Calculus I-Review ProblemsNew York UniversityExam is on Sections 2.3-2.6, 2.8, 3.1-3.5, 3.7, 4.1, 4.2, 4.3, 4.4, 4.5, 4.7, 5.1-5.51. Compute the following limits:(a) limx→3−e2/(x−3)Solution:!Page 11(b) limx→10−ln(100 − x2)Solution:!Page 11(c) limx→∞e−xsin xSolution:!Page 11(d) limx→∞1+2x1−2xSolution:!Page 11You can also apply L’Hospital’s,(e) limx→∞x3e−xSolution:!Page 11(f) limx→1+xx−1−1ln xSolution:!Page 21(g) limx→π/2−tan xcos xSolution:!Page 21(h) lim→∞x[ln(x + 5) − ln x]Solution: lim→∞x lnx+5x= lim→∞lnx+5x1/x= lim→∞−5x2x+5x−1x2= lim→∞−5x/x2(x+5)−1/x2= lim→∞5xx+5= −5.2. Calculate y0.(a) y =2x√x2+1Solution: y0=2√x2+1−2x2(x2+1)−1/2x2+1(b) y =1sin(x−sin x)Solution:!Page 11(c) y =sec 2θ1+tan 2θSolution:!Page 11(d) xy4+ x2y = x + 3ySolution:!Page 11(e) sin(xy) = x2− ySolution:!Page 11(f) y = ln(x2ex)Solution:!Page 11(g) y = (cos x)xSolution:!Page 11(h) y = ln(1/x) +1ln xSolution:!Page 21(i) y = arctan(arcsin√x)Solution:!Page 21(j) xey= y − 1Solution:!Page 21(k) y = (tan x)1/xSolution:!Page 11(l) ex2y= x + ySolution:!Page 113. The graph of y = x3− 9x2− 16x + 1 has a slope of 5 at two points. Find the coordinates of the points.Solution:y0= 3x2− 18x − 165 = 3x2− 18x − 160 = 3x2− 18x − 210 = x2− 6x − 70 = (x + 1)(x − 7)x = −1 or x = 7.When x = −1, y = 7; when x = 7, y = −209.Thus, the two points are (−1, 7) and (7, −209).4. Find the equation of the line tangent to f (x) at x = 2, iff(x) =x32−43x.Solution:The slope of the tangent line is the value of the first derivative at x = 2. Differentiating givesddxx32−43x=ddx12x3−43x−1=12· 3x2−43(−1)x−2=32x2+43x2.For x = 2,f0(2) =32(2)2+43(2)2= 6 +13= 6.333andf(2) =232−43(2)= 4 −23= 3.333.To find the y-intercept for the tangent line equation at the point (2, 3.333), we substitute in the general equation,y = b + mx, and solve for b.3.333 = b + 6.333(2)−9.333 = b.The tangent line has the equationy = −9.333 + 6.333x.5. (a) Find the slope of the graph of f(x) = 1 − exat the point where it crosses the x-axis.(b) Find the equation of the tangent line to the curve at this point.(c) Find the equation of the line perpendicular to the tangent line at this point. (This is the normal line.)Solution:(a) f(x) = 1 −excrosses the x-axis where 0 = 1 −ex, which happens when ex= 1, so x = 0. Since f0(x) = −ex,f0(0) = −e0= −1.(b) y = −x(c) The negative of the reciprocal of −1 is 1, so the equation of the normal line is y = x.6. Suppose f and g are differentiable functions with the values shown in the following table. For each of the followingfunctions h, find h0(2).(a) h(x) = f(x) + g(x)(b) h(x) = f(x)g(x)(c) h(x) =f(x)g(x)xf(x) g(x) f0(x) g0(x)2 3 4 5 −2Solution:(a) We have h0(2) = f0(2) + g0(2) = 5 − 2 = 3.(b) We have h0(2) = f0(2)g(2) + f(2)g0(2) = 5(4) + 3(−2) = 14.(c) We have h0(2) =f0(2)g(2)−f(2)g0(2)(g(2))2=5(4)−3(−2)42=2616=138.7. If you invest P dollars in a bank account at an annual interest rate of r%, then after t years you will have Bdollars, whereB = P1 +r100t.(a) Find dB/dt, assuming P and r are constant. In terms of money, what does dB/dt represent?(b) Find dB/dr, assuming P and t are constant. In terms of money, what does dB/dr represent?Solution:(a)dBdt= P1 +r100tln1 +r100. The expressiondBdttells us how fast the amount of money in the bankis changing with respect to time for fixed initial investment P and interest rate r.(b)dBdr= P t1 +r100t−11100. The expressiondBdrindicates how fast the amount of money changes withrespect to the interest rate r , assuming fixed initial investment P and time t.8. Use the following graph to calculate the derivative.!3.6 THE CHAIN RULE AND INVERSE FUNCTIONS 153246810121416182022242628303234363840424446485052Job: chap3-temp Sheet: 39 Page: 153 (March 8, 2012 10 : 10) [ex-6]ins3-6w51-53In Problems 60–62, use Figure 3.32 to calculate the derivative.ins3-6w51-53fig(2, 5)(2.1, 5.3)f(x)Figure 3.323-6w5160. h!(2) if h(x)=(f(x))33-6w5261. k!(2) if k(x)=(f(x))−13-6w5362. g!(5) if g(x)=f−1(x)3-6w5463. (a) Given that f (x)=x3,findf!(2).(b) Find f−1(x).(c) Use your answer fr om part (b) to find (f−1)!(8).(d) How could you have used your answer from part (a)to find (f−1)!(8)?3-6w5564. (a) For f(x)=2x5+3x3+ x,findf!(x).(b) How can you us e your answer to part (a) to deter-mine if f( x) is invertible?(c) Find f(1).(d) Find f!(1).(e) Find (f−1)!(6).3-6w5765. Use the table and the fact that f(x) is invertible and dif-ferentiable everywhere to find (f−1)!(3).x f(x) f!(x)3 1 76 2 109 3 53-miscw11766. At a particular location, f(p) is the number of gallons ofgas sold when the price is p dollars per gallon.(a) What does the statement f(2) = 4023 tell you aboutgas sales?(b) Find and interpret f−1(4023).(c) What does the statement f!(2) = −1250 tell youabout gas sales?(d) Find and interpret (f−1)!(4023)3-6w5867. Let P = f(t) give the US population6in millions inyear t.(a) What does the statement f(2005) = 296 tell youabout the US population?(b) Find and interpret f−1(296). Give units.(c) What does the statement f!(2005) = 2.65 tell youabout the population? Give units.(d) E valuate and interpret (f−1)!(296). Give units.3-6w5968. Figure 3.33 shows the number of motor vehicles,7f(t),in millions, registered in the world t years after 1965.With units, estimate and interpret(a) f(20) (b) f!(20)(c) f−1(500) (d) (f−1)!(500)3-6w59fig’65’70’75’80’85’90’95 2000200400600800(year)(millions)Figure 3.333-6w6069. Using Figure 3.34, where f!(2) = 2.1, f!(4) = 3.0,f!(6) = 3.7, f!(8) = 4.2,find(f−1)!(8).3-6w60fig246881624xf(x)Figure 3.343-miscw11870. If f is increasing and f(20) = 10, which of the two op-tions, (a) or (b), must be wrong?(a) f!(10)(f−1)!(20) = 1.(b) f!(20)(f−1)!(10) = 2.3-6w6271. An invertible function f (x) has values in the table. Eval-uate(a) f!(a) · (f−1)!(A) (b) f!(b) · (f−1)!(B)(c) f!(c) · (f−1)!(C)x a b c df(x) A B C D3-6w6372. If f is continuous, invertible, and defined for all x, whymust at least one of the statements (f−1)!(10) = 8,(f−1)!(20) = −6 be wrong?3-6w6473. (a) C alculate limh→0(ln(1 + h)/h) by identifying thelimit as the derivative of ln(1 + x) at x =0.(b) Use the result of part (a) to show thatlimh→0(1 + h)1/h= e.(c) Use the result of part (b) to calculate the relatedlimit, limn→∞(1 + 1 /n)n.6Data from