Math 1A: introduction to functions and calculus Oliver Knill, Spring 20113/1/2011: First hourly PracticeYour Name:• Start by writing your name in the above b ox.• Try to answer each question on t he same page as the question is asked. If needed, use theback or the next empty page for work. If you need additional paper, write your name on it.• Do not detach pages from this exam packet or unstaple the packet.• Please write neatly and except for multiple choice problems, give computations. Answerswhich are illegible for the grader can not be given credit.• No notes, books, calculators, computers, or other electronic aids can be allowed.• You have 90 minutes time to complete your work.• All unspecified functions are assumed to be smooth: one can differentiate arbitrarily.• The actual exam has a similar format: TF questions, multiple choice and then problemswhere work needs to be shown.1 202 103 104 105 106 107 108 109 1010 10Total: 110Problem 1) True/False questions (20 points) No justifications are needed.1)T FThe function cot(x) is the inverse of the function tan(x).Solution:No, it is artan(x) which is the inverse.2)T FWe have cos(x)/ sin(x) = cot(x)Solution:That is the definition.3)T Fsin(3π/2) = −1.Solution:Draw the circle. The angle 3π/2 corresponds to 270 degrees.4)T FThe function f(x) = sin(x)/x has a limit at x = 0.Solution:Yes, it is called the sinc function.5)T FFor the function f(x) = sin(sin(exp(x))) the limit limh→0[f(x+h)−f(x)]/hexists.Solution:The function is differentiable. The limit is the derivative.6)T FIf a differentiable function f (x) satisfies f′(3) = 3 and is f′is odd then ithas a critical point.Solution:We have f′(3) = 3 and f′(−3) = −3. The intermediate value theorem assures thatf′(x) = 0 for some x ∈ [−3, 3].7)T FThe l’Hopital rule assures that the derivative satisfies (f /g)′= f′/g′.Solution:The l’Hopital rule tells something about a limit f(x)/g(x) as x → p but does not computederivatives.8)T FThe intermediate value theorem assures that a continuous function has aderivative.Solution:The intermediate value theorem deals with continuous functions.9)T FThe function f(x) = (x + 1)/(x2− 1) is continuous everywhere.Solution:We divide by zero at z = 1.10)T FIf f is concave up on [1, 2] and concave down on [2, 3] then 2 is an inflectionpoint.Solution:Indeed, f′′changes sign there.11)T FThere is a function f which has the property that its second derivative f′′is equal to its negative f .Solution:It is the sin or cos function.12)T FThe function f (x) = [x]4= x(x + h)(x + 2h)(x + 3h) has the property thatD f (x) = 4[x]3= 4x(x + h)(x + 2h), where Df(x) = [f(x + h) − f (x)]/h.Solution:Yes, this is a cool property of the polynomials [x]n.13)T FThe quotient rule is d/dx(f/g) = (f′g −fg′)/g2and holds whenever g(x) 6=0.Solution:This is an impor tant rule to know.14)T FThe chain rule assures that d/ dxf(g(x)) = f′(g(x)) + f(g′(x)).Solution:This is not true. We have f′(g(x))g′(x).15)T FIf f and g are differentiable, then (3f + g)′= 3f′+ g′.Solution:This is called linearity of the differentation.16)T FFor any f unction f, the Newton step T (x) is continuous.Solution:There is trouble at critical points f′(x) = 0.17)T FOne can rotate a four legged table on an arbitrary surface such that all fourlegs are on the ground.Solution:We have seen this in class18)T FThe fundamental theorem of calculus relates integration S with differentia-tion D. The r esult is DSf(x) = f (x), SDf(x) = f(x) − f (0).19)T FThe product rule implies d/dx(f(x)g(x)h(x)) = f′(x)g(x)h(x) +f(x)g′(x)h(x) + f (x)g(x)h′(x).Solution:This was checked in a homework.20)T FEuler and Gauss are the founders of infinitesimal calculus.Solution:It was Newton and Leibniz who are considered the founders, Euler and Gauss came laterProblem 2) Matching problem (10 points) No justifications are needed.Match the following functions with their gra phs.Function Fill in 1-8x2− xexp(−x)sin(3x)log(|x|)tan(x)1/(2 + cos(x))x − cos(6x)sin(3x)/x1) 2) 3) 4)5) 6) 7) 8)Solution:Function Fill in 1-8x2− x 5exp(−x) 4sin(3x) 7log(|x|) 6tan(x) 21/(2 + cos(x)) 3x − cos(6x) 8sin(3x)/x 1Problem 3) Matching problem (10 points) No justifications are needed.Match the following functions with their derivatives.Function Fill in the numbers 1-8graph a)graph b)graph c)graph d)graph e)graph f)graph g)graph h)a)b) c) d)e) f) g) h)1) 2) 3) 4)5) 6) 7) 8)Solution:Function Fill in the numbers 1-8graph a) 3graph b) 5graph c) 4graph d) 2graph e) 6graph f) 7graph g) 1graph h) 8Problem 4) Functions (10 points) No justifications are neededMatch the following functions. In each of the cases, exactly one of the choices A-C is true.Function Choice A Choice B Choice C Enter A-Cx4−1x−11 + x + x2+ x31 + x + x21 + x + x2+ x3+ x42xe2 log(x)ex log(2)2e log(x)sin(2x) 2 sin(x) cos(x) cos2(x) − sin2(x) 2 sin(x)1/x + 1/(2x) 1/(x + 2x) 3x/2 1/(x + 2x)ex+2exe22ex(ex)2log(4x) 4 log(x) log(4) log(x) log(x) + log(4)√x3x3/2x2/33√xSolution:A,B,A,B,A,C,AProblem 5) Roots (10 points)Find the roots of the following functionsa) (2 points) 7 sin(3πx)b) (2 points) x5− x.c) (2 points) log |ex|.d) (2 points) e5x− 1e) (2 points) 8x/(x2+ 4) − x.Solution:a) The function is zero if 3x is an integer. The solutions aren/3 , where n is an integer.b) The function is zero if x is zero or if x4= 1. The later has the solutions x = 1, −1.The roots are0, 1, −1 .c) log(x) = 0 if x − 1. Therefore, the root isx = 1/ e . You might have been tempted totry x = 1 which gives log(e) = 1.d) e5x= 1 forx = 0 .e) One solution is x = 0. We can factor x out. We need to solve then 1 = 8/(x2+ 4)which means x2+ 4 = 8 or x2= 4. We have two more solutions, in totalx = 2, x = −2 .Problem 6) Derivatives (10 point s)Find the derivatives of the following functions:a) (2 points) f(x) = cos(3x)/ cos(10x)b) (2 points) f(x) = sin2(x) log(1 + x2)c) (2 points) f(x) = 5x4− 1/(x2+ 1)d) (2 points) f(x) = tan(x) + 2xe) (2 points) f(x) = arccos(x)Solution:a) Use the quotient rule−3 sin(3x) cos(10x) + 10 sin(10x) cos(3x)cos2(10x).b) Use the product rule2 sin(x) cos(x) log(1 + x2) + sin2(x)2x/(1 + x2) .c) 20x3+ 2x/(x2+ 1)2d) 1/ sin2(x) + ex log(2)log(2).e) Use the chain rule on cos(arccos(x)) = x. This is −sin(arccos(x)) arccos′(x) = 1 Sincesin(x) =q1 − cos2(x) we get the derivative −1/√1 − x2.Problem 7) Limits (10 points)Find the limits limx→0f(x) of the following functions:a) (2 points)