SOLUTIONS TO MIXED REVIEW

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1 Math 16: SOLUTIONS TO MIXED REVIEW R1.. Your graphs should show: (a) downward parabola; simple roots at x = ±1; y-intercept (, 1). (b) downward parabola; simple roots at, 1; maximum at x = 1/, by symmetry. (c) cubic, simple roots at, ±, 1, to as x (d) sinusoidal; amplitude 1 (max=1, min=-1); one period when x π, that is x π/; roots at x = nπ, that is x = nπ/. (e) sinusoidal; amplitude 1 (max=1, min=-1); one period of cosine shape when π x π, that is x 4; roots at π x =..., π/, π/, π/,..., that is x =..., 1, 1,,...; y-intercept (, 1). R1.. t Ra.. (a) quartic with 1 simple roots at x = 1, 1 cubic roots at x =, to + as x ±. So when x (, ] [1, ). (b) cubic with simple roots at x =, ±1 (factor first), to as x. So < when x (, 1) (, 1). (c) cubic with simple roots at x =, 1, 4 (factor first), to as x. So < when x (, ) (1, 4). (d) graph 1/x and the line y = to see that 1/x < when x (, ) (1/, ). Rb.. (a) upward parabola with roots at x ± a; x a or x a; x ( a, a). (b) Absolute value function shifted down, with roots at x = ±a; x > a or x < a; a < x < a. R.. (a) c v, use graph of negative parabola to find v [ c, c] (b) Graph should show correct domain, intercepts (±c, ), (, Ac), It has infinite slopes at ±c since f = A c v c + v c va c for v c, for example, so it looks like a stretched square root function. R.. (a) cubic with double root at, simple root at r, to as r. (b) upward quadratic with roots at x = ±r (c) domain t (, ), roots at (, ), infinite slope at root. (d) asymptotes at x = (n + 1)π/ (odd multiples of π/), roots at x = nπ, periodic with period π, show correct asymptotic behaviour on either side of asymptote. R4a.1. Your graphs should show (a) Domain x 5, that is x 5; shifted square root graph; infinite slope at x = 5 must be clearly visible. (b) Use that x = x. (c) Shifted graph of x to the left, with intercept at x =. (d) Negative cosine with period shifted up by one and then halved in magnitude. So f(x) 1 (max 1, min ), with intercepts when cos(πx) = 1, that is, πx = nπ, that is, x = n. Maximum value of 1 at all odd integers values of x. R4b.1. (a) f (x) = lim f(x+h) f(x) h. Your sketch should show a generic graph of y = f(x), the points (x, f(x)) (x + h, f(x + h)), and a triangle showing slope of secant line. Draw a sketch that illustrates its geometrical significance. (b) f (x) = lim lim (x+h) 1 x 1 ((x+h) 1) (x 1) h = lim = 1 x 1 x 1. (x+h) 1+ x 1 = (c) f (x) = 8x (use Pascal s Triangle to foil (x + h) 4 ) h( (x+h) 1+ x 1) = lim h h( (x+h) 1+ x 1) = R4b.. (a) The change in cost per change in yards produces. in $/yards (b) That when 1 yards have been produced, it will cost 9 per yard for every additional yard produced. R5.1.., : f (x) =. 1: B (y) = 6cy 7, 6: g (t) = 4 t 1 t 5/6 1

2 R5.. (a) f(x) = x, a = (b) f(x) = 1/x, a = 1 (c) f(x) = sin(x), a = (d) f(x) = cos(x), a = R6.1. (a) Your graph should show double root at r = and simple root at r = r. Negative leading coefficients v as r. (b) (, ), (r /, 4r/7) R6.. Volume is increasing for t (, ) (9, 11), decreasing for t (6, 8) (why round and not square brackets?) After minutes water is entering at 4/ gal/min. The graph of the rate of change is piecewise constant, with holes at t =,, 6, 8, 9. make sure its correctly labeled. sin x R6.. You ll need formulas for cos(a + b) and the special limits lim x x cos x 1 = 1, lim x x =. Use algebra to find cos(x + h) cos x cos(x) cos(h) sin(x) sin(h) cos x lim = lim h h [ = lim cos(x) cos(h) 1 h sin(x) sin(h) ] = sin(x) h R7.1. f (x) = x(a + x)/(a + x) = at x =, a. f (x) = (a )/(a + x) = nowhere. R7.. (a) y = g(x) + xg (x) (b) y = (g(x) xg (x))/g (x) (c) y = (xg (x) g(x))/x R7.. Obtain the graphs in steps: first reflect y = e x about y-axis, then divide by and reflect about x-axis, the add 1. Your graph should show: asymptote at y = 1 intercepts (1/, ) and ( ln, ), y as x, concave down throughout. R8.1. (a) f(x) = x sin(x ), (b) g (x) = cos x sin x (c) g (t) = tan(t) + (t 5) sec (t) (d) h (s) = 1(s + ) 7 (s + ) 5 (5s + s + 6) R8.. see book R8.. (a) 1 (b) + R8.4. (a) y = sec ( x)/( x) R11.1. At the points (, ) and (, 1/). R11.. y = (x y cos(xy))/(x cos(xy) + 1) (b) y = sec (x)/( tan(x)) R11.. (a) f(x) = x, a = (b) f(x) = 1/x, a = 1 (c) f(x) = sin(x), a = (d) f(x) = cos(x), a = R1.1. see book. R1.. (a) change in oxygen solubility per change in water temperature, in (mg/l)/ o C. (b) S (16).5 (mg/l)/ o C. At 16 o C, the solubility decreases by.5 mg/l for every degree increase in temperature. R1.1. You can use your graping calculator to check your answers. Some special notes: (b) Use the fact that this is an upward parabola with roots at t =,. (c) This is a shift of y = x to the right. Clearly note intercept and infinite slope at 5. (d) y = x. (e) Shift of y = x to the left. Clearly note intercepts. (f) Rewrite as function defined piecewise. Using x = x if x and x if x < get that f(x) = if x and x if x <. (i) Upper part of circle. (j) First graph y = 1 x, then reflect negative values across x-axis. (k) Clearly mark period, range, intercepts. (l) Start with y = cos(πx) then reflect, shift, compress. (m) Know how to find intercepts, asymptotes. (n) important simple function! (m) shift of (n) R1.. (7π/6, /) and (11π/6, /) r R R14.1. (a) P (R) = E (R+r) (b) g (s) = 1/ x + 1 df R14.. (a) (a) dl = 1 T L ρ df dt = 1 4L T ρ s 1/ 1 s / = 4s s (s 1) (c) f (x) = df dρ = 1 T 4L ρ /

3 (b) Pitch decreases if string is lengthened (by sliding finger up the bar, for example). Pitch increases if tension T is increased. Pitch decreases if density ρ is increased by switching to a thicker (lower) string. R : 6. g(x) L(x) = 1 + x/..95 = g(.5) L(.5) = 1.5/. 1.1 = g(.1) L(.1) = 1 +.1/..9: 41. Use that for x 1, f(x) 5 + (x 1). Thus f(.9) 4.8 f(1.1) 5.. Linear approximation estimates are too large since function is concave down at x = 1 (because f (x) < ). Draw a picture! R15.. f (x 4)(7x 8) (x) =. f (x) = or undefined at critical points x =, 4, 8/7. Comparing 5x 1/5 function values at critical points and endpoints, f( 1) = 5, f() = f(4) =, f(8/7) 9.1 (using calculator), find that absolute max is 5, absolute min is. To graph, use that f(x) everywhere (why?), the values you found above, and the fact that the slope at x = is infinite. f has a cusp at x =. R15.. (a) change in cost of loan per change in interest rate (b) If interest rate increases by 1%, the total cost of repaying the loan will increase by 1. (c) f (r) > always since larger interest rates will yield larger total cost. R15.1. Clearly show intercepts, linear behaviour, behaviour at. R15.. f(x) = tan x 1 + (x π/4). Graph should clearly show tangential behaviour at (π/4, 1). R15...7, 9. For better understanding, here we find absolute max/min over all real values, N R: Y (N) = at N = ±1 with Y (1) = k/, Y ( 1) = k/. Y is increasing on ( 1, 1) and decreasing on (, 1) (1, ). Furthermore Y (N) =. Putting all this together lim N ± gives a plot that shows that Y has an absolute max at N = 1, with max value of k/. R : For 9,4, See book. 46: y = /(e u + e u ). R17.. f (x) = cos x(sin x + 1) = at critical points x = π/, π/. Comparing values at critical points and endpoints, f() = 1 = f(π), f(π/) =, f(π/) = find maximum and minimum values of and, respectively. R17...9, 8: F F (R) R = 4kR R so F F 4KR R kr 4 = 4 R R A 5% increase in radius will yield a % increase in blood flow. More importantly, a 5% decrease in radius will yield a % decrease in blood flow! R18.1. Find linearization L(x) of f(x) = x about x = to obtain f(x) L(x) = 1 + x if x. Us this to approximate 4.9 = f(.5) L(.5) = 1.5 =.75 R18.. Chapter, Review:. f (x) = x = at x = 1 x. Comparing values at critical points and endpoints, f( 1) =, f(1) =, f( ) =, find that abs max is, abs min is. Chapter, Review: 6. f (x) = cos x(1 sin x) = at x = π/6, π/, 5π/6. Comparing values at critical points and endpoints, f() = 1, f(π/6) = 1.5 = f(5π/6), f(π/) = 1, f(π) = 1. find that abs max is 1.5, abs min is 1. R18.. (a) This is a continuous function on a closed interval, with critical points at x = π/6, 5π/6. Values at critical points and endpoints f( π) = π, f(π/6) = π/6 +, f(5π/6) = 5π/6, f(π) = π, and approximate values of pi, 1.7 is enough to sketch a rough graph clearly showing abs max and min. (b) First sketch an approximate graph: quartic, positive leading coefficient, triple root at 1, simple root at. Then heed the hint! to find f (x) = (x 1) (4x 1), f (x) = 6(x 1)(x 1), giving you inflection points at (1, ) and (1/, 1/16), abs min at (1/4, 7/56). Clearly label and finish graph.

4 R18.4. Start with a rough sketch: cubic with negative leading coefficient (since k > ), double root at zero, single root at r. Using product rule and factoring (not foiling!) find v (r) = kr(r r) giving you a local maximum (based on your sketch) at r = r / with value f(r /) = k4ro/7. R18.5. v (L) = K( 1 C + C L )/ ( ) L C + C L = when L = C. Other critical point L =. v increases for L > C and decreases for < L < C. We deduce that v has an absolute minimum at L = C. R18.6. Minimize the distance squared from the point (, ) to the point (x, (x)): dsq(x) = (x ) + x. Since this is an upward parabola, the minimum occurs at the one point where dsq (x) =, that is, x =.5. Answer to question: point (.5,.5) R , 67. See book. R1...7: 14. Surface area S = x + 4xh = x + 4 x since x h =. Here x >. A graph obtained by adding y = x and y = c/x shows that there is one minimum for x >. It is obtained when S (x) = which gives x = 4cm, h = cm. R.1. (a) y = sin(x) + x cos(x) (b) y = (x + x 1)/(x + 1) (c) y = y/(sec y x) R.. 1.1, 49: See book. 5: clearly show linear behaviour and behaviour near breakpoints. R.. (a) 5 x + C (b) tan θ + θ + C R.1. Review Chapter 4, 1: (a) (b) R.. 6.: 7. (a) f (x) = e x (x )/x, increasing on (, ) (, ) decreasing on (, ) (b) f (x) = e x (x 4x + 6)/x 4, concave up (, ) (, ) R4.1. (a) v(t) = t 18t+15 (b) t = 1, 5 (c) t (1, 5) (d) t (, 1) (5, ) (e) s(8) s() = 56ft (f) s(8) s(5) + s(5) s(1) + s(1) s() = = 1ft R4.. P () =, Q () = /8, C () = 6 R5.1. (a) y = tan x sec x (b) y = (y 4 + x y + xy 1)/( 4xy ) (c) y = (x y cos(xy))/(x cos(xy) + y) R5.. 1 R6.1. (a) y = 1 + u is the tangent line to y = e u at u =. Since e u is concave up, the tangent line lies below. y = 1 + (e 1)u is the secant line through y = e u at u = and u = 1. Since e u is concave up, this secant line lies above. (b) In (a), replace u by x /4. (c) Integrate the inequality f(x) g(x) h(x) over [, 1] in (b) to get 1 f(x) dx 1 g(x) dx 1 h(x) dx which results in the given bounds. R6.. (a) t (b) 4 (c) line y = t with hole at t = (d) f(t) = t, a = R6.. (a) h (b) 6 (c) quadratic y = 6 + 1h + h with hole at h = (d) f(x) = x, a = R7.1. (a) f (x)/( f(x)) (b) f ( x)/( x) (c) f(x)/( x) + xf (x) (d) f (g(x))g (x) (e) f (f(x))f (x) (f) g(x)f (x) f(x)g (x) g (x) 1 f(x)/g(x) R7.. (a) IVP: V (t) = r(t), V () = (b) Using Change Theorem V (9) = V () + 9 r(t) dt (c) Solve (a) to get V (t) = (/)t / cos(πt) +, so V (9) = (/)7 + + = 484. (Alternative: evaluate integral in (b)) R7.. 6., :. 4:. 8:. 9: :. R8.1. x(x ) + C R8.. (a) 88/m (b) 16/m R8.. Review Ch4, 4: F (x) = x + sin x 6: g (x) = cos x 1+sin 4 x 8: y = sin((x + 1) 4 ) sin(4x ) R π/4 4

5 R9.. (a) 6 (b) 14 (c).718 R9.. Ch4 Review, 1: T 5 /5 4T + 7T. 1: 1/1. 14: 1/1. 16: (/9)(8 / 1). 18: /(π) R9.4. (a) True. (b) True. (sub u = x ) R9.5. Ch4 Review, :. : x + 4x + C. 4: sin(cos(x)) + C. 6: (1 + tan t) 4 /4 + C. 8:. R.1. (a) water flowed in at t (6, 1). water flowed out at t (, 4) there was no flow at t (4, 6) (b) 4 gallons (c) 4 gallons (d) gallons R.. Ch4 Review, 4: π sin x dx = R.1. (a) g (s) = s /(1 + s ) (b) s (t) = cos(t ) (c) f (x) = cos x/(1 + sin 4 x) R.. (a) π R R ( R y ) dy = π R (R y )dy = 4 πr R.. π/4 (b) π R/ R/ ( R y ) (R/) dy = π R/ (R /4 y )dy = πr (c) /8 65% R.4. Ch4 Review, 8: (a) / (b) (c) sin(x/) cos(x/) (Explain!) R4.1. (a) (P (t+ t) P (t))/ t: change in population per change in time (also referred to as average growth rate over given time interval), in bees/hour. P (t): instantaneous growth rate, in bees/hour. If P () = 1, then at t = the population is increasing at 1 bees/hour. (eg, at t = 1 there will be approximately 1 more bees. (b) (S(r + r) S(r))/ r: change in surface area per change in radius (also referred to as rate of change of surface area with respect to changes in radius), in cm /cm S (r): instantaneous rate of change, in cm /cm If S (1) = 8π, then at r = 1 the surface are is increasing at 8πcm per every cm increase in radius. (eg, at r = 11cm the surface area will be approximately 8πcm larger than at r = 1cm.) (c) S S (1) r = 16π 5.7 S = S(1) S(1) = 55. (Note S(r) = 4πr ). R4.. (a) Midpoint rule: P (4) P () 61. (b) P (4) 7 R4.. The increase in child s weight between ages 5 and 1. R4.4. (a) f(x) = 4x tan x + C (b) f(u) = u / + u + 1/ R5a.1. (a) (b) 9 R5a.. (a) displacement between t = 6 and t = 1. (b) distance travelled between t = 6 and t = 1. (b) change in veloicty from t = 6 to t = 1. R5a.. Min 11, Max 64 R5a.4. (a) Total population at 15 weeks. (b) da/dx is in (lb/ft)/ft = lb/ft. (lb/ft)ft = lb. 8 a(x) dx is in R5b.1. f(x) L(x) = 1 x/,.9 = f(.1) L(.1) =.95,.99 = f(.1) L(.1) =.995. Your graph should clearly show domain, intercepts, and infinite slope at x = 1. R5b.. Volume of paint = V (5) V (4.9995) = V V (5) r where V (r) = πr is the volume of a half-sphere of radius r. So V π m. That is approximately 519 gallons! R5b.. (a) 8.8 m (using 8 intervals) (b) 91. m (using 4 intervals) R5b.4. (a) V V (r) r where V = πr h is the volume of a cylinder of height h, radius r. The change in volume is V πrh r. (b) V = V (r + r) V (r) = πrh r + πh r R5b.5. Use the substitution u = 1 x. 5

6 R6.1. Let f(x) = x. Find linearization about x =. Get L(x) = 1 + x. L(.5) = 1.5 =.975. R6...9, 8: F F (R) R = 4kR R so 4.9 = f(.5) F F 4KR R kr 4 = 4 R R A 5% increase in radius will yield a % increase in blood flow. More importantly, a 5% decrease in radius will yield a % decrease in blood flow! for x (, 1) R6.. (a) Use that F (x) = f(x) to get that f(x) = for x (1, ) 1 for x (, ) (b) f av = 1 f(x) dx = F () = 1 (c) F av = 1 F (x) dx = R : 44. a / R6.5. (a) family of functions. a unique function. a number. (b) none. 6