MATH1131/1141 Calculus Test S1 v8a

Transcription

1 MATH/ Calculus Test 8 S v8a October, 7 Tese solutions were written by Joann Blanco, typed by Brendan Trin and edited by Mattew Yan and Henderson Ko Please be etical wit tis resource It is for te use of MatSoc members, so do not repost it on oter forums or groups witout asking for permission If you appreciate tis resource, please consider supporting us by coming to our events and buying our T-sirts! Also, appy studying :) We cannot guarantee tat our working is correct, or tat it would obtain full marks - please notify us of any errors or typos at unswmatsoc@gmailcom, or on our Facebook page Tere are sometimes multiple metods of solving te same question Remember tat in te real class test, you will be expected to explain your steps and working out Note tat f is a polynomial function, so it is indeed continuous (note tat you sould mention tis or you could lose a mark) on te intervals [, ] and [, ] Note tat f () and f () Since f () < < f (), ten from te Intermediate Value Teorem, we know tat tere exists a real number c (, ) suc tat f (c) Hence, f as a zero on te interval [, ] Similarly, note tat f () and f () Since f () < < f (), ten from te Intermediate Value Teorem, we know tat tere exists a real number d (, ) suc tat f (d) Hence, f also as a zero on te interval [, ]

2 If f (x) x + x, ten from te definition of te derivative, we ave f (x) f (x + ) f (x) (x + ) + (x + ) x + x + x + + x + + x x x x + x + + x x ( x + ) x ( x + ) x + Te area of a rectangle is given by A LW Noting tat bot W and L depend on time, we implicitly differentiate bot sides wit respect to t (using te product rule), dw dl da L +W L W, by substituting in te information given, noting tat dl is negative since L is decreasing Tus, wen L and W, da cm /s Note: Wenever te questions asks for rate of cange, ten you re most likely going to be differentiating wit respect to t, ie time Te Mean Value Teorem states tat if f is continuous on [a, b] and differentiable on (a, b), ten tere exists a number c (a, b) suc tat f (c) f (b) f (a) b a For f (x) x x + 5, we ave f (x) x x Also, note tat f is continuous on te interval [, ] and differentiable on (, ) since it is a polynomial function Tus, from te Mean Value Teorem, we know tat tere exists a number c (, ) suc tat

3 c c f () f () Simplifying tis and solving for c, we get c c c c c (c ) c or But c (, ), so c is te point wic satisfies te conclusions of te Mean Value Teorem 5 Currently, tis it takes te indeterminate form of x sin x x cos x + sin x x cos x x sin x x cos x + x sin x x cos x + x sin x x + So, we can try L Ho pital s rule, (L Ho pital s rule) (by te Algebra of Limits) (L Ho pital s rule)

4 MATH/ Calculus Test 8 S vb October, 7 Tese solutions were written by Joann Blanco, typed up by Brendan Trin and edited by Mattew Yan and Henderson Ko Please be etical wit tis resource It is for te use of MatSoc members, so do not repost it on oter forums or groups witout asking for permission If you appreciate tis resource, please consider supporting us by coming to our events and buying our T-sirts! Also, appy studying :) We cannot guarantee tat our working is correct, or tat it would obtain full marks - please notify us of any errors or typos at unswmatsoc@gmailcom, or on our Facebook page Tere are sometimes multiple metods of solving te same question Remember tat in te real class test, you will be expected to explain your steps and working out Te Mean Value Teorem states tat if f is continuous on [a, b], and differentiable on (a, b), ten tere exists a real number c (a, b) suc tat f (c) Note tat te function f (x) f (b) f (a) b a x is continuous on [, ] and differentiable on (, ) Ten, from te mean value teorem, we know tat tere exists a real number c (, ) suc tat f (c) f () f ()

5 Since f (x), x we substitute tis into te equation above to get c Squaring bot sides, we ave (c ) c c Note tat (, ) Tus, c is te point tat satisfies te conclusions of te mean value teorem Note tat te it takes te indeterminate form, and so we can try L Ho pital s rule, cos x sin x (L Ho pital s rule) x x x x Tis also takes te indeterminate form, so we can try L Ho pital s rule again 9 cos x x 9 (L Ho pital s rule) Note tat f is continuous on te intervals [, ] and [, ] (indeed it is actually continuous everywere, since f is a polynomial function) Note tat f ( ) and f ( ) Since f ( ) < < f ( ), ten from te Intermediate Value Teorem, we know tat tere exists a number c (, ) suc tat f (c) Hence, f as a zero on te interval [, ] Similarly, note tat f () and f () Since f () < < f (), ten from te Intermediate Value Teorem, we know tat tere exists a number d (, ) suc tat f (d) Hence, f also as a zero on te interval [, ] Note: Always remember to ceck or at least mention tat te function is continuous, before you apply te intermediate value teorem In general, wenever you are using a teorem, like te min-max teorem for example, always ceck te conditions before you continue Note : It s important tat you distinguis c and d in tis question, as tey are different 5

6 numbers If f (x) x, ten from te definition of te derivative, we know tat f (x + ) f (x) (x + ) x x + x + x + + x x x ( x x ) f (x) ( x x ) x 5 We want te line tangent to x + y x y at (, ) By differentiating implicitly wit respect to x (applying te product rule to x y), we obtain x + y dy xy + x dy Rearranging, we see tat dy x xy x y So, at (, ), () ()() dy () Using te point-gradient formula at te point (, ), we get y y m (x x ) y (x ) y Terefore, te tangent line at (, ) is y 6

7 MATH/ Calculus Test 9 S v8b October, 7 Tese solutions were written by Joann Blanco, typed up by Brendan Trin and edited by Silong Zu and Aaron Hassan Please be etical wit tis resource It is for te use of MatSoc members, so do not repost it on oter forums or groups witout asking for permission If you appreciate tis resource, please consider supporting us by coming to our events and buying our T-sirts! Also, appy studying :) We cannot guarantee tat our working is correct, or tat it would obtain full marks - please notify us of any errors or typos at unswmatsoc@gmailcom, or on our Facebook page Tere are sometimes multiple metods of solving te same question Remember tat in te real class test, you will be expected to explain your steps and working out Te Mean Value Teorem states tat if f is continuous on [a, b], and differentiable on (a, b), ten tere exists a number c (a, b) suc tat f (c) f (b) f (a) b a Note tat f is continuous on [, ] and differentiable on (, ), as it is a polynomial function Ten, from te Mean Value Teorem, we know tat tere exists a number c (, ) suc tat f (c) f () f () 7

8 Note tat f (x) x x So we ave c c c c c (c ) c or But c (, ), and so c is a point wic satisfies te conclusions of te Mean Value Teorem Te it currently takes te indeterminate form So we try L Ho pital s rule x 5x + x + 9x x + (L Hopital s rule) x x x x + x Tis it also takes te indeterminate form, so we try L Ho pital s rule again and find tat 8x 8 x (L Ho pital s rule) Clearly, te function is continuous at every point except x since it is composed of polynomials We need to be more careful about weter it is continuous at x For f to be continuous at x, x f (x) must exist and equal to f () So first, we will ceck weter te it actually exists Since te function splits up at x, we need to take te two-sided it at x We ave f (x) x x x and f (x) (x ) + x + x + + Since te two sided its do not agree, ten x f (x) doesn t actually exist So, we can safely conclude tat f is not continuous at x 8

9 Tus, f is continuous for x 6 If f (x) x, ten from te definition of te derivative we know tat f (x + ) f (x) (x + ) x x + x + x + + x x x f (x) x x x 9

10 MATH/ Calculus Test 9 S va October, 7 Tese answers were written by Joann Blanco, typed up by Brendan Trin and edited by Aaron Hassan and Henderson Ko Please be etical wit tis resource It is for te use of MatSoc members, so do not repost it on oter forums or groups witout asking for permission If you appreciate tis resource, please consider supporting us by coming to our events and buying our T-sirts! Also, appy studying :) We cannot guarantee tat our answers are correct - please notify us of any errors or typos at unswmatsoc@gmailcom, or on our Facebook page Tere are sometimes multiple metods of solving te same question Remember tat in te real class test, you will be expected to explain your steps and working out Recall te definition tat a function f is continuous at x if f (x) f () () x At te moment, f is not continuous, since f () isn t defined So to find a value tat will make f continuous at, we simply need to just assign f () to be watever f (x) is x We can see tat f (x) x x x x x x + x (x )(x ) x x (since x 6 ) So if we assign f () to be equal to, ten we satisfy te continuity condition () So te

11 answer is ax + b, for x f (x) tan πx, for < x < For te function to be differentiable at x, it must be continuous tere, and te derivative must exist tere First, to ensure continuity at x, te two sided its must agree wit te function value at tis point, ie f (x) f (x) f () x x + But tis is simply a + b a + b So f is continuous at x if a + b Next, to ensure tat te derivative exists at x, we use te definition of te derivative, f ( + ) f () f ( + ) f () + π(+) tan (a + b) a ( + ) + b (a + b) + f () Note tat from te continuity condition, we know tat a + b And so, te LHS simplifies to a Te RHS takes te indeterminate form, so trying L Ho pital s rule π sec a a + π π sec π π a π(+) (L Ho pital s rule) From te continuity condition, we ave b π Hence, we require a π π, b for f to be differentiable at x Note : In tese types of questions, tere are always two tings we must ensure: tat f is continuous at te point and differentiable at te point Note : Because te function is piecewise and splits up at x, wen we are ensuring continuity at x, we need to consider te two-sided its f (x) and f (x), x x +

12 not just f (x) x (i) Tis is most easily done by first sketcing te grap of y f (x) Tis is done by first sketcing y x x and ten reflecting anyting below te x-axis above From tis, note tat in tis case, te critical points are te endpoints of te interval [, 5], te stationary points and te points were te derivative does not exist (te cusps) Te endpoints are clearly (, ) and (5, 5) Te stationary points can be found by noting tat te x-coordinates of te stationary points of y x x and y f (x) are actually te same So we first find te xcoordinate of te stationary point of te grap y x x Letting y, we find tat y x wic gives x Substituting x back into y f (x) gives y f () Terefore, te stationary point is (, ) Te points were te derivative does not exist are te points were te grap cuts sarply From te grap, it is easy to see tat tose points are (, ) and (, ) Hence, te critical points are (, ), (, ), (, ), and (5, 5) Note: Be careful of te internet wen it comes to critical points A lot of places on te internet, including Wikipedia define critical points as just te stationary points However, in your course notes/pack, critical points are actually defined as te endpoints of te interval tat te domain takes (as long as it s not te wole of R), te points were f isn t differentiable and te stationary points (ii) From looking at te grap, we can easily see tat te absolute minimum values of f (x) are attained wen x and at x So te absolute minimum value of f (x)

13 is Since te domain of f is restricted to [, 5], ten we simply look for te greatest value tat f attains on tis interval It is clear tat te absolute maximum value of f (x) is attained at x 5 So te absolute maximum value of f (x) is 5 Te it at te moment takes te indeterminate form So, we can try L Ho pital s rule Have a go at it yourself! Te steps are relatively simple Wit a couple applications of L Ho pital s rule, you sould find tat te it equals to Alternate metod Here is a metod tat does not use L Ho pital s rule We notice by inspection, polynomial long division, or syntetic division (aka Horner s Metod) tat x x +x x x x x Let f (x) x x x Note tat f () Now, denoting te it by L, we ave x x + x L x x (x ) x x +x x x x x x x x f (x) f () x x f () def (as we noted f () ) (recalling one of te definitions of te derivative) But f () exists as f (x) x x x is a polynomial and is tus differentiable everywere So L f () 6 L Refer to Question in Test 8 S vb

14 MATH/ Calculus Test 9 S va October, 7 Tese answers were written by Joann Blanco, typed up by Brendan Trin and edited by Henderson Ko and Aaron Hassan Please be etical wit tis resource It is for te use of MatSoc members, so do not repost it on oter forums or groups witout asking for permission If you appreciate tis resource, please consider supporting us by coming to our events and buying our T-sirts! Also, appy studying :)! We cannot guarantee tat our answers are correct - please notify us of any errors or typos at unswmatsoc@gmailcom, or on our Facebook page Tere are sometimes multiple metods of solving te same question Remember tat in te real class test, you will be expected to explain your steps and working out To find te line tangent to te curve at (, ), we first need to find dy, in order to obtain te gradient of te tangent at (, ) It can be easily found troug implicit differentiation tat of te tangent at (, ) is From ere, it can be found dy y(x+) x(+y) So te gradient tat te equation of te line tangent is y x+ We first need to interpret te problem in a matematical setting Drawing a diagram will make tis muc easier Let b denote te (perpendicular) distance from te wall to te base of te ladder in metres Also, let denote te distance between te top of te ladder and te floor in metres and t denote time in seconds Te question is asking us to find d Te variables b and are related by te equation b + 9 (Pytagoras Teorem) Refer to Question 5 in Test 8 S vb

15 Implicitly differentiating bot sides wit respect to t, gives b But we know tat db db d + () 5 and b Substituting tis into equation () gives + d Tere s one more ting to do before we can get d : we need to find wat is We use Pytagoras Teorem to see tat 9 b 9 And so, we can conclude tat d m/s 8 Tat is, te top of te ladder is dropping at a rate of 8 m/s Note: tat te answer was negative because te top of te ladder was dropping down te wall It can be found tat te point c satisfies te conclusions of te Mean Value Teorem (i) Critical points are end points, points were te function is not differentiable and were te derivative is zero It can be found tat te critical points on te given interval are (, ), (, ) and (, ) (ii) As f is continuous on [, ], te max-min teorem guarantees tat f attains a global maximum and global minimum on [a, b], on a critical point By looking at te critical points we found in te previous part, we can easily see tat te absolute minimum value of f (x) is and te absolute maximum value of f (x) is 5 Since te it as it is now seems to be in te indeterminate form, we can try L Ho pital s Rule and it can be found tat te it is equal to Refer to Question in Test 8 S vb Refer to Question in Test 9 S va Refer to Question 5 in Test 8 S v8a 5

16 MATH/ Calculus Test S va October, 7 Tese solutions were written and typed up by Gary Liang and edited by Henderson Ko Please be etical wit tis resource It is for te use of MatSoc members, so do not repost it on oter forums or groups witout asking for permission If you appreciate tis resource, please consider supporting us by coming to our events and buying our T-sirts! Also, appy studying :) We cannot guarantee tat our working is correct, or tat it would obtain full marks - please notify us of any errors or typos at unswmatsoc@gmailcom, or on our Facebook page Tere are sometimes multiple metods of solving te same question Remember tat in te real class test, you will be expected to explain your steps and working out For f to be differentiable at x, f ( + ) f () f () must exist and it must be continuous at x For f () f () to exist, it must be true tat f () f () sin + sin + + f () f () ae + b a b ae a LHS is a standard it, wile RHS is in te indeterminate form, so we try L Ho pital s 6

17 Rule and see tat a e a For te function to be continous at x, we require tat f () f (x) f (x) x x + a+b b x tan x ex is in te indeterminate form, so we try L Hopital s rule, tan x sec x x ex x ex Differentiating implicitly: x + y dy dy y dy (y y) x dy x y y At (,), dy Using te point-gradient formula, y (x ) y x + 7

18 Let p(x) x 6x + p (x) x x Stationary points are at (, ) and (, ) Note tat p( ) 6 p() p() p() Since polynomials are continuous on R, ten by te Intermediate Value Teorem, tere is at least one root in eac of te intervals [, ], [, ] and [, ] since te polynomial canges signs at te endpoints of eac of tese intervals As it is a cubic polynomial, it only as at most tree roots Terefore p(x) as exactly real roots 5 From te cain rule, we know tat d (tan (x + )) + (x + ) 6x + 8x + 8x + x + 8