculate the Taylor polynomials P ) an P ) centere at = 7 for f) = +. f) = + f ) = + ) f ) = f ) = + ) + ) 4 an P ) = A + B 7) + C 7) where A = f7) = 8) = 0.5 B = f 7) = 4) 0.05 C = f 7)! = 5) = 0.0095 Thus, or, P ) = 8) 7) + 7) 4) 5) P ) = 0.5 0.05 7) + 0.0095 7) Similarly, P ) = D + E 7) + F 7) + G 7)
where D = f7) = 8) = 0.5 E = f 7) = 4) = 0.05 F = f 7)! = 5) = 0.0095 G = f 7)! = 409) = 0.00044 Thus, P ) = 8) 4) 7)+ 5) 7) 409) 7) or, P ) = P ) 0.00044 7) 5. Calculate the Taylor polynomials P ) an P ) centere at = for f) = e + e. f) = e + e f ) = e e f ) = e + 4e f ) = e 8e an P ) = A + B ) + C ) where A = f) = e + e 4 B = f ) = e + e 4) C = f ) Thus,! = e +4e 4 P ) = e +e 4 e + e 4) )+ e + 4e 4 ) Similarly, P ) = D + E ) + F ) + G ) where D = f) = e + e 4 E = f ) = e + e 4) F = f )! G = f )! Thus, P ) = = e + 4e 4 = e + 8e 4 = e + e 4 e + e 4) ) + e + 4e 4 ) e + 8e 4 ). Calculate the Taylor polynomials P ) an P ) centere at = π 4 for f) = tan). f) = tan) tan) = cos ) sin) tan) = cos ) tan) = So, in this case, P ) = + π 4 sin ) + ) cos 4 ) ) + π ) 4 an P ) = + π ) + π ) 8 + 4 4 π 4 ) 7. Fin the secon-egree Taylor polynomial for f) = + 8 about = 0. What o you notice about your polynomial? We note that f0) = 8; f ) = 4, so that f 0) = ; an f ) = 4, so that f 0) = 4. Thus P ) = 8 + 4! = 8 +. We notice that f) = P ) in this case, which makes sense because f) is a polynomial. 8. Consier the simple function a) f) = a) Fin the fourth egree Taylor polynomial for f), centere at a =. b) Does the function f) = equal its fourth egree Taylor polynomial P 4 ) centere at a =? Hint: Graph a few points for each of them. If it looks like they are equal, then show they are equal algebraically the algebra. So f) = f ) = ) = 7 f ) = f ) = ) = 7 f ) = f ) = f ) = f ) = 0 f ) = 0 f ) = ) = 8 P 4 ) = 7 + 7 + ) 8 + ) + + ) = 7 + 7 + ) 9 + ) + + )
b) If you plug in values to both f) = an P 4 ), you ll see you get the same y values back. This seems to inicate they are really the same function. You can confirm this by epaning the terms in P 4 ). -0.5in P 4 ) = 7 + 7 + ) 9 + ) + + ) = 7 + 7 + 8) 9 + + 9) + + 9 + 7 + 7) = 7 + 8 8 + 7) + 7 54 + 7) + 9 + 9 ) + = = f) 4. Suppose that P ) = a + b + c is the secon egree Taylor polynomial for the function f about = 0. What can you say about the signs of a, b, c if f has the graph given below? Note that the central lines are the an y aes. In other wors, when we buil an approimating polynomial to another equal or lower egree) polynomial, the two functions will en up being the same, even if they on t look like it initially. 9. a) Fin the egree Taylor polynomial P ) centere at a = of the function f) = + ) 5/4 b) Does the function f) = + ) 5/4 equal its secon egree Taylor polynomial P ) centere at a =? Hint: Graph a few points for each of them. If it looks like they are equal, then confirm they are equal with algebra. a) P ) = 04 5 ) + 45 048 ) b) No: the Taylor polynomial woul have non-zero terms for the cubic an higher-egree powers of, so we are only approimating with the secon egree polynomial. 40. a) Fin the Taylor polynomial of egree 4 for cos), for near 0. b) Approimate cos) with P 4 ) to simplify the ratio: cos) c) Using this, evaluate the it cos). 0 a) We can fin the Taylor polynomial of egree 4 by ifferentiating cos) an evaluating it at zero: f ) = sin), f ) = cos), f ) = sin), an f 4) ) = cos), so at = 0 we have f 0) = 0, f 0) =, f 0) = 0 an f 4) 0) =. We also have f0) =, so that the Taylor polynomial is P 4 ) =! + 4 4!. b) Using the Taylor polynomial, we have cos) =! 4!. c) Because the approimation is vali for small values of, this lets us calculate the it: ) cos) 0 0! = 0. 4! Since P ) is the secon egree Taylor polynomial for f) about = 0, P 0) = f0), which says a = f0). Since P ) =0 = f 0), b = f 0); an since P ) =0 = f 0), c = f 0). In other wors, a is the y-intercept of f), b is the slope of the tangent line to f) at = 0 an c tells us the concavity of f) near = 0. Thus a < 0; b > 0; an c > 0 4. The function f) is approimate near = 0 by the secon egree Taylor polynomial P ) = + 8. Give values the values of f0), f 0), an f 0). Using the fact that f) P ) = f0) + f 0) + f 0)! an ientifying coefficients with those given for P ), we obtain: f0) = the constant term, which equals, so f0) =. f 0) = the coefficient of, which equals, so f f 0) =. 0)! = the coefficient of, which equals 8, so f 0) =. 4. Fin the secon-egree Taylor polynomial P ) for the function f) = 5 + at the number =. P ) = 4 + /4) ) + 5/)4)) ) 44. Fin the egree Taylor polynomial approimation to the function f) = lnsec)) about the point a = 0. P ) = 45. Fin P ), the Taylor polynomial of egree, for the function f) = arctan5) at a = 0.
At the point = 0, f0) = 0 f 0) = 5 f) = arctan5) f ) = f ) = 5 + 5 50 + 5 ) f 0) = 0 So now we can piece together the power series: P ) = 00 0! + 5! + 0! = 0 + 5 + 0 = 5 Note that, even though we trie to construct a quaratic, the best quaratic is a simpler linear function, because arctan5) has f 0) = 0. l Hôpital s Rule 4. Evaluate the it h 0 sin 4h h This it gives an ineterminate 0 form, so we can 0 use l Hôpital s rule. sin4h) = h 0 h h 0 h sin4h) h h = h 0 4 cos4h) = 4 47. Fin the it. Use l Hospital s Rule if appropriate. 7e This is an ineterminate form, /, so we can use l Hôpital s rule. 7e = 7e = 7e = The new it is still of ineterminate form, so we use l Hôpital s rule again. 7e = an again: = 7e = 7e = 7e = ) 7e = This is a eterminate form, so the the it is infinity, or unefine. 48. Fin the following its, using l Hôpital s rule if appropriate a) b) arctan 5 ) 8 8 ln) 0 + a) This question is a reminer that you nee to check whether l Hôpital s rule actually applies first. If you review graph of arctan), you will see that it approaches the value of π as. Thus the it is arctan 5 ) 8 π/ 0 There is no nee to apply l Hôpital s rule here. b) This is an ineterminate form 0. To use l Hôpital s rule though, we nee to rewrite it as a ratio/fraction. 0 + 8 ln) ln) = = ) 0 + or ineterminate /8 = 0 + ln) /8 = 0 + / /8) 9/8 89/8 re-organizing: = 0 + = 8/8 = 0 0 + The overall value of the it is zero. 4
49. Evaluate the it 0 tan 5 4 The it is 0/0, an ineterminate form, so we can use l Hôpital s rule. tan 5 = 0 4 0 tan 5 4 = 0 5 sec 5) 4 So the original it s value is 9 b) This is an ineterminate it, 4 can use l Hôpital s rule. 4 h = h 4 ) ) = 0 0, so we = h 4 ln4) ln)) Since sec0) = / cos0) =, tan 5 = 5 0 4 4 50. Evaluate the it z 0 sinz/) sin z We note that this it gives 0/0, an ineterminate form, so we can use l Hôpital s rule. sinz/) = z 0 sin z z 0 z sinz/) z sin z = cosz/) z 0 cos z = 5. Compute the following its using l Hôpital s rule if appropriate. cos4) a) 0 cos) 4 b) a) This is an ineterminate it, use l Hôpital s rule. cos4) h 0 cos) = h 0 = 0 0 cos4) cos)) = h 0 4 sin4) sin)), so we can Taking the it now still gives an ineterminate form 0/0), so we apply l Hôpital s rule again. h 0 4 sin4) sin)) = h 0 4 sin4) sin)) = h 0 cos4) 9 cos)) = 9.778 Taking the it now we get a eterminate form: 4 ln4) ln).47 5. Use L Hôpital s Rule possibly more than once) to evaluate the following it cot0) ) 0 0 Note that the ouble-angle formula may be helpful: sinθ) cosθ) = sinθ) The it is of an ineterminate form,, but it is not a ratio fraction) which is neee for l Hôpital s rule. cot0) 0 0 ) = 0cot0) ) 0 0 L Hôpital s Rule applies because the functions in the numerator an enominator are ifferentiable, an the it is ineterminate of type 0/0. Hence, we can replace the functions by their erivatives, to obtain: 0cot0) 0 0 = 0 0cot0) ) 0 sin 0) 0 cot0) 0 = 0 0 sin0) cos0) 0 = 0 sin 0) Here, to simplify, it will really help if we can simplify the sine-times-cosine, so later erivatives become simpler. We can o that using the ouble-angle formula mentione in the question, using 0 as θ: 0 sinθ) cosθ) {}}{ sin0) cos0) 0 sin = 0) 0 sinθ) {}}{ sin0) 0 sin. 0) 5
This it is again an ineterminate form of type 0/0, so we apply L Hôpital s Rule again: sin0) 0 0 sin 0) 0 cos0) 0 = 0 0 sin0) cos0) = 0 sin0) This it is again an ineterminate of type 0/0, so we apply L Hôpital s Rule again: cos0) 0 sin0) = 0 sin0) 0 0 cos0) = 0 0 = 0. Finally!) 5. Apply L Hôpital s Rule to evaluate the following it. It may be necessary to apply it more than once. e e e ln) e ln) = e =.788. 54. Use L Hôpital s Rule possibly more than once) to evaluate the following it 5 sint) lnt)) t 0 5 sint) lnt)) = 5 lnt) t 0 t 0 csct) Applying L Hôpital s Rule, we get: 5 lnt) t 0 csct) = t 0 5 t sin t) cost) = 5 sin t) t 0 t cost) This it is again ineterminate of type 0/0, so we apply L Hôpital s Rule again: sint) cost) ) = 5 t 0 cost) t sint) = 5 0 0 = 0 55. Fin the it 0 4 4 ) sin) To get this it into a form to which l Hôpital s rule applies, we combine fractions. Then we have 4 sin) 4 0 sin) 4 cos) 4 = 0 sin) + cos) 4 sin) = 0 cos) sin) = 0 5. Fin the it. Use l Hospital s Rule if appropriate. e/ The it is an ineterminate form, ), but we nee to write it as a ratio before we can use l Hôpital s rule. e/ e/ = = 00 ) / ineterminate = e/ / = / )e / / = e/ = e 0 = The final value of the it is. 57. Fin the it. Use l Hospital s Rule if appropriate. 7 7 ln Rewrite as 7 7/) ln). / You have to be careful here: it is not clear that the it as a whole is actually ineterminate now in fact, it won t be as, we shall see). Insie this new form, the it has an ineterminate ln) it in its numerator though: = ). After using l Hôpital s rule, 0. ln) Going back to the original it then, =0 {}}{ 7 7 /) ln) / = / = 7 / = 7 = 58. Fin the it. Use l Hospital s Rule if appropriate. 7 tan4/) Rewrite as 7 tan4/). After using l Hôpital s rule, the / it will be 8. =