Math 105: Review for Final Exam, Part II - SOLUTIONS

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1 Math 5: Review for Fial Exam, Part II - SOLUTIONS. Cosider the fuctio f(x) = x 3 lx o the iterval [/e, e ]. (a) Fid the x- ad y-coordiates of ay ad all local extrema ad classify each as a local maximum or local miimum. f (x) = 3x lx + x 3 x = x (3 lx + ) x = (ot i our domai) or lx = /3, which meas x = e /3 < x < e /3 e /3 < x f egative positive f y-value: f(e /3 ) = (e /3 ) 3 l(e /3 ) = (e )( /3) = ( /3e) So, f has a local miimum at (e /3, /3e). (b) Fid the x- ad y-coordiates of ay ad all global extrema ad classify each as a global maximum or global miimum. We check the y-values at the local extrema ad the edpoits. f(/e) = (/e) 3 l(/e) = (/e 3 )( ) = ( /e 3 ) f(e /3 ) = ( /3e) from above f(e ) = (e ) 3 l(e ) = (e 6 )() = e 6 So, f has a global miimum at (e /3, /3) ad a global maximum at (e, e 6 ). (c) Fid the x-coordiate(s) of ay ad all iflectio poits. f (x) = x(3 lx + ) + x (3 x + ) = 6x lx + x + 3x = x(6 lx + 5) x = (ot i our domai) or lx = 5/6, which meas x = e 5/6 < x < e 5/6 e 5/6 < x f egative positive f cocave dow cocave up So, the x-value of the iflectio poit of f is x = e 5/6.. Your compay is mass-producig a cylidrical cotaier. The flat portio (top ad bottom) costs 3 cets per square ich ad the curved (lateral) portio costs 5 cets per square ich. If your budget is $9. per cotaier, what dimesios will give the largest volume? area of circle = r Objective fuctio: volume = V = r h lateral area of cylider = rh We eed to get this dow to a fuctio of just oe variable, so we use the costrait equatio : cost = 9 = 3 r + 5 rh 9 = 6r + rh 9 6r = rh 9 6r = h r volume of cylider = r h

2 Substitutig this back ito the objective fuctio gives V = r h = r 9 6r 9 6r = r = r (9r 6r3 ). Now that we have V as a fuctio of just oe variable, we fid its maximum. V (x) = (9 8r ) = (9 8r ) 8r = 9 r = 5 5 r = < x < 5/ 5/ < x f positive egative f Thus, we have i fact foud the global maximum at r = 5/. 9 6r 7 Ad h = =...much simplifyig... = iches. r 3. You are stadig o a pier, 6 feet above the deck of a boat. Attached to the boat is a lie, which you are pullig i at a rate of 3 feet per secod. Whe there are feet of lie betwee your had ad the boat, at what rate is the boat movig across the water? You b 6 Boat a We kow db da, ad we wat to fid dt dt. So, we write a equatio that relates a ad b ad the differetiate implicitly with respect to time t. a + 6 = b a da dt + = bdb dt da dt = b db a dt At the momet i questio, b =, a = 8 (by the Pythagorea Theorem), ad db dt = 3. So, da dt = ( 3) = 3.75 feet per secod, meaig the boat is movig toward the dock at 3.75 feet 8 per secod. 4. Use the Itermediate Value Theorem to show that f(x) = x 3 x has a root o [, ]. IVT: If f is cotiuous o [a, b] ad y is a umber betwee f(a) ad f(b), the there is a umber c betwee a ad b such that f(c) = y. For the fuctio give above, f() = ad f() = 3. Sice is a umber betwee ad 3, the IVT says there is a umber c betwee ad such that f(c) = ; this c is the desired root.

3 5. What (if aythig) does the Extreme Value Theorem say about f(x) = x o each of the followig itervals? EVT: If f is cotiuous o [a, b], the f has both a maximum ad a miimum o [a, b]. (a) [, 4] f has a maximum ad a miimum o [, 4] (b) (, 4) The EVT does t apply because (, 4) is ot a closed iterval sice its edpoits are ot icluded. 6. Fid the value of the costat c that the Mea Value Theorem specifies for f(x) = x 3 + x o [, 3]. MVT: If f is cotiuous o [a, b] ad differetiable o (a, b), the there is a umber c betwee a ad b such that f f(b) f(a) (c) =. b a f(3) f() For our fuctio, we have = 3 =. 3 3 Ad f (x) = 3x +, so f (c) = 3c +. So, we solve 3c + =, which meas c = Fid the followig. (a) all atiderivatives of + x + x x + x 5 (b) (c) Ay such atiderivative will take the form x + x + x x3/ 3/ + x C. Note that we have used the facts that x = x / ad /x 5 = x x dx = 3 l x 7 = 3 l7 3 l = 3 l 7 4 x dx = () = This itegral represets the area of a semicircle of radius. (d) d dx (e) lim si t dt = si x ( ) k si k= The derivative of the area fuctio is the origial fuctio. This represets the limit of a right-had sum as the umber () of rectagles goes to ifiity. ( ) k ( ( ) ( ) As k goes from to, the expressio takes o the values,,..., ; the ) first of these values is just to the right of ad the last is equal to, so we see that we are lookig at the fuctio o the iterval [, ]. The out frot is our x, which cofirms that we are dealig with a iterval of legth beig subdivided ito equal subitervals.

4 Fially, we are takig the sie of each of our x-values, so the fuctio i questio must be f(x) = si x. Thus, we see that the expressio is equal to the area uder f(x) = si x o [, ]. Its value is si x dx = cosx = cos() ( cos()) = ( ) ( ) =. 8. Water is leakig out of a tak at a decreasig rate r(t) as show below. time (mi) rate (gal/mi) (a) Fid a overestimate ad uderestimate for the total amout that leaked out durig these 8 miutes. (b) Iterpret the expressio overestimate = L 4 = ( )() = 76 uderestimate = R 4 = ( )() = 5 6 r(t)dt i terms of the situatio described above. This itegral gives the amout (i gallos) of water that leaked from the tak o the iterval [, 6] miutes. 9. Cosider the graph of f(t) show. It is made of straight lies ad a semicircle. f(t) t 4 - Let G(x) = f(t) dt ad H(x) = 3 f(t) dt. (a) Compute G(), G(4), ad H(4). G() is the area uder f betwee t = ad t =. This is a rectagle plus a triagle ad has area + = 3. Similarly, G(4) = + + () = 3 +. H(4) is the area uder f betwee t = 3 ad t = 4. Remember that area below the t-axis couts as egative. H(4) = ( + ) + + [area uder f from to 4, foud above as G(4)] [ = ] = + (b) Where is G icreasig? Where is G decreasig? G is icreasig where f is positive: (, 4]. Note that G has a horizotal slope at x = but sice f is positive o each side of t =, we say G is icreasig at x =. G is decreasig where f is egative: [ 4, ).

5 (c) Where is G cocave up? Where is G cocave dow? G is cocave up where f is icreasig: (, ) (, 3). G is cocave dow where f is decreasig: (, ) (3, 4]. (d) At what x-value(s) does G have a local maximum? At what x-value(s) does G have a local miimum? G has a local maximum where f chages from positive to egative: ever. G has a local miimum where f chages from egative to positive: x =. (e) Fid a formula that relates G ad H. From their defiitios, H(x) = 3 f(t) dt + G(x) = + G(x). (f) How would your aswers to (b), (c), ad (d) chage if the questios were about H istead of G? They would ot chage at all because H (x) = G (x).. (a) Use sigma otatio to express L ad M as approximatios to We re subdividig the iterval ito pieces, so each piece has width x = L = [f() + f(4) + f(8) f(5) + f(56)] x = [l() + l(4) + l(8) l(5) + l(56)] 4 9 = l( + 4k) 4 k= 6 lx dx. 6 = 4. M = [f() + f(6) + f(3) f(54) + f(58)] x = [l() + l(6) + l(3) l(54) + l(58)] 4 9 = l( + 4k) 4 k= (b) Draw a sketch that represets the sum M 4. 6 Now we re subdividig the iterval ito 4 pieces, so each piece has width x = =. 4 Note that the height of each rectagle is determied by the y-value of the curve at the middle x-value of the rectagle (that is, at x = 5, 35, 45, 55). y=l x

Math 21B-B - Homework Set 2

Math 21B-B - Homework Set 2 Math B-B - Homework Set Sectio 5.:. a) lim P k= c k c k ) x k, where P is a partitio of [, 5. x x ) dx b) lim P k= 4 ck x k, where P is a partitio of [,. 4 x dx c) lim P k= ta c k ) x k, where P is a partitio