Mat - Final Exam August 3 rd, Name: Answer Key No calculators. Sow your work!. points) All answers sould eiter be,, a finite) real number, or DNE does not exist ). a) Use te grap of te function to evaluate te limits. Ten assuming te function is continous everywere outside of te region graped, determine were te function is continuous. lim fx) = 5 Te limit from te left is clearly 5. x lim fx) = 5 Te limit from te rigt is 5. x + lim fx) = DNE Te limit does not exist because te left and x rigt limits don t matc. lim fx) = 5 Notice tat tis does not matc f) =. x fx) is continuous except at x = and x =, so f is continuous on te interval, ), ), ). b) Use te limit definition of te derivative to find te derivative of fx) = x + 3x +. [Do not use L Hopital s rule wile evaluating tis limit.] f fx + ) fx) x + ) + 3x + ) + ) x + 3x + ) x) x + x + + 3x + 3 + x 3x x + + 3 x + + 3 = x + 3 x + + 3 c) lim x cosx) x = Notice tat cosx) cos) = as x. On te oter and /x blows up to. As x gets very small x gets very small), /x gets very large. Essentially + = +. d) x 3 + 5x + 7x + 3 lim x x + x + L H = / x 3x + x + 7 x + L H = / x 6x + = 6 ) + = e) lim x x e x x x e x L H == / x x e x L H == / x e x = For te last step, essentially =.. 9 points) Let F x) = x ft) dt were te grap of y = fx) is given below. Note: Te lines continue forever in bot directions and te grap between and 6 is consists of alf circles. a) Find F ) = b) Find F ) = c) Find F ) = fx) dx =. fx) dx = 6 fx) dx + 6 fx) dx = π + 4 4 = 8 π. fx) dx = d) Were is F x) increasing? fx) dx = 4 π = π. F is increasing wen F x) = fx) is positive. Looking at te grap we get tat F x) as a positive derivative i.e. it is increasing) wen x is between and and also after 6., ) 6, )
e) Locate all of te critical points of F x). Label eac as a relative minimum, relative maximum, bot, or neiter. Critical points occur were te derivative eiter does not exist wic isn t te case ere since F x) = fx) is defined everywere) or te derivative is. We want to solve F x) = fx) =. Tese are precisely te x-intercepts of fx). Tus x =,, 6. Next, minimums occur wen we switc from decreasing to increasing i.e. F x) = fx) switces from negative to positive) tis appens wen x = and 6. Maximums occur wen we switc from increasing to decreasing i.e. F x) = fx) switces from positive to negative) tis appens wen x =. f) Were is F x) concave down? x =, 6 are local minimums and x = is a local maximum. F x) is concave down wen F x) = f x) is negative. f x) is negative wen fx) is decreasing. Looking at te grap, tis occurs wen < x < 4., 4) g) Locate all of te inflection points of F x). Inflection points occur wen concavity switces. From te last part we ave tat F is concave down on, 4). Similarly we ave tat F is concave up on, ) 4, ) tis is were f is increasing). So we get canges in concavity at x = and 4. Answer F x) as inflection points at x = and x = 4. ) Locate all of te critical points of fx). Critical points occur wen a derivative is zero or does not exist. Keep in mind tat te derivative f x) is te slope of ta tangent to fx) at x. So te derivative does not exist wen f as no tangent or if f s tangent does not ave a well defined slope i.e. it is vertical). fx) as orizontal tangents i.e. derivative = zero) at x = and x = 4. Also, fx) as sarp corners were tangent lines are not well defined at x = and x = 6. Finally, fx) as a vertical tangent at x = so its derivative does not exist ere eiter). fx) as critical points at x =,,, 4, 6. 3. points) Suppose tat we know f) =, f ) =, f ) =, and f x) < for all x. a) Find te equation of te tangent to y = fx) at x =. Sketc tis tangent along wit some of y = fx). Te tangent passes troug te point, ) since f) = ) and as slope m = f ) =. Tus y = x ) so tat y = x + 3. Wen sketcing te grap of f we sould keep in mind tat f is concave down since f x) < ). Answer y = x + 3 b) Is it possible tat f3) =? Wy or wy not? Notice tat f3) must be less tan y = x + 3 = 3 + 3 = te value of te tangent found in x=3 x=3 No. part a)) since f is concave down te grap lies below its tangents). Tus f3) < so f3) = is impossible. c) Wat can be said about te grap of y = fx) at x =? We ave been told tat f ) = so x = is a critical point). We also know tat f ) < since f x) < everywere. Terefore, by te second derivative test, fx) as a local maximum at x =. fx) as a relative maximum at x =.
4. 4 points) Differential Equations. a) Is y = e x a solution of y + y = e x? Wy or wy not? Notice tat y = e x so y = e x and y = e x. Terefore, y + y = e x + e x = e x e x te equation is not satisfied). No. y = e x is not a solution of y + y = e x. [Note: On te oter and, y = e x is a solution of te equation y + y = e x.] b) Verify tat y = t + Ce t is a solution of y + y = t +. Ten determine C given te initial condition y) = 3. Notice tat y = Ce t so y + y = Ce t ) + t + Ce t ) = Ce t + t + Ce t = + t so te equation is satisfied). Next, 3 = y) = + Ce = C = C. y = t + 3e t 5. points) Taylor s Revenge. a) Find te Taylor polynomial of fx) = lnx) of order centered at x =. We need to compute te first derivatives of fx) and evaluate f and its derivatives at x = to find te second order Taylor polynomial at x =. f x) = /x = x and f x) = x = /x. Plugging in x =, we get: f) = ln) =, f ) = / =, and f ) = / =. Finally, recall tat te formula for suc a Taylor polynomial is P x) = f) + f )x ) + f ) x )! P x) = + x ) +! x ) = x ) x ) b) Find te fift order Maclaurin polynomial of gx) = cosx). Recall tat a MacLaurin polynomial is just a Taylor polynomial centered at x =. We need to find te first 5 derivatives of gx) and ten plug in x =. Derivatives: g x) = sinx), g x) = cosx), g x) = sinx), g 4) x) = cosx), and g 5) x) = sinx). Evaluated at x = : g) = cos) =, g ) = sin) =, g ) = cos) =, g ) = sin) =, g 4) ) = cos) =, and g 5) ) = sin) =. P 5 x) = + x +! x + 3! x3 + 4! x4 + 5! x5 = x + 4 x4 6. points) For eac part: You sould i) Draw a picture, ii) Model te situation by writing down some equations, and iii) Solve te problem using Calculus. a) Find te points on te grap of y = 3 x wic are closest to te origin i.e.,)). We are seeking to minimize te distance to te origin. Te distance from a point x, y) to te origin is x ) + y ) = x + y. Using a common trick we ll instead minimize te square of te distance: x + y since te square root function is an increasing function tis won t effect our answer minimizing te distance is te same as minimizing te square distance). So we wis to minimize D = x + y. Our next problem is tat te above objective function involves variables x and y). We need to eliminate one of tese variables. At tis point we sould recall tat we are minimizing te distance between points on te parabola and te origin. To be a point x, y) on te parabola we must satisfy te equation y = 3 x. Terefore, D = x + y = x + 3 x ). We need to minimize D = x + 3 x ) = x + 9 6x + x 4 = x 4 5x + 9. To do tis we ll compute te derivative and look for critical points: D = 4x 3 x =. So tat x4x ) = tus x = or 4x = so tat x = 4 = 5. 5 Tus our critical points are located at x =, ±. Let s plug tese in and find our minimums). Wen x = : D = 4 5 )+9 = 9 and wen x = ± 5/: D = ± 5/) 4 5± 5/) +9 = 5 4 55 Since.75 < 9, we ave tat te minimum occurs at x = ± 5/ so y = 3 x = 3 ± 5/) = 3 5 =. Distance is minimized at x, y) = 5/, / +9 = 9 5 4 = 36 4 5 4 = 4 =.75. ) and 5/, / ). 3
b) Find te points on te grap of y = 3 x wic are closest to te origin if we restrict x to te interval [, ]. We repeat te previous part except tat we must enforce te conditions tat x. Notice tat x = ± 5/ fall outside tese bounds since 5/ > so 5/ > = ), so we sould only consider te critical point x = and te end points x = ±. Plugging tese in, we get: D) = 9 and D±) = ±) 4 5±) + 9 = 5 + 9 = 5. Since 5 < 9, we ave tat te distance is minimized at te end points. Quickly compute: y = 3 x = 3 ±) = 3 = and get... Distance is minimized at x, y) =, ) and, ). c) Two concentric circles are growing. Wen te inner circle s radius is 3 meters it is growing at a rate of meter per second. Wen te outer circle s radius is 4 meters it is growing at a rate of meter per second. At wat rate is te area of te annulus te ring bounded by tese circles) canging? Is tis area increasing or decreasing? Te area of te annulus is te difference between te area of te outer circle say wit radius R): π R and te area of te inner circle say wit radius r): π r. So te area of te annulus is A = πr πr. Next, A, R, and r are functions of time say t). We wis to know ow tis area is canging so we need to differentiate wit respect to time. For example: te derivative of Rt)) wit respect to t is Rt)) R t) by te cain rule). All togeter we get: A t) = πrt)r t) πrt)r t). At our current time, te outer radius is R = 4 meters, te inner radius is r = 3 meters, and tese radii are canging at rates of R = meter/second and r = meters/second. Terefore, A = π 4 π 3 = 8π π = 4π meters /second). Te area is decreasing at a rate of 4π meters per second. 7. 4 points) Approximating I = 4 x 3 + 8 dx. a) Find te left and rule approximation of I if we use n = 3 rectangles. [Dont worry about simplifying.] We are integrating over te interval [, 4]. Splitting tis interval into 3 pieces, we get x = 4 ) =. So our partition points are x =, x = + =, 3 x = + =, and x 3 = + = 4. Te left end points are x =,, and. I = 4 x 3 + 8 dx L 3 = x f ) + f) + f)) = ) 3 + 8) + 3 + 8) + 3 + 8) ) = + 8 + 6) = 48 b) Find te midpoint rule approximation of I if we use n = 3 rectangles. [Dont worry about simplifying.] Same setup as in te previous part except we need to use te midpoints. Te midpoint between and is, te midpoint between and is and te midpoint between and 4 is 3. If we couldn t just see tis, we could ave computed midpoints by adding x/ to eac left end point. For example: + x/ = + / = our first midpoint). I = 4 x 3 + 8 dx M 3 = x f ) + f) + f3)) = ) 3 + 8) + 3 + 8) + 3 3 + 8) ) = 7 + 9 + 35) = c) If we computed some rigt and rule approximation of I, would it be an underestimate, overestimate, or does it depend on te number of rectangles? Draw a picture to back up your claim. Notice tat fx) = x 3 + 8 is an increasing function for example, f x) = 3x ). Rigt and rule approximations always give overestimates for integrals of increasing functions. Overestimate 4
8. 54 points) Compute te derivative. a) y = 3 x + lnx) + arcsinx) + tanx) + 5x + π = x /3 + lnx) + arcsinx) + tanx) + 5x + π After rewriting 3 x as x /3, we just use basic formulas. Keep in mind π is just a number i.e. a constant) so its derivative is. y = 3 x /3 + x + x + sec x) + 5 x 5 x 3 + 5)e 6x+ ) b) y = ln x + 5) [You must use te laws of logs to get credit.] y = lnx 5 x 3 + 5)e 6x+ ) lnx + 5) ) = lnx 5 ) + lnx 3 + 5) + lne 6x+ ) lnx + 5) = 5 lnx) + lnx 3 + 5) 6x + lnx + 5) Now use standard formulas as well as te cain rule. You need te cain rule for lnx 3 + 5) and lnx + 5).) y = 5 x + 3x x 3 + 5 6 x + 5 c) y = x e 3x sinx) Here we need te product rule twice or te extended product rule once). y = xe 3x sinx)) + x e 3x sinx)) = xe 3x sinx) + x 3e 3x sinx) + e 3x cosx)) d) y = x + cosx) + e x + 3x 9 + x y = xe 3x sinx) 3x e 3x sinx) + x e 3x cosx) We simply need to apply te quotient rule. [Don t boter simplifying tis one.] y = x sinx) )e x + 3x 9 + x) x + cosx) + ) e x + 7x 8 + ) e x + 3x 9 + x) e) y = e sin5x+) Here we need to apply te cain rule twice. y = e sin5x+) cos5x + ) 5 f) y = x +3 sint) t dt Here we need to apply te fundamental teorem of calculus plus te cain rule. Keep in mind tat dy du = sinu) u u sint) y = dt. Ten use dy t dx = dy du du dx wit u = x + 3. y = sinx + 3) x + 3) x) if g) y = x 3 + x + = x 3 + x + ) / Cain rule. y = x 3 + x + ) / 3x + ) 5
) y = sinx lnx)) Tis one requires te cain rule and ten te product rule for te inside part i.e. x lnx)). y = cosx lnx)) i) y = x arctanx ) lnx) + x ) x = + lnx)) cosx lnx)) Tis one involves an application of te product rule and ten te cain rule to andle arctanx ) wile doing te product rule). y = arctanx ) + x 9. 7 points) Indefinite Integrals. a) x + + x + x dx = x + + x + x dx + x ) x = arctanx ) + x + x 4 b) c) After rewriting /x, we can just apply basic formulas. x + x 3 + 3x + 5 dx x + x + ln x + x + C Here we need to use a u-substitution. Let u = x 3 + 3x + 5 so tat du = 3x + 3) dx and so du = 3x + ) dx tus 3 du = x + ) dx. x + x 3 + 3x + 5 dx = u 3 du = 3 ln u + C 3 ln x3 + 3x + 5 + C cosx) + sin x) dx Just a u-substitution is needed. Let u = sinx) ten du = cosx) dx. arctansinx)) + C cosx) + sin x) dx = du = arctanu) + C + u. 8 points) Definite Integrals. 4 4 a) x dx = x / dx = x3/ 4 3/ = 3 43/ 3 3/ = 3 3 = 6 3 b) c) π/ sinx) dx Unless we see ow to integrate directly tis one isn t too ard sinx) = cosx) + C), we ll need a u-substitution. Don t forget to cange te limits as well! Let u = x so tat du = dx and tus du = dx. Also, x = u = ) = and x = π/ u = π/) = π. π/ π sinx) dx = sinu) π du = cosu) = cosπ) cos) = ) + ) = xx + ) 99 dx Again we need to use a u-substitution to do tis one. Let u = x + so tat du = x dx and tus du = x dx. Also, x = u = ) + = and x = u = + = 5. xx + ) 99 dx = 5 u 99 u du = 5 = 5 6