MATH1151 Calculus Test S1 v2a

Transcription

1 MATH5 Calculus Test 8 S va January 8, 5 Tese solutions were written and typed up by Brendan Trin 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 epected to eplain your steps and working out ( ) First, we let L + 7 ( + ) Now, in order to elp us investigate te beaviour of tis it, multiply te numerator and denominator by te conjugate, ie by ( + ) Simplifying, + 7 ( + ) L ( + + )

2 Dividing te numerator and te denominator by will elp us see te iting beaviour as, q By te Algebra of Limits, we can cange tis to, q L To determine te number of real roots of a cubic polynomial, we can consider te product of te y-values of te stationary points If tis value is negative, tat means te y-values of te stationary points are opposite in sign, implying tat tere must be real roots (try draw sketces of tis scenario if you re unsure about tis) First, we let f () 6 + Differentiating, f () Now, for stationary points, let f () So, ( ) or Consider te product of te y-values of te two stationary points, f () f () 6 () + 6 () + ( ) < Tus, bot stationary points lie on opposite sides of te -ais and ence tere must be real solutions (roots) to te function Alternatively, note tat tis is a cubic wit a positive leading coefficient We can coose suc tat we ave te cange of signs + + wic would allow us to use te Intermediate Value Teorem and te Fundamental Teorem of Algebra to argue tat

3 tere are eactly tree real roots An eample of possible values were we would see tese sign canges are,,, ( f () for a + b, tan, for < < For te function to be differentiable at, it must first be continuous, and te derivative must eist First, ensuring continuity at, te two sided its must agree wit te function evaluated at tis point, ie f () f () f () + So, a + b a + b implying tat a + b Now, ensuring differentiability, we use te definition of te derivative, f ( + ) f () f ( + ) f () + (+) tan a ( + ) + b (a + b) + Note tat te RHS takes te form, so sec (+) a + a sec () a (L Hopital s rule) From te first condition, we ave b Hence, a,b Mean Value Teorem: If a function f is continuous on [a, b] and differentiable on (a, b), ten tere eists a constant c (a, b) suc tat f (c) f (a) f (b) a b Now, let f : [, ] R, f () tan were > We coose tis domain for f by first inspecting te required answer Note tat f is continuous on [, ] and differentiable on (, ) Also, f () +

4 Tus, by te Mean Value Teorem, tere eists c (, ) suc tat f () f () f () + c f () tan f (c) < +c tan < for Since c >, tis implies tat Hence, tan At, < and so tan > Terefore, tan for

5 MATH5 Calculus Test 9 S vb December, Tese solutions were written and typed up by Brendan Trin 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 epected to eplain your steps and working out f () { a + b, for ln, for > For te function to be differentiable at, it must be bot continuous and differentiable at tis point First, ensuring continuity, we want te two sided its to agree wit te function evaluated at tis point, ie Tis implies tat a + b f () f () f () + a + b ln a + b

6 Now, ensuring differentiability, we use te definition of te derivative, f ( + ) f () f ( + ) f () + a ( + ) + b (a + b) ln ( + ) ln () + a ln ( + ) + Since te RHS takes te indeterminate form, a + + (L Hopital s rule) Hence, we ave a, and from te first equation, b Mean Value Teorem: If f is continuous on te interval [a, b] and differentiable on (a, b), ten tere eists a constant c (a, b) suc tat f (c) f (b) f (a) b a Since f () ( ) is continuous and differentiable on R (as it is a polynomial), it must be continuous on [, ] and differentiable on (, ) Also, note tat f () ( ) By te Mean Value Teorem, tere eists a constant c (, ) suc tat f () f () 9 Solving for c, we find tat c ± But c (, ), so c + (c ) Let L e +e Note tat tis takes te indeterminate form e e Tis also takes te indeterminate form, L e + e +, so we can apply L Hopital s rule, (L Hopital s rule) (L Hopital s rule)

7 Factorising te denominator, f () ( ) ( ) Case :, Wen, we know tat Rewriting f (), f () So, as +, f () ( ) ( ) Case : <, Wen <, we know tat ( ) Rewriting f (), f () ( ) ( ) ( ) So, as, f () Tus, since te -sided its do not agree, a f () does not eist 5 Let f () + cos We coose two simple -values tat would result in opposite signs of f () and would allow to use te Intermediate Value Teorem Coosing and, and evaluating te function at tese points, f () <, f + > Tese points were cosen to obtain simple function values Since f is continuous on,, ten by te Intermediate Value Teorem, tere eists c, suc tat f () f (c) f Tus, we can coose c, suc tat f (c) Tis implies tat tere eists at least one positive root

8 MATH5 Calculus Test 9 S vb January 8, 5 Tese answers were written and typed up by Brendan Trin 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 epected to eplain your steps and working out Recall te definition of te derivative, f f ( + ) f () () c L Hopital s rule may or may not elp Limit is Multiply te numerator and denominator by te conjugate and do someting similar as in Test 8 S va Limit is 5 Consider cases wen > and wen < Te two sided its do not agree, ie f () + f () implying tere is a jump discontinuity (an essential discontinuity) Tis means tat it cannot be removed by simply defining a function value for

9