1Practice Problems A. Miller Spring 97 Math 221The final exam will beMonday December 15 at 5:05pmin 494 Van Hise.These problems were taken from exams given byJoel Robbin in 1995-1996.He will be the one writing your final exam.For more see http://math.wisc.edu/∼robbin/1.1Find the constant c which makes g continuous on (−∞, ∞),g(x) =(x2if x < 4,cx + 20 if x ≥ 4.Answer:By the limit laws, g(x) is continuous at any x 6= 4.limx→4−g(x) = limx→4−x2= 16, limx→4+g(x) = limx→4+cx + 20 = 4c + 20.The function g(x) is continuous when these are equal, i.e. when c = −1.1.2Find limt→−∞t3− 1t2− 1Answer:limt→−∞t3− 1t2− 1= limt→−∞t −1t21 −1t2= −∞1.3Find limt→3t3− 1t2− 1Answer:limt→3t3− 1t2− 1=33− 132− 1=2681.4Find limt→1t3− 1t2− 1Answer:limt→1t3− 1t2− 1= limt→1(t − 1)(t2+ t + 1)(t − 1)(t + 1)= limt→1(t2+ t + 1)(t + 1)=32Practice Problems A. Miller Spring 97 Math 221 22.1Consider the curvey2+ xy − x2= 11.(1) Find the equation of the tangent line to the curve at the point P (2, 3). (2) Findd2ydx2at the point P (2, 3).Answer:(1)Differentiate:2ydydx+ y + xdydx− 2x = 0.Evaluate at the point P (2, 3):6dydx(x,y)=(2,3)+ 3 + 2dydx(x,y)=(2,3)− 4 = 0.Solve:dydx(x,y)=(2,3)=18.This is the slop e of the tangent line. The point P(2, 3) lies on the tangent line so theequation of the tangent line is(y − 3) =(x − 2)8.Answer:(2)Differentiate again:2 dydx!2+ yd2ydx2+dydx+"dydx+ xd2ydx2#− 2 = 0.Evaluate at (x, y) = (2, 3):2182+ 3d2ydx2(x,y)=(2,3)+18+18+ 2d2ydx2(x,y)=(2,3)− 2 = 0.Solve:d2ydx2(x,y)=(2,3)=−2182−18−18+ 26 + 2.2.2(a) Find the equation for the tangent line to y = ln(x) at the point x = 1.Answer:(a) The equation for the tangent line is y = L(x) where L(x) is the linearapproximation of y = ln(x) at x = 1. Using the formulaL(x) = f(a) + f0(a)(x − a)Practice Problems A. Miller Spring 97 Math 221 3with f(x) = ln x and a = 1 and f0(x) = 1/x we get f(1) = ln(1) = 0, f0(1) = 1 and soL(x) = 0 + 1 · (x − 1) = (x − 1).(b) Find the quadratic approximation Q(x) to the function f(x) = ln x at 1.Answer:(b) Using the formulaQ(x) = f (a) + f0(a)(x − 1) +f00(a)(x − a)22with f(x) = ln(x). we get f0(x) = 1/x, f00(x) = −1/x2so f00(1) = −1 and henceQ(x) = (x − 1) −(x − 1)22.(c) Estimate ln(1.1) without a calculator.Answer:(c) ln(1.1) ≈ Q(1.1) = 0.1 − (0.1)2/2 = 0.095.2.3Findddx3sin x.Answer:ln(3) cos(x)3sin x2.4Findddxesin−1(x).Answer:esin−1(x)√1−x22.5FindddxF (ex). where the derivative of the function F is F0(u) = sin(u2).Answer:exsin(e2x)2.6Find f0(x) and f00(x) if f(x) = sin(x3− 2).Answer:f0(x) =cos(x3− 2)3x2, f00(x) = −sin(x3− 2)3x22+cos(x3− 2)6x.2.7Find g0(3) if h0(9) = 17 and g(x) = h(x2).Answer:g0(x) = h0(x2)2x, g0(3) = h0(9)6 = 102.2.8Consider the function y = f(x) whose graph is shown below. Match the expression inthe left column with the correct corresponding value in the right column.Practice Problems A. Miller Spring 97 Math 221 4f0(0) −6f0(0.9) 0f0(1) 3f0(1.732) 0.6-2-1.5-1-0.500.511.52-2 -1.5 -1 -0.5 0 0.5 1 1.5 2Answer:For positive x, the slope of the tangent line is decreasing as x increases, sof0(0) > f0(0.9) > f0(1) > f0(1.732). The only possibility is f0(0) = 3, f0(0.9) = 0.6,f0(1) = 0, f0(1.732) = −6.2.9Prove the product rule (fg)0= f0g + fg0using the definition of the derivative, highschool algebra, and the appropriate limit laws. Justify each step.3.1Find limx→∞sin(3x)xAnswer:Since −1 ≤ sin(3x) ≤ 1 for all x we have−1x≤sin(3x)x≤1xfor x > 0. Hence limx→∞sin(3x)x= 0 by the Squeeze theorem.3.2Find limx→0sin(x2)xPractice Problems A. Miller Spring 97 Math 221 5Answer:limx→0sin(x2)x= limx→0sin(x2)x2· x = limx→0sin(x2)x2· limx→0x = limu→0sin(u)u· limx→0x = 1 · 0 = 0.3.3Find limx→π/3sin(x) − sin(π/3))x − π/3Answer:Let f(x) = sin(x) and a = π/3. Then f0(x) = cos(x) andlimx→π/3sin(x) − sin(π/3)x − π/3= limx→af(x) − f(a))x − a= f0(a) = cos(π/3) =12.3.4Find limn→∞1 +2nn. Justify your steps.Answer:Using the formulae = limm→∞1 +1mm(♥)impliese = limn→∞1 +2nn/2=limn→∞1 +2nn1/2.so squaring gives:limn→∞1 +2nn= e2The wording of the question leaves some doubt as to whether the professor wants aproof of (♥). To be on the safe side we’ll provide it:lnlimm→∞1 +1mm= limm→∞ln1 +1mm=limm→∞ln1 +1m− ln(1)1/m= limh→0ln(1 + h) − ln(1)h.The limit on the right is the derivative of the ln(x) evaluated at x = 1. This isln0(1) = 1/1 = 1 solnlimm→∞1 +1mm= 1 so limm→∞1 +1m= e.You can also use l’Hospital’s rule.3.5Suppose g is the inverse function to f(x) = x5+ x + 1. Find g0(1) and g0(3)Practice Problems A. Miller Spring 97 Math 221 6Answer:x = g(y) ⇐⇒ y = f(x). Since f(0) = 1 we have g(1) = 0. Since f (1) = 3we have g(3) = 1. Now g(f(x)) = 1 so by the chain ruleg0(f(x)) =1f0(x)=15x4+ 1.When x = 0 this givesg0(1) = g0(f(0)) =15 · 04+ 1= 1When x = 1 this givesg0(3) = g0(f(1)) =15 · 14+ 1=16.3.6(a) If f(x) = xln xwhat is f0(x)?Answer: (a) x = eln xso xln x= e(ln x)2. Hence by the Chain Rulef0(x) =ddxe(ln x)2= e(ln x)2ddx(ln x)2= e(ln x)22(ln x)ddxln x =e(ln x)22(ln x)x.(b) Expressdudtin terms of t and u if u = t ln(t + u).Answer:(b) Use implicit differentiation:dudt= ln(t + u) +tt + u· 1 +dudt!.Solve for du/dt:1 −tt + ududt= ln(t + u) +tt + u;sodudt=1 −tt + u−1ln(t + u) +tt + u.3.7(a) Find limt→0sin tt3− t.Answer: (a) By l’Hospital’s Rule:limt→0sin tt3− t= limt→0cos t3t2− 1= −1.Practice Problems A. Miller Spring 97 Math 221 7(b) Find limx→∞ln(2x + 1) − ln(x).Answer: (b) ln(2x + 1) − ln(x) = ln2x + 1x= ln2 +1x. Hencelimx→∞ln(2x + 1) − ln(x) = lnlimx→∞2 +1x= ln(2).3.8Find limx→32x− 13x− 1Answer:7263.9Find limx→02x− 13x− 1Answer:ln(2)ln(3)3.10Find limx→∞2x− 13x− 1Answer:03.11Find limx→01xZx0et2+1dtAnswer:e3.12Find limx→∞3 + xxx.Answer:e33.13Find limx→0.6sin(cos−1x).Answer:.83.14Find the equation for the tangent line to y = ex−2yat the point (x, y) = (2, 1).Answer:3(y − 1) = x − 23.15The population of the country of Slobia grows exponentially.(a) If its population in the year 1980 was 1,980,000 and its population in the