December 09, 20 Calculus PracticeTest s Name: (4 points) Find te absolute extrema of f(x) = x 3 0 on te interval [0, 4] Te derivative of f(x) is f (x) = 3x 2, wic is zero only at x = 0 Tus we only need to ceck x = 0 and x = 4 We see tat f(0) = 0, wile f(4) = 54 Tus tere is a maximum of 54 at x = 4 and a minimum of 0 at x = 0 2 (5 points) If f(x) = sin(2x) (cos x) 2, find f ( π 4 ) Using te quotient rule and te cain rule, we ave Evaluating at π/4 yields wic simplifies to 4 Tus f (π/4) = 4 f (x) = (cos x)2 cos(2x) 2 sin(2x) 2 cos x ( sin x) (cos x) 4 f (π/4) = ( 2/2) 2 0 2 2 ( 2/2) ( 2/2) ( 2/2) 4 3 (7 points) Sketc a function f wic meets all of te following criteria: f(x) is continuous on (, ), f( 2) = 9, f(0) = 5, f(2) = 2, f (2) = f ( 2) = 0, f (x) < 0 for x < 2, wile f (x) > 0 for x > 2 Also, f (x) < 0 for x < 0 and f (x) > 0 for x > 0 Tere is more tan one possible correct answer, but any function sketced sould be increasing on (, 2), decreasing on ( 2, 2) and increasing on (2, ) Tere is a local maximum of 9 at x = 2 and a local minimum of 2 at x = 2 Te function sould be concave down on (, 0) and concave up on (0, ), wit an inflection point at (0, 5)
4 (6 points) Wat is te domain of te function f(x) = sec(2x)? Te function f(x) = sec(2x) = / cos(2x) is defined werever cos(2x) 0 But cos(2x) = 0 exactly were 2x = π/2 + kπ, wic is were x = π/4 + kπ/2 Tus te domain of sec(2x) is {x : x π/4 + kπ/2 Some examples of numbers not in te domain include π/4, 3π/4, 5π/4, 5 (7 points) Find all relative maximums and minimums for f(x) = x 2 x2 + Differentiating, we obtain f x2 + 2x x 2 (/2)(x 2 + ) ( /2) 2x (x) = x 2 Tis expression always exists (note tat te + denominator is defined everywere and is never zero and tat te numerator is defined everywere) so to find stationary points we consider wen te numerator is equal to zero Tis is wen 2x x 2 + x3 = 0 Since x2 x 2 + is never + zero, we can clear denominators by multiplying bot sides of te above equation by it, yielding 2x(x 2 + ) x 3 = 0 Tis is satisfied wen x 3 + 2x = 0 or x(x 2 + 2) = 0, wic only occurs wen x = 0 Cecking to see if f canges sign on eiter side of x = 0, we see tat wen x < 0 we ave tat f (x) < 0, wile if x > 0, f (x) > 0 Tus tere is a local or relative minimum of 0 at x = 0
6 (5 points) Does te function y = x 3 + 2 ave any inflection points? If so, wat are tey? We must analyze y Computing, we see tat y = 3 x( 2/3) and y = 2 9 x( 5/3) Clearly y is never zero, but it fails to exist at x = 0 Note owever tat 0 is in te domain of te original function Cecking te sign of y on eiter side of zero, we see tat y > 0 for x < 0 wile y < 0 for x > 0 Tis, coupled wit te fact tat zero is in te domain of y sows tat tere is an inflection point at (0, 2) 7 (3 points) Is te following statement true or false? Explain Every function wic is continuous at x = is also differentiable at x = Tis is false For example, f(x) = x is continuous at x = but is not differentiable tere 8 (7 points) Sow ow to get te trigonometric identity for sin(2x) from te addition identity for sin(x + y) Do te same for te identity involving cos(2x) First recall tat sin(x + y) = sin(x) cos(y) + cos(x) sin(y) Tus, sin(2x) = sin(x + x) = sin(x) cos(x) + sin(x) cos(x) = 2 sin(x) cos(x) Likewise, recall tat cos(x + y) = cos(x) cos(y) sin(x) sin(y) Tus, cos(2x) = cos(x + x) = cos(x) cos(x) sin(x) sin(x) = cos 2 (x) sin 2 (x)
9 (6 points) Suppose tat f is a function satisfying f (x) = 4, wit f(0) = 2 and f() = 5 Find f Since f (x) = 4, f (x) = 4x + C and tus f(x) = 2x 2 + Cx + D for some constants C and D Since f(0) = D = 2, we ave D = 2 Since f() = 2 + C + D = 4 + C = 5, we must ave C = Tus f(x) = 2x 2 + x + 2 0 (6 points) Evaluate x 2 sin x 3 dx We will use substitution wit u = x 3 Ten du dx = 3x2, so du 3 3 = x2 dx Substituting, we ave sin u du = cos x3 ( cos u) + C = + C 3 3 (7 points) Use te definition of derivative to sow tat te derivative of f(x) = /x is f (x) = /x 2 No credit will be given if you don t use te definition of derivative Te definition of derivative says tat f f(x + ) f(x) (x) = lim Applying tis definition to te function f(x) = /x 0 yields: lim 0 x+ x If we simplify te numerator of tis expression by finding a common denominator and subtracting, we obtain lim 0 x (x + ) (x)(x + )() = lim 0 and tis last limit is clearly equal to /x 2 Tus f (x) = /x 2 (x)(x + )() = lim 0 (x)(x + ),
2 (5 points) Find f (x) for f(x) = sin(x 2 + ) tan(x) You do not need to simplify f (x) = sin(x 2 + ) sec 2 (x) + tan(x) cos(x 2 + ) 2x 3 (7 points) For f(x) = x 3 + x, find and simplify f(2+) f(2) We ave f(2+) f(2) equal to = (2+)3 +2+ 0 If we expand (2 + ) 3 it becomes 8 + 2 + 6 2 + 3 Tus our difference quotient is 8 + 2 + 6 2 + 3 + 2 + 0 = 3 + 62 + 3 If we factor an out of te numerator and cancel wit te factor in te denominator, we ave 3 + 6 + 2 4 (8 points) Prove rigorously tat lim x 4 3x = Note: te following is te polised proof, wic is given after some work on scratc paper to figure out te proper value of δ Let ɛ > 0 be given Let δ = ɛ/3 Ten if 0 < x 4 < δ, we would ave x 4 < ɛ/3 and tus 3x 2 < ɛ Tis can be written as (3x ) < ɛ, wic as te form f(x) L < ɛ were f(x) = 3x and L = Tus we ave sown tat for every ɛ > 0, tere exists a δ > 0 so tat f(x) < ɛ wenever 0 < x 4 < δ
5 (5 points) Find any oblique (slant) asymptotes for f(x) = 2x2 +4 x Since f can be written as f(x) = 2x + 4 x, and since lim x 4 x = 0, we must ave tat y = 2x is an oblique asymptote 6 (5 points) Compute ( x + x) 2 dx Expand te integrand Tis yields x 2 + 2 + x 2 dx = x3 3 + 2x x + C 7 (7 points) Solve te equation 2 sin(x) + = 0 for x Tis equation would be valid werever sin(x) = /2 Using our cart and our memory, we recall tat tis occurs at te point directly across from te point associated wit x = π/6 Tus one solution of tis equation is π + π/6 = 7π/6 Anoter solution would be at te oter point on te unit circle wic intersects te line y = /2, namely te point associated wit π/6 Tus te set of all solutions is {x : x = 7π/6 + 2kπ} {x : x = π/6 + 2lπ}, were k and l are any integers
8 (5 points) Wen a certain company produces x widgets, it finds tat te demand is given by p = 30000 x Wat is te 0000 marginal revenue wen x = 0, 000? Te revenue function is given by Tus R(x) = xp = x 30000 x 0000 Wen x = 0, 000, we compute tat R (0, 000) = = 0000 (30000x x 2) R (x) = (30000 2x) 0000 9 (6 points) For some coice of f(x), a and b, te quantity is equal to b a lim n n i= f(x) dx Find a suitable suc f(x), a and b (( sin + 2i ) ) + 2 n n It appears as toug x = 2 n and tat x i = + i x Tus it appears tat we could let a =, b = 3 and f(x) = sin(x + ) 20 (5 points) Find te average value of f(x) = x(x + ) over te interval [0, 4] We need to compute 4 0 Using te Fundamental Teorem, we ave 4 0 x(x + ) dx = 4 4 0 x 3/2 + x /2 dx [ 2 4 5 x5/2 + 2 ] 4 3 x3/2 = ( 64 0 4 5 + 6 ) = 68 3 5