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MAT 137Y: Calculus with proofs Assignment 3 Due on Thursday, Nov 25 by 11:59pm via GradeScope Instructions This problem set is based on Unit 3 and 4: Derivatives and Transcendental functions. Please read the Problem Set FAQ for details on submission policies, collaboration rules, and general instructions. Remember you can submit in pairs or individually. • Submissions are only accepted by Gradescope. Do not send anything by email. Late submissions are not accepted under any circumstance. Remember you can resubmit anytime before the deadline. • Submit your polished solutions using only this template PDF. You will submit a single PDF with your full written solutions. If your solution is not written using this template PDF (scanned print or digital) then you will receive zero. Do not submit rough work. Organize your work neatly in the space provided. • Show your work and justify your steps on every question, unless otherwise indicated. 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Signatures: 1) 2) https://q.utoronto.ca/courses/237708/pages/policies-and-faq#problemsets https://www.gradescope.ca https://q.utoronto.ca/courses/237708/pages/policies-and-faq#problemsets http://www.governingcouncil.utoronto.ca/policies/behaveac.htm 1. Let a, b, c ∈ R and a < c="">< b.suppose="" that="" f="" :="" (a,="" b)→="" r="" is="" differentiable="" at="" c="" and="" f="" ′(c)=""> 0. (a) Prove that there exists δ > 0 such that f(x) > f(c) for all x in (c, c + δ) and f(x) < f(c)="" for="" all="" x="" in="" (c−="" δ,="" c).="" (b)="" does="" f="" have="" to="" be="" increasing="" in="" some="" neighborhood="" of="" c?="" if="" so,="" prove="" it.="" if="" not,="" find="" a="" counterex-="" ample="" and="" justify.="" remark:="" we="" say="" a="" function="" f="" is="" increasing="" on="" an="" inverval="" i="" if="" ∀x1,="" x2="" ∈="" i,="" x1="">< x2="⇒" f="" (x1)="">< f (x2) . 2. assume h is continuous everywhere and h(0) = 0. here is the graph of its derivative: −7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7 −4 −3 −2 −1 1 2 3 4 y = h′(x) x y use the space below to sketch the graph of h. remember h(0) = 0 and h is continuous everywhere. −7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7 −4 −3 −2 −1 1 2 3 4 x y 3. let a ∈ r. we want to study the curve with equation y2(y2 − a) = x2(x2 − a− 1) notice that for each value of a we get a different curve. you can see the graph on desmos.com/calculator/fuxts9nua4. you will find one slider that allows you to change the value of a. note that the curve is differentiable at every point for every value of a, except at the origin, which is always a singular point.you can use this information without justification for this problem. (a) fix a = 4. find out how many points on the curve have a horizontal tangent line. find all of their coordinates. hint: use implicit differentiation. think of y as a function of x. https://www.desmos.com/calculator/fuxts9nua4 (b) fix a = 4. find out how many points on the curve have a vertical tangent line. find all of their coordinates. hint: use implicit differentiation. think of x as a function of y. (c) now play with the slider and try different values of a. you will notice that sometimes the curve has exactly 2 points with a vertical tangent line, sometimes it has no points with a vertical tangent line and sometimes it has exactly 10 points with a vertical tangent line. for which values of a does it have 10, for which values does it have 2, and for which values does it have 0? prove it. 4. (a) one may avoid the absolute values that appear in the derivative of arccsc(x) by changing the definition of this functions. in defining arccsc(x), we restricted the domain of sin(x) to [−π2 , π 2 ] in order to ensure injectivity. alternatively, we could have restricted the domain of sin(x) to (0, π/2] ∪ (π, 3π/2], where it is still injective. prove that if we define arccsc(x)using the new choice of domain: arccsc : (−∞,−1] ∪ [1,∞)→ (0, π 2 ] ∪ (π, 3π 2 ], then the derivative is given by ddx arccsc(x) = 1 −x √ x2−1 . (b) the following “theorem” is not quite true as stated: flawed “theorem”: for the function arctan : r→ (−π2 , π 2 ), for every x, y ∈ r we have arctan(x) + arctan(y) = arctan( x+ y 1− xy ) fake “proof”: proof. let θ1 = arctan(x) and θ2 = arctan(y). then we get tan(θ1) = x and tan(θ2) = y. now, tan(θ1 + θ2) = tan(θ1)+tan(θ2) 1−tan(θ1) tan(θ2) = x+y 1−xy . θ1 + θ2 = arctan( x+y 1−xy ). thus, arctan(x) + arctan(y) = arctan( x+y 1−xy ). explain the problem with the statement of the theorem and the errors in the proof. then fix them: correct the statement, and write a correct proof. f="" (x2)="" .="" 2.="" assume="" h="" is="" continuous="" everywhere="" and="" h(0)="0." here="" is="" the="" graph="" of="" its="" derivative:="" −7="" −6="" −5="" −4="" −3="" −2="" −1="" 1="" 2="" 3="" 4="" 5="" 6="" 7="" −4="" −3="" −2="" −1="" 1="" 2="" 3="" 4="" y="h′(x)" x="" y="" use="" the="" space="" below="" to="" sketch="" the="" graph="" of="" h.="" remember="" h(0)="0" and="" h="" is="" continuous="" everywhere.="" −7="" −6="" −5="" −4="" −3="" −2="" −1="" 1="" 2="" 3="" 4="" 5="" 6="" 7="" −4="" −3="" −2="" −1="" 1="" 2="" 3="" 4="" x="" y="" 3.="" let="" a="" ∈="" r.="" we="" want="" to="" study="" the="" curve="" with="" equation="" y2(y2="" −="" a)="x2(x2" −="" a−="" 1)="" notice="" that="" for="" each="" value="" of="" a="" we="" get="" a="" different="" curve.="" you="" can="" see="" the="" graph="" on="" desmos.com/calculator/fuxts9nua4.="" you="" will="" find="" one="" slider="" that="" allows="" you="" to="" change="" the="" value="" of="" a.="" note="" that="" the="" curve="" is="" differentiable="" at="" every="" point="" for="" every="" value="" of="" a,="" except="" at="" the="" origin,="" which="" is="" always="" a="" singular="" point.you="" can="" use="" this="" information="" without="" justification="" for="" this="" problem.="" (a)="" fix="" a="4." find="" out="" how="" many="" points="" on="" the="" curve="" have="" a="" horizontal="" tangent="" line.="" find="" all="" of="" their="" coordinates.="" hint:="" use="" implicit="" differentiation.="" think="" of="" y="" as="" a="" function="" of="" x.="" https://www.desmos.com/calculator/fuxts9nua4="" (b)="" fix="" a="4." find="" out="" how="" many="" points="" on="" the="" curve="" have="" a="" vertical="" tangent="" line.="" find="" all="" of="" their="" coordinates.="" hint:="" use="" implicit="" differentiation.="" think="" of="" x="" as="" a="" function="" of="" y.="" (c)="" now="" play="" with="" the="" slider="" and="" try="" different="" values="" of="" a.="" you="" will="" notice="" that="" sometimes="" the="" curve="" has="" exactly="" 2="" points="" with="" a="" vertical="" tangent="" line,="" sometimes="" it="" has="" no="" points="" with="" a="" vertical="" tangent="" line="" and="" sometimes="" it="" has="" exactly="" 10="" points="" with="" a="" vertical="" tangent="" line.="" for="" which="" values="" of="" a="" does="" it="" have="" 10,="" for="" which="" values="" does="" it="" have="" 2,="" and="" for="" which="" values="" does="" it="" have="" 0?="" prove="" it.="" 4.="" (a)="" one="" may="" avoid="" the="" absolute="" values="" that="" appear="" in="" the="" derivative="" of="" arccsc(x)="" by="" changing="" the="" definition="" of="" this="" functions.="" in="" defining="" arccsc(x),="" we="" restricted="" the="" domain="" of="" sin(x)="" to="" [−π2="" ,="" π="" 2="" ]="" in="" order="" to="" ensure="" injectivity.="" alternatively,="" we="" could="" have="" restricted="" the="" domain="" of="" sin(x)="" to="" (0,="" π/2]="" ∪="" (π,="" 3π/2],="" where="" it="" is="" still="" injective.="" prove="" that="" if="" we="" define="" arccsc(x)using="" the="" new="" choice="" of="" domain:="" arccsc="" :="" (−∞,−1]="" ∪="" [1,∞)→="" (0,="" π="" 2="" ]="" ∪="" (π,="" 3π="" 2="" ],="" then="" the="" derivative="" is="" given="" by="" ddx="" arccsc(x)="1" −x="" √="" x2−1="" .="" (b)="" the="" following="" “theorem”="" is="" not="" quite="" true="" as="" stated:="" flawed="" “theorem”:="" for="" the="" function="" arctan="" :="" r→="" (−π2="" ,="" π="" 2="" ),="" for="" every="" x,="" y="" ∈="" r="" we="" have="" arctan(x)="" +="" arctan(y)="arctan(" x+="" y="" 1−="" xy="" )="" fake="" “proof”:="" proof.="" let="" θ1="arctan(x)" and="" θ2="arctan(y)." then="" we="" get="" tan(θ1)="x" and="" tan(θ2)="y." now,="" tan(θ1="" +="" θ2)="tan(θ1)+tan(θ2)" 1−tan(θ1)="" tan(θ2)="x+y" 1−xy="" .="" θ1="" +="" θ2="arctan(" x+y="" 1−xy="" ).="" thus,="" arctan(x)="" +="" arctan(y)="arctan(" x+y="" 1−xy="" ).="" explain="" the="" problem="" with="" the="" statement="" of="" the="" theorem="" and="" the="" errors="" in="" the="" proof.="" then="" fix="" them:="" correct="" the="" statement,="" and="" write="" a="" correct="">