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MAT1330A_Ver1_82copies.pdf LastyearMidterm2VerB.pdf (Replacing a question about continuity) Find the limit of f(x) as x goes to 0 from the left, justifying all steps fully and carefully. In particular, if you apply any theorems from class, you must verify that the hypothesis is met. Then answer: for what value of a are the two one-sided limits equal, if any? — CALCULUS FOR THE LIFE SCIENCES I — MAT1330 DGD WORKBOOK - ANSWER KEY TO SELECT QUESTIONS • If you detect or suspect errors in the answers to this document, please let your professor know! 1 LEC 1 – High School Review. 1.3 # 20 Find the domain of f(x) = √ 2x− 7. Ans: [72 ,∞) like 1.4 # 17 Find the formulas for the composition f ◦ g and g ◦ f and the product of the functions f and g; simplify where possible. f(x) = x− 1 x+ 1 g(x) = 1/(2x) Ans: f ◦ g(x) = 1−2x1+2x g ◦ f(x) = x+1 2x−2 fg(x) = x−1 2x2+2x 1.4 #33 Sketch the graph, state the domain and range, decide if the function has an inverse and if so, find it: f(x) = 3 √ x+ 4. Ans: D : (−∞,∞) R : (−∞,∞) f−1(x) = (x− 4)3 2.2 # 24 Solve 4e2x+1 = 20. Ans: x = 12(ln(5)− 1) 2.2 #26 Solve 4e2x+3 = 7e3x−2. Ans: x = ln 4 + 5− ln 7 2.2 #32 Express y = 0.27x in base e. Ans: y = e(ln(0.27)x 2.2 #42 Solve ln(ln(x)) = 0. Ans: x = e LEC 2 – High School Review. 2.3 # 68 Graph the function. Give the average, max, min, amplitude, period and phase and mark them on the graph: f(x) = 3 + 4 cos ( 2π ( x− 1 5 )) . Ans: period P = 5 phase φ = 1 amplitude A = 4 mean M = 3 min M −A = −1 max M +A = 7 Inequalities Find all solutions to the inequality: x+ 2 2x− 1 < 1="" x+="" 7="" .="" ans:="" all="" x="" ∈="" (−7,="" 12)="" lec="" 3="" –="" intro="" to="" dtds.="" 3.1="" #2="" write="" the="" updating="" function="" f="" associated="" with="" the="" following="" dtds,="" and="" evaluate="" it="" at="" the="" given="" arguments.="" is="" it="" a="" linear="" dtds?="" mt+1="m2t" mt="" +="" 2="" ;="" evaluate="" f="" at="" mt="0,mt" =="" 8,mt="20." ans:="" up.="" fun.="" f(x)="x" 2="" x+2="" 0="" 6.4="" 200="" 11="" 1last="" updated:="" july="" 22,="" 2021.="" 1="" 3.1="" #12="" write="" the="" updating="" function="" f="" associated="" with="" the="" following="" dtds.="" is="" it="" a="" linear="" dtds?="" mt+1="0.75Mt" +="" 2="" determine="" the="" backward="" dtds="" associated="" to="" the="" above="" dtds.="" use="" the="" backward="" dtds="" to="" find="" the="" value="" m0="" (the="" value="" at="" the="" previous="" time="" step)="" given="" that="" m1="16." ans:="" up.="" fun.="" f(x)="0.75x" +="" 2="" (linear)="" f−1(x)="43x" −="" 8="" 3="" (up.="" fun.="" for="" backward="" dtds)="" m0="56" 3="" 3.1="" #18="" &="" 22="" graph="" some="" values="" of="" the="" following="" dtds,="" starting="" with="" the="" given="" initial="" condition:="" `t+1="`t" −="" 1.7="" with="" initial="" value="" `0="13.1" cm="" write="" down="" a="" formula="" for="" the="" general="" solution="" and="" sketch="" its="" graph.="" sketch="" the="" graph="" of="" the="" updating="" function.="" label="" the="" axes="" for="" each="" graph!="" ans:="" general="" solution="" lt="13.1−" 1.7t="" graphical="" answers="" omitted="" 3.1="" #19="" &="" 23="" graph="" some="" values="" of="" the="" following="" dtds,="" starting="" with="" the="" given="" initial="" condition:="" nt+1="0.5nt" with="" initial="" value="" n0="1200" write="" down="" a="" formula="" for="" the="" general="" solution="" and="" sketch="" its="" graph.="" sketch="" also="" the="" graph="" of="" the="" updating="" function.="" ans:="" up.="" fun.="" f(x)="0.5x" general="" solution="" xt="(0.5t)(1200)" graphical="" answers="" omitted="" lec="" 4="" –="" fixed="" points="" and="" cobwebbing.="" 3.2="" like="" #6="" given="" the="" dtds="" governing="" the="" daily="" dose="" of="" a="" drug,="" mt+1="0.75Mt+2," do="" three="" iterations="" of="" a="" cobweb="" starting="" at="" m0="16" mg/l.="" then="" plot="" the="" solution="" you="" found="" this="" way="" on="" a="" graph="" of="" t="" vs="" mt.="" compare="" with="" the="" general="" solution="" formula="" we="" proved="" in="" class.="" ans:="" omitted.="" 3.2="" #8="" &="" 26="" graph="" the="" updating="" function="" underlying="" the="" dtds="" zt+1="0.9zt" +="" 1.="" then="" cobweb="" four="" steps,="" starting="" from="" z0="3." label="" the="" axes!="" next:="" solve="" for="" all="" fixed="" points,="" and="" classify="" their="" stability="" using="" cobweb="" diagrams.="" ans:="" fixed="" point:="" x∗="10" (stable)="" graphical="" answers="" omitted.="" ex.="" graph="" the="" updating="" function="" underlying="" the="" dtds="" zt+1="1.1zt" −="" 1.="" then="" cobweb="" four="" steps,="" starting="" from="" z0="3." label="" the="" axes!="" next:="" solve="" for="" all="" fixed="" points,="" and="" classify="" their="" stability="" using="" cobweb="" diagrams.="" ans:="" fixed="" point:="" x∗="10" (unstable)="" graphical="" answers="" omitted.="" 3.2="" #12="" &="" 30="" using="" the="" graph="" of="" the="" updating="" function="" underlying="" the="" dtds="" xt+1="xt" xt="" −="" 1="" ,="" cobweb="" four="" steps,="" starting="" from="" x0="3." (restrict="" your="" updating="" function="" to="" the="" domain="" x=""> 1.) Label the axes! Next: solve for all fixed points, and classify their stability using cobweb diagrams. Ans: one fixed point: x∗ = 2 (keep in mind: domain is restricted to x > 1) Based on cobweb, solutions that start near x∗ = 2 do not approach x∗ = 2, so we consider this an unstable fixed point. Note: textbook calls this “stable” because nearby solutions are not moving away from x∗ = 2, but in MAT1330, we defined “stable” as “all nearby solutions must move closer to the fixed point”. 2 Ex. Consider the DTDS xt+1 = −15x 2 t + 2xt. Cobweb this DTDS starting at x0 = 2. Next: solve for all fixed points, and classify them according to their stability. Ans: two fixed points: x∗ = 0 (unstable) x∗ = 5 (stable) LEC 5 – Stability of Fixed Points. 3.2 like #14 (or 3.4 #16) Given the following graph of f , which is the updating function of a DTDS, determine the number of fixed points of the DTDS and determine their stability using cobwebbing. Write you conclusions in full sentences. Ans: two fixed points x∗1 ≈ 1.3 (stable) x∗2 ≈ 5.5 (unstable) 3.2 like #30 Find the equilibria of the following DTDS. Use cobwebbing to check each equilibrium for stability. xt+1 = 2xt xt − 1 (x > 1). Ans: fixed point x∗ = 3 (stable) 3.3 #28 In 1990 there were about 5000 southern mountain caribou in BC. In 2009, only about 1900 remained. Assume the annual per capita decline is constant. How long until the popula- tion falls below m = 500 (which is a level, below which it is expected the species will go extinct)? Ans: when t > 19 ln(1/10)ln(19/50) years have passed since 1990 (1990 is the year for which t = 0) 3.4 #14 Find all nonnegative equilibria of the following DTDS, where a is some real positive pa- rameter: xt+1 = xt a+ xt . (Extra bizarre: what happens if a = 0?) Ans: fixed point(s) x∗ = 0 and x∗ = 1−a. Since a > 0, we would need a ≤ 1 in order for x∗ = 1−a to be non-negative If a = 0, then f(x) would degenerate to f(x) = 1 for all x. LEC 6 – Limits and Continuity. 4.2 (typical question) Sketch the graph and decide if the left-hand and right-hand limits at a = 0 exist, and then decide if the limit at a = 0 exists. f(x) = { x+ 1 if x ≤ 1 x2 if x > 1 Ans: limx→ 0−f(x) = 1 limx→ 0+f(x) = 1 limx→ 0f(x) = 1 Ans: limx→ 1−f(x) = 2 limx→ 1+f(x) = 1 limx→ 0f(x)DNE 4.2 like #13 Sketch the graph and decide if the left-hand and right-hand limits at a = 0 exist, and then decide if the limit exists. f(x) = { |x+ 1|+ 1 if x ≤ 0 |x− 2| if x > 0 Ans: limx→ 0−f(x) = 2 limx→ 0+f(x) = 2 limx→ 0f(x) = 2 3 Evaluate each of the following limits, showing all your steps: 4.2 #48 lim x→0 x2 − 3x x3 − 9x Ans: 13 4.2 #48 lim x→4 √ x− 2 4− x Ans: −14 4.4 #36 Sketch the graph and discuss continuity of f(x) = x2 − 4 x− 2 if x 6= 2 0 if x = 2 Ans: f is discontinuous at x = 2 because lim x→2 f(x) = 4 but f(2) = 0. LEC 7 – Infinite Limits & Limits at Infinity. Evaluate each of the following limits, showing all of your steps: 4.3 #12 lim x→1+ x x2 − 1 Ans: +∞ (DNE) 4.3 #14 lim x→−7+ √ 1 x+ 7 Ans: +∞ (DNE) 4.3 #36 lim x→∞ 0.7x Ans: 0 4.3 #43 lim x→∞ x3 − 6x+ 4 3− x3 Ans: −1 4.3 like #45 lim x→∞ (x− 1)(x− 3)(x− 5) x2 − 4 Ans: ∞ (DNE) 4.3 #50 lim x→−∞ ln(3− x3) Ans: ∞ (DNE) 4.4 # 6 Find a formula for a function g(x) that makes the composition sin(g(x)) discontinuous at x = π. Ans: many answers possible. One possibility: g(x) = { 0 if x > π π/2 if x ≤ π Ex. (like Course Guide Lecture 7 question 7) Let f(x) = x2 − 4x+ 3 (x− 1)3 if x 6= ±1 x+ b if x = ±1 Find the limit of f as x → 1 and as x → −1. Is there a value of b that makes f continuous at x = 1? Is there a value of b that makes f continuous at x = −1? Ans: lim x→1− f(x) = −∞ so lim x→1 f(x) DNE. f(1) = 1 + b. There is no b that makes f continuous at x = 1. lim x→−1− f(x) = −1 lim x→−1+ f(x) = −1 lim x→−1 f(x) = −1 f(−1) = −1 + b If b = 0, then f would be continuous at x = −1. LEC 8 – The Derivative: Definition & Basic Rules. Let f be a function defined on a interval around x. The derivative of f at x is, by definition, 4 Ans: f ′(x) = lim h→0 f(x+ h)− f(x) h if this limit exists. The derivative of f(x) at a point a is, by definition, Ans: f ′(a) = lim h→0 f(a+ h)− f(a) h if this limit exists. Using the definition, compute the derivative of each of the following functions: a. f(x) = 2 + √ 3x+ 1 Ans: f ′(x) = 3 2 √ 3x+1 b. g(x) = 3 4 + 2x Ans: g′(x) = − 6 (4+2x)2 c. h(x) = √ x2 + 1 Ans: h′(x) = x√ x2+1 Using the rules of differentiation and simplifications where appropriate, compute the de- rivative