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v2.0 Solutions* to the Review Problems for the Final Exam, Math 151 Disclaimer: We believe the answers to be correct, but we assume no responsibility to any damage due to possible errors and/or typos. 1. For what values of B is the following function continuous at x = 1? 3x − x − Bx, if x > 1; f(x) = Bx − 2, if x ≤ 1. Solution to 1. The limit from the left (at x = 1), lim x→1− f(x) is 3 ▯ (1) − 1 − B ▯ 1 = 2 − B, whereas the limit from the right lix→1 + f(x) is B▯1−2 = B−2. In order for f(x) to be continuous at x = 1 we set these equal and solve. 2 − B = B − 2 getting B = 2. Answer to 1. B = 2 2. Find the equation of the tangent line to the curve y = x−1 at each point where the tangent is x+1 parallel to the line x − 2y = 2. Solution to 2. The line x − 2y = 2, can be written as y = x/2 − 1, hence its slope is 1/2. Now the slope of the tangent to the curve y = x−1 is x+1 dy (x − 1) (x + 1) − (x − 1)(x + 1) 1 ▯ (x + 1) − (x − 1) ▯ 1 2 = 2 = 2 = 2 . dx (x + 1) (x + 1) (x + 1) Setting this equal to 1/2 yields that the points we are interested in satisfy 2 1 2 = . (x + 1) 2 Solving this we get (x+1) −4 = 0 hence, x +2x−3 = (x+3)(x−1) = 0, whose roots are x = −3 x−1 and x = 1. Plugging-in into y = x+1, gives, when x = −3 y = 2, and when x = 1, y = 0. So the points we need are (−3,2) and (1,0). Answer to 2. The tangent at the point (−3,2) is (y − 2) = (1/2)(x + 3), which is x − 2y = −7, and the tangent at the point (1,0) is (y − 0) = (1/2)(x − 1), which is x − 2y = 1. 3. Find the equation of the tangent line to the curve deﬁned by the equation ln(xy)+2x−y+1 = 0 1 at the point (2,2). Solution to 3. First simplify to get lnx + lny + 2x − y + 1 = 0 . By implicit diﬀerentiation 1 y′ ′ + + 2 − y = 0 . x y * Written by D. Zeilberger. Adapted by F. Knop. 1 Now plug-in x = 1/2 and y = 2, to get 1 y′ ′ + + 2 − y = 0 , 1/2 2 ′ ′ getting 4 = y /2, and so the slope y (at the designated point) is 8. The equation of the tangent line at the designated point ( ,2) is (y − 2) = 8(x − ), which is y = 8x − 2. 2 2 Answer to 3. y = 8x − 2 4. Find the following limits: 2 3x − 7 sin3x a) lim (x − 2x + 1) b) lim√ 2 c) lim x→−1 x→∞ x + 2 x→0 2sin5x x −x d) lim 3e + 4e e) lim |x − 8| . x→∞ 5e + 4e −x x→8 − x − 8 Solution to 4. (a) Plugging in we get (−1) − 2(−1) + 1 = 4. Answer to 4a) 4 (b) “forgetting about the little ones” we get 3x − 7 3x2 3x 2 lim √ 2 = lim √ = lim = lim 3x = ∞ . x→∞ x + 2 x→∞ x 2 x→∞ x x→∞ Answer to 4b) ∞ (c) When the argument of sin goes to 0, you can replace sinw by w, hence sin3x 3x 3 3 lim = lim = lim = . x→0 2sin5x x→0 2(5x) x→0 10 10 Answer to 4c) 3/10 x d) It is better to replace e by another variable, let’s call it y. x −x lim 3e + 4e = lim 3y + 4/y x→∞ 5e + 4e −x y→∞ 5y + 4/y Now ‘forgetting about the little ones’ this equals 3y 3 y→∞ 5y = 5 . Answer to 4d) 3/5 2 e) Recall that |z| = −z for z < 0, so |x − 8| = −(x − 8) for x < 8. We have lim |x − 8|= lim −(x − 8) = lim (−1) = −1 . x→8− x − 8 x→8− x − 8 x→8− Answer to 4e) −1 5. Find the following limits √ sin x (lnx)2 a) lim xlnx b) lim x c) lim . x→0+ x→0+ x→∞ x Solution to 5 a) We use l’Hopital: ′ √ lnx (lnx) 1/x lim+ xlnx = lim + −1/2 = lim+ −1/2 ′= lim + −3/2 = x→0 x→0 x x→0 (x ) x→0 (−1/2)x 1/2 √ √ lim x = lim −2 x = lim −2 0 = 0 . x→0+ (−1/2) x→0+ x→0+ Answer to 5a) 0 sin x ln x b) We must ﬁrst take the ln, getting ln() = sinxlnx, and writing it as a quotient we csc x Taking the limit of this we have: lnx (lnx)′ 1/x lim = lim = lim = x→0+ cscx x→0+ (cscx)′ x→0+ −cotxcscx tanxsinx sinxsinx x2 lim = lim = lim = lim x = 0 . x→0+ x x→0 + (cosx)x x→0+ x x→0+ Now the limit of the original is the exponential of this Answer to 5b) e = 1 Shortcut: When x is near 0, sinx can be replaced by x, so we get the simpler limix→0imx x whose ln is xlnx, and the limit ′ lnx (lnx) li+ xlnx = lim + −1 = lim+ −1 ′= x→0 x→0 x x→0 (x ) 1/x lim = lim (−x) = 0 . x→0+ −x −2 x→0 + and we get quicker Answer e = 1. c) By l’Hopital again: (lnx)2 ((lnx) ) 2(lnx)(1/x) 2lnx lim = lim = lim = lim . x→∞ x x→∞ x′ x→∞ 1 x→∞ x 3 Using l’Hopital once again, we have that this equals ′ 2 lim (lnx) = 2 lim 1/x = 2 lim 1 = 0 . x→∞ x′ x→∞ 1 x→∞ x Answer to 5c) 0 6. Find y in each case. Do not simplify your answer. 7 1 5x 2 2tanx a) y = x − 3x + 6 − 4 b) y = x sin(x ) c) y = √ 2 x 1 − x 4 6 3 xy+3y 1/x d) x y + 5y x = 8 e) xe = y f) f(x) = (1 + 2x) . Solution to 6. a) 1 4 y = (x − 3x + 6 − ) = (x − 3x + 6 − x −4 ) = 7x − 3 − (−4)x −5 = 7x − 3 + . x4 x5 6 4 Answer to 6a) 7x − 3 + x5 b) y = (x 5xsin(x )) = (x ) sin(x ) + x (sin(x )) = (e 5x ln) sin(x ) + x (sin(x )) = (5xlnx) (e 5x ln)sin(x ) + x (2x)cos(x ) = (5lnx + 5)(e 5x ln)sin(x ) + x (2x)cos(x ) = 5x 2 5x 2 (5lnx + 5)x sin(x ) + x (2x)cos(x ) . 5x 2 5x 2 Answer to 6b) (5lnx + 5)x sin(x ) + x (2x)cos(x ) c) √ 2tanx (2tanx) ′ 1 − x − (2tanx)((1 − x )2 1/2)′ y = ( √ ) = √ = 1 − x2 ( 1 − x )2 √ 2 2 2 −1/2 2 1 − x sec x − (2tan x)(1/2)(1 − x ) ▯ (−2x) 2 = 1 − x √ 2 2 −1/2 2 2 1 − x sec x + 2x(1 − x ) tanx 2sec x 2xtanx 2 = √ 2 + √ 3 1 − x 1 − x 1 − x 2 2 sec x 2x tan x Answer to 6c) √ 1−x2 + √ 23 1−x d) 4 6 3 ′ (x y + 5y x ) = 0 , 4 so 4 ′ 4 ′ 6 3 ′ 6 ′ 3 (x ) y + x y + 5y (x ) + 5(y ) x = 0 , 3 4 ′ 6 2 5 ′ 3 4x y + x y + 5y (3x ) + 5(6y y )x = 0 , 4x y + x y + 15y x + 30y y x = 0 ′ 3 . Moving all the terms not involving y to the right gives x y + 30x y y = −4x y − 15y x 6 2 . 4 3 5 ′ 3 6 2 (x + 30x y )y = −4x y − 15y x . Hence 3 6 2 ′ −4x y − 15y x y = 4 3 5 x + 30x y Since we are not supposed to simplify, we have ′ −4x y−15y x6 2 Answer to 6d) y = x +30x y 5 e) xy+3y ′ ′ (xe ) = y , ′ xy+3y xy+3y ′ ′ x e + x(e ) = y , exy+3y + xe xy+3y ▯ (xy + 3y) = y ′ , exy+3y + xe xy+3y ▯ (x y + xy + 3y ) = y ′ , xy+3y xy+3y ′ ′ e + xe ▯ (y + (x + 3)y ) = y , xy+3y xy+3y ′ xy+3y ′ e + xye + (x + 3)xy e = y (1 + xy)e xy+3y = y (1 − x(x + 3)e xy+3y ) Hence exy+3y (1 + xy) y = . 1 − x(x + 3)e xy+3y xy+3y Answer to 6e) y = ′ e (1+xy+3y 1−x(x+3)e f) ′ ln(1 + 2x) ′ (lnf(x)) = ( x ) ′ ′ ′ f (x) = (ln(1 + 2x)) x − ln(1 + 2x) ▯ x = (2x/(1 + 2x)) − ln(1 + 2x) f(x) x 2 x 2 Hence (2x/(1 + 2x) − ln(1 + 2x) 2x/(1 + 2x) − ln(1 + 2x) f (x) = f(x) = (1 + 2x) 1/x . x2 x 2 5 2x/(1+2x)−ln(1+2x) Answer to 6f) (1 + 2x) 1/x x2 7. State the formal deﬁnition of the derivative of the function f(x). Use the deﬁnition to calculate ′ √ f (x) for f(x) = 3 − 5x. Solution to 7. The formal deﬁnition of the derivative of a function f(x) is ′ f(s) − f(x) f (x) =s→xm s − x . Note: There is another version with h and lim, but for this problem it is more convenient to use h→0 this one. √ √ √ √ √ √ f (x) = lim 3 − 5s − 3 − 5x = lim ( 3 − 5s − 3√− 5x)( 3 √ 5s + 3 − 5x) s→x s − x s→x (s − x)( 3 − 5s + 3 − 5x) ( 3 − 5s) − ( 3 − 5x) 2 (3 − 5s) − (3 − 5x) = lim √ √ = lim √ √ s→x (s − x)( 3 − 5s + 3 − 5x) s→x (s − x)( 3 − 5s + 3 − 5x) −5(s − x) = lim √ √ = s→x (s − x)( 3 − 5s + 3 − 5x) −5 −5 −5 s→xm√ √ = √ √ = √ . 3 − 5s + 3 − 5x 3 − 5x + 3 − 5x 2 3 − 5x −5 Answer to 7. 2√3−5x (but it had to be derived from the deﬁnition!). √ 8. Suppose that S(x) = x for x ≥ 0, and let f and g be diﬀerentiable functions about which the following is known f(3) = 2, f (3) = 7, g(3) = 4, g (3) = 5 . Compute the following f ′ f ▯ g ′ (f + g) (3) (f ▯ g) (3), (3), (S ◦ g) (3), (3) . g f − g Solution to 8. (f + g) (3) = f (3) + g (3) = 7 + 5 = .2 (f ▯ g) (3) = f (3) ▯ g(3) + f(3) ▯ g (3) = 7 ▯ 4 + 2 . 5 = 38 ′ ′ ′ f f (3)g(3) − f(3)g (3) 7 ▯ 4 − 2 ▯ 518 9 g (3) = g(3) = 42 = 16 = 8 . ′ ′ ′ ′ −1/2 (S ◦ g) (3) = S (g(3))g (3) = S (4) ▯ 5 = 5 ▯ (1/2)(= 5/4 . ′ f ▯ g (3) = f − g 6 (f ▯ g) (f − g) − (f ▯ g)(f − g)′ (3) = (f − g) 2 (f g + fg )(f − g) − (f ▯ g)(f − g )′ (3) = (f − g) 2 f g − f g′ 2 (3) = (f − g) 2 2 ′ ′ 2 f(3) g (3) − f (3)g(3) = (f(3) − g(3)) 2 2 2 2 ▯ 5 − 7 ▯ 4 = −23 . (2 − 4)2 Answer to 8. 12, 38, 9/8, 5/4, −23 ′′ ′ 9. Suppose f (x) = −3x + cos(πx) and f(1) = 2 and f (1) = −1. What is f(5)? ′ 2 Solution to 9. The general antiderivative of −3x + cos(πx) is f (x) = −3x /2 + sin(πx)/π + A, ′ 2 where A is a general constant. To ﬁnd A, we plug-in x = 1, and get f (1) = −3▯1 /2+sin(π)/π+A = −3/2 + 0 + A. On the other hand, in this speciﬁc problem, we are told that f (1) = −1, so we get the equation −1 = A − 3/2, and solving for A, we get A = 1/2. So f (x) = −(3/2)x + sin(πx)/π + 1/2 . To get f(x), we take the antiderivative of the above, getting f(x) = −(1/2)x −cos(πx)/π +x/2+B, where B is a constant yet to be determined. Plugging-in x = 1 into this yields f(1) = −1/2 − 2 2 cos(π)/π + 1/2 + B = 1/π + B. On the other hand, in this speciﬁc problem, we are told that f(1) = 2, so we get the equation 2 = 1/π + B, so B = 2 − 1/π . Hence2 3 2 2 f(x) = −(1/2)x − cos(πx)/π + x/2 + 2 − 1/π 3 2 2 Finally, plugging-in x = 5, we get f(5) = −(1/2)5 − cos(5π)/π + 5/2 + 2 − 1/π = −125/2 + 2 2 1/π + 5/2 + 2 − 1/π = −120/2 + 2 = −60 + 2 = −58. Answer to 9. -58 10. A farmer with 450 feet of fencing wants to enclose the four sides of a rectangular region into four pens of equal size with fencing parallel to one side of the rectangle. What is the largest possible total area of the four pens? Solution to 10. Let’s call the length of the side parallel to the three internal walls x, and the length of the other side (perpendicular to these walls) y. We have to maximize the area. So the Goal function is xy. We also have the constraint: 2y + 5x = 450. Solving for y, we get y = (450 − 5x)/2, and plugging this into the goal function, 7 2 we express it in terms of x alone, getting f(x) = x(450 − 5x)/2 = (450/2)x − (5/2)x . To ﬁnd the maximum, we take the derivative f (x) = 450/2 − 5x, and then set it equal to 0. Solving 450/2 − 5x = 0, we get x = 45. Plugging this into f(x) we get that the maximal total area is f(45) = 45 ▯ (450 − 5 ▯ 45)/2 = 45 ▯ 5/2 = 10125/2. Ans to 10. 10125/2. Note: In some problems, they might ask you about the dimensions of the optimal rectangle. Then the answer would be 45 × (225/2). 11. A ladder of length 15 feet is leaning against a wall when its base begins to slide along the ﬂoor, away from the wall. By the time the base is 12 feet away from the wall, the base is moving at the rate of 5 ft/sec. How fast is the top of the ladder sliding down the wall at that moment? How fast is the area of the triangle formed by the ladder, the wall, and the ﬂoor changing at that time? Solution to 11. This is a related rates problem. Let’s call the height of the top of the ladder y, and the distance of the foot of the ladder from the wall x. By Pythagoras, x + y = 15 2 . Taking derivatives with respect to time (using the chain rule in its abstract setting, a.k.a. implicit diﬀerentiation), we get dx dy 2x + 2y = 0 . dt dt Right now x = 12 and dx/dt = 5. We also need y right now. Using the relation, x + y = 15 we 2 plug-in x = 12 and solve for y: 12 + y = 15 , getting y = 9. Substituting these values into the derived relation we get 2 ▯ 12 ▯ 5 + 2 ▯ 9= 0 , dt and solving for dy/dt, we get dy/dt = −20/3. Since the sign is negative, it is sliding down (which is also obvious from common sense). Answer to the ﬁrst part of 11. The top of the ladder is sliding down at a speed of 20/3 ft/sec. The area, let’s call it A, is given by the relation A = xy/2. Taking derivative with respect to time (using the product rule), we get dA 1 dy dx = x + y , dt 2 dt dt and we plug-in the above numbers, getting dA 1 dt = 2 (12 ▯ (−20/3) + 9 ▯ 5) = −35/2 , Answer to second part of 11. The area is decreasing at a rate of 35/2 square feet per second. 8 3x 12. Let f(x) = x −1 . Find the domain of the function, the intervals where f(x) is increasing or decreasing, maximum and minimum, the concavity and inﬂection points, horizontal and vertical asymptotes of the graph. Then sketch the graph of f(x). Solution to 12. Domain: {x|x ▯= −1,1}. Even though the horizontal, and vertical asymptotes are mentioned last in the statement of the problem, it is best to start with them, since they contain the most important information. horizontal asymptotes: lim f(x) = lim 3x = lim 3x = 0 . x→∞ x→∞ x − 1 x→∞ x2 So y = 0 (alias the x-axis) is a horizontal asymptote. vertical asymptotes: Set the denominator equal to 0, we get x = −1 and x = 1. Let’s consider each vertical asymptote on its own. Plugging-in a value of x slightly less than −1, say −1.01, we get −BIG, so this means that lim x→−1 − f(x) = −∞, and the curve would drop down to “hell” (−∞) as it approaches the vertical asymptote x = −1. Plugging-in a value of x slightly bigger than −1, say −.99 we get that f(−.99) = BIG, so the curve “comes down from heavens” right after x = −1. Similarly the curve goes down to hell as x approaches x = 1, and emerges from heavens right after x = 1. It is also a good idea to ﬁnd the x-intercept (solve y = 0, getting (0,0)) and the y-intercept, plugging-in x = 0, getting, once again (0,0). So the curve intersects the x-axis only at the origin (0,0). Notice that we haven’t taken derivatives yet, but we can already pretty much draw the function. When x is very negative, it is almost coincident with the x-axis (the horizontal Asymptote) but it is below the x-axis, since f(x) is negative there, and then it is dropping down-down right before x = −1. By inspection you can see that the curve is concave down at the interval −∞ < x ≤ −1 Then from x = −1 to x = 1 it is a steady decrease (since at x = −1 it is coming from y = ∞ dropping to y = −∞ at x = 1. Somewhere between these, there should be an inﬂection point, that by symmetry is (0,0). Also for x > 1 it dropping from ∞ to 0. So you can draw the curve without any calculus! Also it is clear that there are no local maxima and minima. But now we can do some calculus. By the quotient rule ′ x + 1 f (x) = −3 2 2 , (x − 1) this is always negative, so intervals of increase: none. intervals of decrease: (−∞,−1),(−1,1), (1,∞). 9 To ﬁnd potential (local) max or min. points, solve f (x) = 0, which has no roots. Hence max and min. points: none. Using the quotient rule one more time, we get ′′ 6x(x + 3) f (x) = 2 3 . (x − 1) ′′ The inﬂection point(s) are where f (x) = 0. Setting the numerator equal to 0, we get x = 0. The y coordinate is f(0) = 0. So inﬂection point: (0,0). The transition points in the life of this function are x = −1,x = 0, and x = 1. Plugging-in random values in each interval we get The interval (−∞,−1) is concave down (since. e.g. f (2) < 0). ′′ The interval (−1,0) is concave up (since. e.g. f (− ′′ 1) > 0). 2 The interval (0,1) is concave down (since. e.g. f ( ′′ 1) < 0). 2 The interval (1,∞) is concave up (since. e.g. f (2) > 0). Answer to 12. max. and min. points: none; intervals of increase: none; intervals of decrease: (−∞,−1),(−1,1),(1,∞); inﬂection point: (0,0); intervals of concave down:(−∞,−1), (0,1); inter- vals of concave up:(−1,0), (1,∞); horizontal asymptote: y = 0; vertical asymptotes: x = −1 and x = 1. 2 13. Repeat the same problem for y = 2 . x +3 Solution to 13. Horizontal Asymptotes: y = 1. Vertical asymptote: none. By the quotient rule (do it!) ′ 6x f (x) = 2 x + 3 ′′ −18(x − 1) f (x) = 2 x + 3 ′′ So potential max or min.: x = 0. Plugging into the second derivative, we get f (0) = 6 > 0 so x = 0 is a minimum. Plugging x = 0 into f(x), we get that when x = 0, y = 0, so the point (0,0) is a local minimum, and that’s the only one. To get the inﬂection points solve f (x) = 0, and get x = −1 and x = 1. When x < −1, f (x) is ′′ negative so in the interval −∞ < x < −1 the curve is concave down. In the interval −1 < x < 1 10 f (x) is positive, so the curve is concave up, and in the interval 1 < x < ∞ the curve is concave ′′ down since f (x) is negative there. Now draw it! Answer to 13. Horizontal asymptote y = 1; Vertical asymptotes: none. local min. (0,0); inﬂection points: (−1,1/4) and (1,1/4); interval of decrease: (−∞,0); interval of increase: (0,∞); intervals of concave down:(−∞,−1),(1,∞); interval of concave up: (−1,1). 14. Sketch the graph of f(x) which satisﬁes the following conditions f (1) = f (−1) = 0, f (x) < 0 ′ ′ ′′ if |x| < 1, f (x) > 0 if 1 < |x| < 2, f (x) = −1 if |x| > 2, f (x) < 0 if −2 < x < 0, inﬂection point (0,1). Solution to 14. First let’s spell everything out w/o the annoying absolute value symbol ||. For −∞ < x < −2 : f (x) = −1. For −2 < x < −1 : f (x) > 0. For −1 < x < 1 : f (x) < 0. For 1 < x < 2 : f (x) > 0. ′ For 2 < x < ∞ : f (x) = −1. Now we are almost ready to sketch, noting that at x = −2 and x = 1 we have local minima, since the derivative changes sign from negative to positive, and at x = −1 and x = 2 we have local maxima, since f (x) changes sign from positive to negative. We also have an inﬂection point at ′′ (0,1) and since f (x) < 0 between x = −2 and x = 0, it is concave down there, and hence between x = 0 and x = 2 it is concave up. To actually draw a curve ﬁtting all these speciﬁcations, draw a straight line with slope −1 coming from −∞ to the point (−2,0), then draw a hat-like curve that peaks at x = −1 and inﬂects at the point (0,1) turning into a cup-like arc that bottoms at x = 1 and goes up to x = 2 and then it suddenly becomes a straight line with slope −1. √ 15. Find the linearization of f(x) = √ x + 1 at a = 15 and use it to ﬁnd an approximation to 15 √ and 17 . Use the second derivative to determine whether the estimate is greater or less than the actual value. ′ Solution to 15. The linearization of a function f(x) at x = a is L(x) = f(a) + f (a)(x − a). In this problem f(x) = (x + 1) 1/2, so ′ −1/2 1 f (x) = (1/2)(x + 1) = √ . 2 x + 1 √ √ We have f(a) = f(15) = 1 + 15 = 16 = 4, and f ′(a) = f (15) = √ 1 = 1/8. Hence 2 15+1 11 Answer to ﬁrst part of 15. L(x) = 4 + (x − 15) 8 √ Now 15 = f(14), and plugging x = 14, we get L(14) = 4 + 1(14 − 15) = 31, hence 8 8 √ Answer to second part of 15. 15 is approximately 81 √ 1 33 Also 17 = f(16), and plugging x = 16, we get L(16) = 4 + 816 − 15) = 8, hence √ 33 Answer to third part of 15. 17 is approximately 8 f (x) = (−1/4)(x + 1) −3/2= −1 . √ 3 4 x + 1 ′′ √ √ Since f (15) = −1/256 is negative, the above approximations to 15 and 17 are larger than the actual value. Answer to fourth part of 15. Both are larger than the actual values. 16. Find the derivative of the following functions: Z Z x cos p a) F(x) = tan(s )ds b) g(x) = 1 − t dt . π 1 Solution to 16.a) Since the upper-limit of integration is identical to the argument of F, (both are x), the derivative is simply the integrand with s replaced by x. ′ 2 Answer to 16a) F (x) = tan(x ) Solution to 16b) Now the upper-limit of integration, cosx is not identical with the argument of g, which is x. We must ﬁrst use the chain rule, multiplying top and bottom by dcosx (so to speak!). d Z cos p dcos x d Z cos p g (x) = 1 − t dt = ▯ 1 − t dt . dx 1 dx cosx 1 The ﬁrst piece is a straightforward diﬀerentiation, while the second is done exactly as in a): replace ′ √ t by cosx in the integrand. Hence F (x) = −sinx 1 − cosx . Using the famous trig. identity this 2 is −sin x. ′ 2 Answer to 16b) g (x) = −sin x 17. Suppose that f is continuous on [0,4], f(0) = 1 and 2 ≤ f (x) ≤ 5 for all x ∈ (0,4). Show that 9 ≤ f(4) ≤ 21. Solution to 17. We must use the Mean Value Theorem. If f(x) is continuous in [a,b] and diﬀerentiable in (a,b) then there exists a c in (a,b) such that ′ f(b) − f(a) f (c) = . b − a 12 In this problem a = 0 and b = 4 so, there exists a c in (0,4) such that f(4) − f(0) f(4) − 1 f (c) = = 4 − 0 4 − 0 But we know that 2 ≤ f (x) ≤ 5 for every x ∈ (0,4), in particular for the c, whatever it is. Hence f(4) − 1 2 ≤ ≤ 5 . 4 Multiplying by 4, we get (since 4 is positive, the inequalities are preserved) 8 ≤ f(4)− 1 ≤ 20, and adding 1 everywhere, we get 9 ≤ f(4) ≤ 21. 18. Let f(x) = 3x − 2x + x − 1. Show that f(x) must have a real root in [0,1]. Solution to 18. The given function f(x) is continuous (because it is a polynomial), Plug-in the leftmost end, 0, into f(x), getting f(0) = −1. Plug-in the rightmost end, 1, into f(x), getting 3 2 f(1) = 3 ▯ 1 − 2 ▯ 1 + 1 − 1 = 1. Since these two numbers −1 and 1 have opposite signs (one is positive and the other is negative), it follows by the intermediate value theorem, that there must be a c ∈ (0,1) such that f(c) = 0. In other words, f(x) = 0 must have a real root in [0,1]. 101 51 19. Show that the equation x + x + x − 1 = 0 has exactly one real root. 101 51 Solution to 19 Let f(x) = x +x +x−1. Since lim x→−∞ f(x) = −∞ and lim x→∞ f(x) = ∞ (because the polynomial f(x) has odd degree), and these have opposite signs (like in problem 18), there must be at least one root in (−∞,∞). Taking derivative, f (x) = 101x 100+ 51x 50 + 1. Since all the powers are even, and the coeﬃcients ′ are positive and the constant term (1) is strictly positive, f (x) is strictly positive. If there would have been more than one root, by Rolle’s theorem there would have been a c such that f (c) = 0, contradicting the fact that f (x) is always positive. Hence there is exactly one real root. 20. Use two steps of Newton’s method to approximate the root of the equation x + x − 4 = 0 in the interval [1,2]. Solution to 20, Recall that Newton’s method is: f(xo) xn= x −o f (x ) , o where x is the old approximation and x is the new approximation. Implementing this here, we 0 n get f(x) = x + x − 4, and f (x) = 4x + 1, so for this problem we have x4 + xo− 4 x n x −o o . 4xo + 1 f(1) = −2 and f(2) = 14. Since −2 is so much closer to 0 than 14, it is reasonable to take as the initial approximation x = 1o We get 4 1 + 1 − 4 −2 xn= 1 − 4 ▯ 1 + 1 = 1 − 5 = 7/5 . 13 Now let x = 7/5 and we get 0 (7/5) + 7/5 − 4 x n 7/5 − . 4 ▯ (7/5) + 1 Since you don’t have a calculator, you leave it like this. (It turns out to be 9703/7485). (7/5) +7/5−4 Answer to 20. 7/5 − 4▯(7/5) +1 = 9703/7485 21. If a stone is thrown vertically upward from the surface of the moon with a velocity of 10m/s, its height (in meters) after t seconds is h(t) = 10t − t . (a) What is the velocity of the stone after 3 seconds? (b) What is the maximal height of the stone? Solution to 21. The velocity is the derivative of the position (in this problem height). So v(t) = h (t) = 10 − 2t . The velocity after 3 seconds is v(3) = 10 − 2 ▯ 3 = 10 − 6 = 4. Answer to 21a) 4 m/s b) When the stone reaches its maximal height its velocity is zero. This is at time t where t solves v(t) = 10 − 2t = 0, i.e., t = 5. Thus, the solution is h(5) = 10 ▯ 5 − 5 = 50 − 25 = 25. Answer to 21b) 25 m 22. Find the absolute maximum and the absolute minimum values of f(x) = 2 in [0,2]. x +1 Solution to 22. Using the quotient rule, we have ′ 2 2 ′ 2 2 2 2 ′ x (x + 1) − x(x + 1) 1 ▯ (x + 1) − x ▯ 2x x + 1 − 2x 1 − x f (x) = 2 2 = 2 2 = 2 2 = 2 2 . (x + 1) (x + 1) (x + 1) (x + 1) We have to solve f (x) = 0. Setting the numerator equal to 0, we get 1−x = 0, so (1−x)(1+x) = 0 and we get the two roots x = −1 and x = 1. But x = −1 does not satisfy the residency requirement, i.e. it does not belong to the interval [0,2], so we have the three ﬁnalists: 0,1,2. (The endpoints of the interval always participate in the ﬁnal contest). x Now let’s plug these numbers in f(x) = x +1: 0 1 1 2 2 f(0) = = 0 f(1) = = = .5 f(2) = = = .4 . 0 + 1 1 + 1 2 2 + 1 5 Answer to 22. Absolute minimum value is 0, Absolute maximum value is 1/2. 14 23. Find the most general antiderivative of the functions 3 6 a) f(x) = 1 − x + 5x − 3x 7 b) g(x) = 5 − 4x + 3x x 6 1 c) h(x) = 3e + 7sec x + 5(1 − x )2 − 2. Solution to 23. Recall the formula Z n+1 n x x = n + 1 + C . a) Z Z 4 6 8 f(x)dx = (1−x +5x −3x )dx = x− x +5 x −3 x = x−(1/4)x +(5/6)x −(3/8)x +C 8 . 4 6 8 Answer to 23a)x − (1/4)x + (5/6)x − (3/8)x + C 8 b) First break it up, then integrate term by term. Z Z 5 − 4x + 3x 6 Z g(x)dx = dx = (5 ▯ x6 − 4x−3 + 3)dx = 5x −5 /(−5) − 4(x−2 /(−2)) + 3x = x 6 −5 −2 −1 2 −x + 2x + 3x = 5 + 2 + 3x + C . x x −1 2 Answer to 23b) x5+ x2 + 3x + C ′ 2 2 c) Recall (or look up in the formula sheet) that (tanx) = sec x, hence the antiderivative of sec x −1 ′ 2 −1 2 −1 −1 is tanx + C. Also (sin x) = (1 − x ) 2 , hence the antiderivative of (1 − x ) is sin x + C. Of course the antiderivative of e is e + C. So Z Z x 2 2 − 1 x −1 f(x)dx = (3e + 7sec x + 5(1 − x ) 2)dx = 3e + 7tan x + 5sin x + C . −1 Answer to 23c) 3e + 7tanx + 5sin x + C 24. Evaluate the integral if it exists. Z 9√ Z 2 Z 5 u − 2u 2 2p dt a) du. b) y 1 + y dy. c) 2 . 1 u 0 1 (t − 4) Z 1 Z √x 2 3 e d) t cos(t )dt. e) √ dx. 0 x 15 Solution to 24. a) Z 9 √ Z 9 u − 2u2 − 1 1 1 2u2 9 √ 29 du = (u 2 − 2u)du = u /( ) − |1= 2 u − u 1 = 1 u 1 2 2 √ √ [2 9 − 9 ] − [2 1 − 1 ] = −76 . Answer to 24a) −76 b) We make the substitution u = 1 + y which entails du/dy = 3y hence dy = du/(3y ). Also the limits of the new integral must be computed. When y = 0, u = 1, and when y = 2, u = 9. So Z Z Z Z 2 2 9 2√ du 1 9√ 1 9 1 1 1 3 9 b) y 1 + ydy = y u 2 = udu = u2du = u2|1= 0 1 3y 3 1 3 1 3 3/2 √ √ 2√ 3 9 2 3 3 2 3 3 52 9 u |1= 9( 9 − 1 ) = 9 (3 − 1 ) = 9 . Answer to 24b) 52/9 c) Beware! The integrand blows up (and hence is undeﬁned) at t = 4, that lies within the interval of integration. Answer to 24c Undeﬁned! d) Here the substitution is u = t giving du/dt = 3t , and so dt = du/(3t ). As for the limits of integration, when t = 0 u = 0 and when t = 1, u = 1. Z Z Z 1 2 3 1 2 du 1 1 1 1 1 t cos(t )dt = t cos(u) 2= cos(u)du = sinu|0= (sin1 − sin0) = (sin1)/3 . 0 0 3t 3 0 3 3 Answer to 24d) (sin1)/3 e) This is an indeﬁnite integral since there are no limits of integration. The answer should be an expression in x. In the method of substitution, we must translate back at the end. √ 1 − 1 1 √ Take u = x = x 2, so du/dx = (1/2)x 2and dx = 2x du = 2 xdu. We have Z √x Z u Z e e √ u u √ x e) √ dx = √ 2 xdu = 2 e du = 2e = 2e + C . x x √ Answer to 24e) 2e x+ C R 5 R 5 R 4 25. If 1 f(x)dx = 12 and 4 f(x)dx = 3.6 , ﬁnd 1 f(x). 16 Solution to 25. We use the obvious relation Z b Z c Z b f(x)dx = f(x)dx + f(x)dx , a a c or rather its companion Z c Z b Z b f(x)dx = f(x)dx − f(x)dx . a a c So Z 4 Z 5 Z 5 f(x)dx = f(x)dx − f(x)dx = 12 − 3.6 = 8.4 . 1 1 4 Answer to 25. 8.4 2 26. Find the area bounded by the two curves (a) y = 2x1and y = 8x. (b)2y = sinx and 1 π y2= cosx, 0 ≤ x ≤ 2 Solution to 26. Recall that the area is (if the two curves meet only at two points) Z b (TOP − BOTTOM)dx , a 2 where a and b are the two roots of y 1 y . 2olving 2x = 8x yields 2x(x − 4) = 0 so x = 0 and x = 4, and indeed there are only two meeting points. To ﬁnd who’s on top just plug-in any x in the interval (0,4). x = 1 is a good choice. When x = 1, y =12 and y = 82 since 2 < 8 we know 2 that y2= 8x is on top and y =12x is at the bottom. So the area is Z 4 2 2 3 4 2 3 64 (8x − 2x )dx = 4x − (2/3)x | =04 ▯ 4 − (2/3) ▯ 4 = . 0 3 64 Answer to 26a) 3 b) The curves y 1 sinx and y = 2osx meet when sinx = cosx, and that happens when x = π/4. There are two parts (equal by symmetry). The one bounded by y = sinx, y = cosx, and the y-axis (alias x = 0), and the one bounded by y = sinx, y = cosx, and x = π/2. The ﬁrst one happens between x = 0 and x = π/4 and the second between x = π/4 and x = π/2. In the ﬁrst half y = cosx is on top and y = sinx is at the bottom, and at the second half it is the other way. Hence the area is Z π/4 Z π/2 π/4 π/2 (cos x − sinx)dx + (sinx − cosx)dx = (sinx + cosx)| 0 + (−cosx − sinx)| π/4 = 0 π/4 √ (sin(π/4)+cos(π/4))−(sin(0)+cos(0))+(−cos(π/2)−sin(π/2))−(−cos(π/4)−sin(π/4)) = 2 2−2. √ Answer to 26b) 2 2 − 2 17 27. Evaluate the limit n X iπ π lim sin( ) . n→∞ i=1 n n This is an instance of the formal deﬁnition of the deﬁnite integral Z b n X f(x)dx = ln→∞ f(xi)h , a i=1 where h = (b − a)/n and i = a + ih, i = 1,2,... ,n. The name of the game is to ﬁrst decide on the function f(x), which in this problem is obviously f(x) = sinx. The argument of sin is iπ/n soix = iπ/n. To get a we plug-in i = 0 getting a = 0, and to get b we plug in i = n, getting b = π. Hence h = π/n, which is exactly as in the problem. Hence this complicated-looking limit is nothing but Z π π sinx = −cosx| 0 −cos(π) − (−cos(0)) = −(−1) + 1 = 2 . 0 Answer to 27. 2 28. a) Use the deﬁnition of the natural logarithm as an integral and compare areas to prove that ln2 < 1 < ln3. b) Use a) to deduce that 2 < e < 3. Solution to 28 a) Recall that Z x 1 lnx = dt . 1 t Hence ln2 is the area under the curve y = 1/t between t = 1 and t = 2. To get an upper bound for ln2 note that it is enclosed in the square bounded by the t-axis, the horizontal line t = 1, and the vertical lines t = 1 and t = 2. Its area is 1 ▯ 1 = 1, so ln2 < 1. The lower bound 1 < ln3 is more tedious. Take the inscribed rectangles with n = 8, getting that ln3, the area under the curve y = 1/t between t = 1 and t = 3 is deﬁnitely more than 1 1 1 1 1 1 1 1 1 28271 + + + + + + + = > 1 . 4 5/4 6/4 7/4 8/4 9/4 10/4 11/4 12/4 27720 b) Since e is an increasing function we can exponentiate the inequality ln2 < 1 < ln3 getting eln < e < e ln 3 . But eln w= w, so 2 < e < 3 . 18

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