SOLUTION BANK PEYAM RYAN TABRIZIAN

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1 SOLUTION BANK PEYAM RYAN TABRIZIAN This document contains solutions to various Math A homework problems I ve compiled, from Fall 00, Spring 0, and Fall 03. At first, the solutions might seem disappointingly short, but don t worry! As the course progresses, the solutions become longer and more detailed! It contains solutions to the following problems from the 7th edition of the book (see next page): Date: Friday, December 3, 03.

2 PEYAM RYAN TABRIZIAN Section Problems. 7, 8,, 38, 5, 63, 69., 4, 8, 6.3, 7, 4, 30, 36.5, 4, 7, 9, 7, 0,.6 3, 5, 6, 7, 8, 6, 5, 63, 69, 70., 6, 3, 33, , 0, 3, 8, 0, 6, 9, 3, 37, 40, 60, 6, 64.4, 4,, 9, 3, 37, 4, 43, ,, 0, 8, 38, 44, 5, 54, 60, , 6, 6, 38, 43, 6.7 6,, 7, 8, 9, 3, 36, 40, 4, , 3, 40, 43, , 3, 35, 47, , 33, 4, 49, , 39, 40, , 5, 39, 45, 49, 63, 66, 67, 68, 7, 80, 85, 86, , 9, 9, 30, 3, 44, 45, 46, 5, 53, 54, 63, 77, ,, 4, , 8, 7, 6, , 9,, , 3, 5, 7, 40, 45, ,, 5,, 35, 40, ,, 5,, 3(a), 6, 9(a)(b), 3, 47 3 Review T/F, 68, 87, 94, 99, 0, 05, 4. 6, 38, 39, 4, 43, 5, ,, 7, 8, 3, 9, 3, 34, , 9, 3, 7, 33, 43, 45, 49, , 4,, 6, 7, 7, 9, 45, 49, 53, 58, 6, 76, , 3, 43, 45, 49, , 3, 4,, 3, 3, 48, 57, 6, 67, , 35, 4, 63, 76 5., 5, 3, 7, 9, 0,, 3, ,, 3, 30, 34, 37, 43, 44, 47, 54, 56, 69, , 5, 7, 3, 35, 37, 39, 43, 45, 57, 67, ,, 3, 5, 3, 37, 49, 54, 6, 6, 63, , 3, 33, 48, 59, 6, 77, 86, 9 6., 3, 3,, 4, 43, , 3, 7, 47, 49, 55, 65, , 3, 5, 9, 46, 48

3 SOLUTION BANK 3 Section.: Four ways to represent functions..7. No (by the vertical line test)..8. Yes (by the vertical line test), Domain = [, ], Range = [, ]... (a) The graph of x(t) should just be a line going through the origin (b) The graph of y(t) should look at first like the right half of a parabola, then should be constant for a while, and then look like the left half of a parabola (c) The graph of the horizontal velocity looks like a horizontal line (d) See announcement on bspace for a detailed solution! The picture you get is: A/Math A Summer/Solution Bank/Vertical Velocity.png..38. Domain = [, ], Range = [0, ], Graph is just the upper-half of the circle centered at 0 of radius...5. f(x) = 5 x..63. V (x) = x(0 x)( x) (no need to expand the answer!)..69. f is odd, g is even

4 4 PEYAM RYAN TABRIZIAN Section.: Mathematical models: a catalog of essential functions... (a) Exponential function (b) Power function (c) Polynomial of degree 5 (d) Trigonometric function (e) Rational function (f) Algebraic function..4. (a) G (b) f (c) F (d) g..8. (a) y = (x 3), (b) y = x 5 x (a) C(x) = 3x (C is the cost and x is the number of chairs produced) (b) 3; Cost per chair (c) 900; Start-up cost (i.e. money needed to buy machines in order to start producing chairs).3.. (a) y = f(x) + 3 (b) y = f(x) 3 (c) y = f(x 3) (d) y = f(x + 3) (e) y = f(x) (f) y = f( x) (g) y = 3f(x) (h) y = 3 f(x) Section.3: New functions from old functions.3.7. y = 3(x + 4) (x + 4).3.4. Basically compress the graph of sin(x) horizontally by a factor of 3 (notice that the new period now is π 3 and then stretch the resulting graph vertically by a factor of 4 (so the new graph has range [ 4, 4] instead of [, ]) (a) (f + g)(x) = 3 x + x (b) (f g)(x) = 3 x + x (c) (fg)(x) = 3 x x (d) ( f g ) (x) = 3 x x All of those functions have domain (, ] [, 3] EXCEPT for (d), which has domain (, ) (, 3]

5 SOLUTION BANK sin(x) (a) (f g)(x) = +sin(x) ; Dom = all odd multiples of π ( ) (b) (g f)(x) = sin ; Dom = all real numbers except - (c) (f f)(x) = x +x + x +x x +x = x +x ; Dom = all real numbers except (d) (g g)(x) = sin ( sin(x)); Dom = all real numbers.5.. (a) 6; (b) 7x (a) x 4n 3 ; (b) a 6 b Section.5: Exponential Functions and.5.7. Basically, the larger the base, the faster the function is increasing.5.9. Notice that ( ) x 3 = 3 x, which means that ( x 3) is the reflection of 3 x across the y-axis! Similarly with 0 x (a) y = e x (b) y = e x (c) y = e x (d) y = e x (e) y = e x.5.0. (a) All real numbers ; (b) All 0 real numbers.5.. f(x) = 3 x Section.6: Inverse functions and logarithms.6.3. No; For example, even though 6, f() = f(6) =.6.5. No (by the horizontal line test).6.6. Yes (by the horizontal line test) (You want to find x such that g(x) = 4, that is, find x such that x+e x =. Here, just guess!).6.8. (a) By the horizontal line test (b) Domain of f = Range of f = [, 3]; Range of f = Domain of f = [ 3, 3] (c) 0 (d).8 (.6.6. f (x) = ln x x ) ( = ln x x )

6 6 PEYAM RYAN TABRIZIAN.6.5. (a) x = ± + e 3 (b) x = 0, ln() (Let X = e x and solve the equation X 3X + = 0 (by using the quadratic formula), then solve for x using e x = X) (a) π 3 (b) π If θ = sin (x), then sin(θ) = x, then draw a triangle with hypothenuse, and opposite side x, and then the adjacent side becomes x, and so our answer becomes: cos(sin (x)) = cos(θ) = adjacent x hypotenuse = = x See the handout Proof of the derivative of arccos for a similar problem; Or look at your notes taken in section! tan(sin x) = sin(sin (x)) cos(sin (x)) = x x by the result of number 69! Section.: The limit of a function... If x approaches from the left, then f(x) approaches 3; If x approaches from the right, then f(x) approaches 7. No, left-hand-limits and right-hand-limits must be equal!..6. (a) 4 (b) 4 (c) 4 (d) Undefined (e) (f) - (g) Does not exist (left and right-side limits not equal) (h) (i) (j) Undefined (k) 3 (l) Does not exist (h does not approach one fixed value as x approaches 5 from the left)..3. (numerator approaches e 5 > 0 while denominator approaches (x 9 approaches 0 + and ln (0 + ) =..46. The mass blows up to ( v c goes to, so the denominator of the fraction goes to 0 +, and so the whole fraction goes to ) Section.3: Calculating limits using the limit laws.3.9. Just plug in x =

7 SOLUTION BANK (a) If you plug in x =, then the left hand side is not defined, but the right hand side is (b) The above equation holds if x, but the point of limits is that in this case you don t care about the value at! So in this case, the equality is correct!.3.3. Does not exist (left-hand-limit is because the numerator tends to 4 and the denominator tends to 0 while the right-hand-limit is because the numerator tends to 4 and the denominator tends to 0 + ).3.8. (Use the formula (A + B) 3 = A 3 + 3A B + 3AB + B 3 ) (use the fact that x 3 = (x )(x + x + )).3.6. (put under a common denominator t + t = t(t + ) and cancel out).3.9. (put under a common denominator and multiply by the conjugate form).3.3. x 3 (put under a common denominator and expand the numerator out) (use the squeeze theorem) (by squeeze theorem, because sin ( ) π x ) Let a = 0 and f(x) = sin ( ) x (or x.3.6. Let a = 0 and f(x) = sin ( x Hints: Use the following steps: ), and g(x) = f(x). ) (or x ), and g(x) = f(x) (a) Find the coordinates of Q. For this, solve for x and y in the system of equations: { (x ) + y = x + y =r For this, plug in y = r x in the first equation and solve for x, then solve for y in y = r x ; remember that you want x > 0 and y > 0, according to the picture). The answer gives you the coordinates of Q (b) Now that you know the coordinates of P and Q, find the equation of the line going through P and Q (c) Find the x-intercept of that line (set y = 0 and solve for x) (d) Finally, take the limit as r 0 + of the answer you found in (c). To do this, multiply as usual by the conjugate form. Answers: (a) Q = ( r, r r 4 ( ) ) (b) y = r r 4 x + r r (c) x intercept = ( (d) 4 r 4 )

8 8 PEYAM RYAN TABRIZIAN Section.4: The precise definition of a limit.4.. δ = δ = min {, 6} = 6 (there are many answers to this question, so don t worry if yours is different from mine).4.. Note: If this problem is too confusing for you, skip it! (it does more harm than good) 000 (a) r = π (from now on, let s call this a) { } 5 (b) r a < min π(a+), 5 = π(a+) (for this, first set r a <, solve for r to get a < r < a +, then r + a = r + a < a +, so π r a r + a < 5 π r a (a + ) < 5, hence r a < π(a+) { } 000 (c) x = r = radius, a = π, L = 000, ɛ = 5, δ = min 5, π(a+).4.9. This is an example of the easy case with δ = 3ɛ This is an example of the complicated case with δ = min {, ɛ 9}. To get this δ, notice that if x <, then < x < 3, and so 7 < x +x+4 < 9, so x + x + 4 < This is again an example of the complicated case with δ = min { )} a, ɛ a ( + To get this δ, notice that if x a < a, then a < x < 3a, and so in particular ( ) a x + a > + and then: x a < x a ) x + a a < ɛ ( + which gives: x a < ɛ ( a + ) The next two are optional, but good for practice:.4.4. δ = 4 M δ = e M (where M is negative)

9 SOLUTION BANK This is absolutely ridiculous, so feel free to skip it if you want! (a) Let M > 0. We want to find δ such that if x a < δ, then f(x)+g(x) > M. However, by replacing M by M + ( c) in the definition of the limit of f, there exists δ such that if x a < δ, then f(x) > M + ( c) Moreover, by letting ɛ = in the definition of the limit of g, we know that there exits δ such that if x a < δ, then g(x) c <. In particular, we get g(x) c >, so g(x) > c. Hence, if you choose δ = min {δ, δ }, then if x a < δ, we have f(x) + g(x) > M + ( c) + (c ) = M. (b) Let M > 0. We want to find δ > 0 such that if x a < δ, then f(x)g(x) > M. By replacing M by M ( ) c in the definition of the limit of f, we know that there exists δ such that if x a < δ, then f(x) > M ( ) c Moreover, by letting ɛ = c > 0 in the definition of the limit of g, we know that there exists δ such that if x a < δ, then g(x) c < c. In particular, g(x) c > c, so g(x) > c c = c > 0. Hence, if you choose δ = min {δ, δ }, then if x a < δ, we have f(x)g(x) > ( M c ) ( c ) = M. (c) Let M < 0. We want to find δ > 0 such that if x a < δ, then f(x)g(x) < M. By replacing M by M ( ) c in the definition of the limit of f, we know that there exists δ such that if x a < δ, then f(x) > M ( ) c Moreover, by letting ɛ = c > 0 in the definition of the limit of g, we know that there exists δ such that if x a < δ, then g(x) c < c. In particular, g(x) c < c, so g(x) < c c = c < 0. Hence, if you choose δ = min {δ, δ }, then if x a < δ, we have f(x)g(x) < ( M c ) ( c ) = M. (Note that in fact the > in the first part, becomes a < here precisely because g(x) < 0!) Section.5: Continuity (f not defined at 4; neither), (left-hand-side and right-hand-side limits not equal; continuous from the left), (ditto; continuous from the right), 4 (left-hand-side limit does not exist; continuous from the right).5.. f() = 4 (You get this by solving for f() in 3f() + f()g() = 36).5.0. Not continuous because the limit as x equals (factor out the numerator and denominator), whereas f() =

10 0 PEYAM RYAN TABRIZIAN.5.8. Continuous because it s a ratio of two continuous functions (and the numerator/denominator are continuous because of composition of continuous functions), Dom = R tan ( ) Yes, you can check that the left-hand-side-limits and the right-hand-side limits are equal! Plug in values for G, M, R if you want to, for example G =, M = 5, R = Define f(x) = x 4 + x 3, then f() = < 0, f() = 5 > 0, so by IVT, there is one number c such that f(c) = Let f(x) = sin(x) x + x Then f() = sin() + = sin() > 0, whereas f() = sin() 4 + = sin() < 0. Moreover, f is continuous on [, ], hence by the Intermediate Value Theorem there exists one c in (, ) such that f(c) = 0, that is, sin(c) = c c Use the fact that sin(a + h) = sin(a) cos(h) + sin(h) cos(a) Define f(t) to be the altitude of the monk on the first day, g(t) to be the altitude of the monk on the second day, and let h(t) = f(t) g(t). Then h(0) > 0, h() < 0 (where 0 means 7AM and means P M), then by IVT, there is one number c such that h(c) = 0, i.e. f(c) = g(c).6.4. Section.6: Limits at Infinity; Horizontal Asymptotes (a) (b) (c) (d) (e) (f) Horizontal asymptotes: y =, y = ; Vertical asymptotes: x = 0, x = (factor out x 3 from the numerator and the denominator)

11 SOLUTION BANK.6.6. lim x + x + x = lim x + x + x x x = lim x + x x = lim x x x ( = lim x x = lim x x = lim x x = lim x x = lim x x = lim x x = lim x = + = ( + x + x + x + x ( + ) ) ( ) + + x ( ) + + x + x + x ( ) + x + + x x + + x x + + x + + x ) tan ( ) = π (by continuity of tan ) y = (at ± ), x =, x = (factor out the denominator) (by the squeeze theorem).7.6. y = 9x Section.7: Derivatives and Rates of Change (a) A runs with constant speed, while B is slow at first and then speeds up (b) 8.5 seconds (c) 9 seconds

12 PEYAM RYAN TABRIZIAN.7.7. g (0) < 0 < g (4) < g () < g ( ).7.8. y = 4x 3 (y + 3 = 4(x 5) is also acceptable).7.9. f() = 3, f () = f(x) = 4 x, a = f(x) = tan(x), a = π f (t) = t (show this, using the definition of the derivative) Velocity = f (5) = meters per second, Speed = 5 meters/second F/min (slope of the red line) (a) Rate of bacterias/hour after 5 houts (b) f (0) > f (5) (basically, the more bacteria there are, the more can be produced). But if there s a limited supply of food, we get that f (0) < f (5), i.e. bacterias are dying out because of the limited supply.8.3. (a) II (b) IV (c) I (d) III Section.8: The derivative as a function.8.3. f (t) = 5 8t (not continuous there); (graph has a kink) (a) Acceleration (b) Velocity (c) Position Not differentiable at the integers, because not continuous there; f (x) = 0 for x not an integer, undefined otherwise. Graph looks like the 0-function, except it has holes at the integers. Section 3.: Derivatives of polynomials and exponential functions S (R) = 8πR y = e x y = 4x 3 + e x, so y (0) =, and so the equation of tangent line is y = (x 0), i.e. y = x + and equation of normal line is y = (x 0), i.e. y = x + (remember that the normal line still goes through (0, ), but has slope = the negative reciprocal of the slope of the tangent line)

13 SOLUTION BANK (a) v(t) = s (t) = 3t 3; a(t) = v (t) = 6t (b) a() = (c) v(t) = 0 if t = or t =, but t > 0 (negative time doesn t make sense), so t =, and a() = First of all y = 3 x and first find a point x where y (x) = 3 (remember that two lines are parallel when their slopes are equal, and the slope of y = +3x is 3). So you want 3 x = 3, so x =, so x = 4. Now all that you need to find out is the slope of the tangent line to the curve at 4. The equation is: y 8 = 3(x 4) (because from the above calculation the slope is 3, and the tangent line goes through (4, f(4)) = (4, 8)) Section 3.: The product and quotient rules y = t(t4 3t +) (t +)(4t 3 6t) (t 4 3t +) y (x) = e x + xe x, so y (0) =, and so the tangent line has equation: y 0 = (x 0),i.e y = x, and the normal line has equation: y 0 = (x 0), i.e. y = x f (x) = x(+x) x (+x) = x +x x +x+, so f (x) = (x+)(x +x+) (x +x)(x+) (x +x+), and so f () = (+)(++) (+)(+) (++) = (4)(4) (3)(4) (4)(4) = 6 6 = 4 6 = = (a) u () = f ()g() + f()g () = ()() + ()( ) = 0 (b) v (5) = f (5)g(5) f(5)g (5) = ( 3)() (3)( 3) (g(5)) 4 = 3 4 = 8 = (900)(30593) + (96400)(400) =,67,000 Section 3.3: Derivatives of trigonometric functions We have sin(θ) = x 0, so x = 0 sin(θ), so x (θ) = 0 cos(θ), and x ( ) π 3 = 0 cos ( ) π 3 = 0 = (multiply the fraction by 3 3 and use the fact that lim x 0 sin(3x) 3x = ) = 3 (multiply the numerator by 4 4 and the denominator by 6 6 sin(4x) sin(6x) the facts that lim x 0 4x = and lim x 0 6x = ) The problem is to calculate: and use lim θ 0 + = s d First of all, let s call the radius of the circle r. Now, let s divide this into three simple sub-problems:

14 4 PEYAM RYAN TABRIZIAN A/Math A Summer/Solution Bank/Arclength.png () Calculate s But this is not very hard! Either you know about the formula for the length of an arc, or you can easily derive it! Namely, if an angle of π radians corresponds to a length of πr (i.e. the circumference of a circle), then an angle of θ radians corresponds to a length of s. Now, because the length of an arc is proportional to the angle, we have the following equality: s θ = πr π = r So s = θr Notice that the use of radians makes this calculation particularly simple! () Calculate d This is a bit harder than the above step, but actually not that bad! Label the triangle in the figure ABC, and let H be the midpoint of BC.

15 SOLUTION BANK 5 Then, the triangle AHB is right in A, and we can use our regular definition of sin to find a relationship between r and d, namely: sin( BAH) = BH AB But you can easily check that BAH = θ, that BH = d the midpoint of BC), and that AB = r! Hence, we get: (because H is That is, d = rsin( θ ). sin( θ ) = d r (3) Compute the limit The last thing we need to do is to calculate the required limit! But this is easy, since we know the values of s and d in terms of r: s lim θ 0 + d = lim θ 0 + rθ r sin( θ ) = lim θ 0 + θ sin( θ ) = lim θ 0 + θ sin( θ ) Finally, let t = θ becomes: and notice that, as θ 0+, t 0 +, then, our limit s lim θ 0 + d = lim t t 0 + sin(t) = lim t 0 + sin(t) t = lim t 0 + lim t 0 + sin(t) t = = And again, the next-to-last step is justified because both limits (numerator and denominator) exist and the limit of the denominator is nonzero! Hence, we get: lim θ 0 + s d = f (t) = cos (e t ) e t + e sin(t) cos(t) y = e kx kxe kx Section 3.4: The Chain Rule f (t) = sec (e t ) + e tan(t) sec (t) y = sin( sin(tan(πx))) sin(tan(πx)) cos(tan(πx)) sec (πx)π y = αe αx sin(βx)+e αx β cos(βx); y = e αx ((α β ) sin(βx)+αβ cos(βx)) (a) h () = f (g())g () = f ()g () = 5 6 = 30 (b) H () = g (f())f () = g (3)f () = 9 4 = 36

16 6 PEYAM RYAN TABRIZIAN (a) h () = f (f())f () = f ()f () = = (b) g () = f (4)4 = 4 = g (3) = f(3) f (3) = (a) F (x) = f (x α )αx α (b) G (x) = α (f(x)) α f (x) ( ) 3 = 3 (which you can rewrite as 6 ) (a) f (x) = g(x ) + xg (x )x = g(x ) + x g (x ) (b) f (x) = g (x )(x) + 4xg (x ) + x g (x )x = 6xg (x ) + 4x 3 g (x ) (a) v(t) = s (t) = Aω sin(ωt + δ) (b) We want v(t) = 0, so ωt + δ = πm, so t = πm δ ω a(t) = dv dt = dv ds ds dt = dv ds v(t) (m is an integer) (a) dv dv dr is the rate of change of V as the radius r changes, and dt of change of V as the time t changes (b) V (t) = 4 3 πr3, so dv dt = dv dr dr dt = 4 dr dr 3π3r dt = 4πr dt is the rate (f(x)[g(x)] ) = f (x)[g(x)] + f(x)( )[g(x)] g (x) =f (x)g(x)[g(x)] f(x)g (x)[g(x)] = f (x)g(x) f(x)g (x) (g(x)) Section 3.5: Implicit differentiation y = y x y = ey sin(x)+cos(xy)y e y cos(x) x cos(xy) y = x (y ) = 3(x + 3 3), or y = 3x y =

17 SOLUTION BANK First of all, by implicit differentiation: x a + yy b =0 ) y ( y b = x a y = b x a y y = b x a y It follows that the tangent line to the ellipse at (x 0, y 0 ) has slope b a x 0 y 0, and since it goes through (x 0, y 0 ), its equation is: ) y y 0 = ( b x 0 a (x x 0 ) y 0 And the rest of the problem is just a little algebra! First of all, by multiplying both sides by a y 0, we get: Expanding out, we get: Now rearranging, we have: (y y 0 )(a y 0 ) = b x 0 (x x 0 ) ya y 0 a (y 0 ) = b x 0 x + b (x 0 ) ya y 0 + b x 0 x = a (y 0 ) + b (x 0 ) Now dividing both sides by a, we get: yy 0 + b a x 0x = (y 0 ) + b a (x 0) And dividing both sides by b, we get: yy 0 b + x 0x a = (y 0) b + (x 0) a But now, since (x 0, y 0 ) is on the ellipse, (y0) b Whence, Which is what we want! yy 0 b x 0 x a + x 0x a = + y 0y b = + (x0) a =, we get:

18 8 PEYAM RYAN TABRIZIAN Slope: x a yy y ( y b b =0 ) = x a y = b x a y y = b x a y Equation: At (x 0, y 0 ), the slope is b a x 0 y 0, so the equation of the tangent line at (x 0, y 0 ) is: y y 0 = ( ) b x 0 a (x x 0 ) y 0 Simplification: First of all, by multiplying both sides by a y 0, we get: Expanding out, we get: Now rearranging, we have: (y y 0 )(a y 0 ) = b x 0 (x x 0 ) ya y 0 a (y 0 ) = b x 0 x b (x 0 ) ya y 0 b x 0 x = a (y 0 ) b (x 0 ) Now dividing both sides by a, we get: And dividing both sides by b, we get: yy 0 b a x 0x = (y 0 ) b a (x 0) yy 0 b x 0x a = (y 0) b (x 0) a But now, since (x 0, y 0 ) is on the hyperbola, (x0) a and we get: (y0) b =, so (y0) b (x0) a =, Whence, yy 0 b x 0x a = x 0 x a y 0y b =

19 SOLUTION BANK Slope: ( ) x + y =0 y ( ) y = y x y = x y y = y x y y = x Equation: At (x 0, y 0 ), the slope is y 0 x0, and so the equation of the tangent line at (x 0, y 0 ) is: y0 y y 0 = (x x 0 ) x0 y intercept: To find the y intercept, set x = 0 and solve for y: y0 y y 0 = (0 x 0 ) x0 y0 y y 0 = ( x 0 ) x0 y y 0 = y 0 x0 y =y 0 + y 0 x0 x intercept: To find the x intercept, set y = 0 and solve for x: y0 0 y 0 = (x x 0 ) x0 y0 y 0 = (x x 0 ) x0 x0 x x 0 = ( y 0 ) y0 x =x 0 + x 0 y0 Sum: The sum of the y and x intercepts is: But the trick is that: (y 0 + y 0 x0 ) + (x 0 + x 0 y0 ) = x 0 + x 0 y0 + y 0

20 0 PEYAM RYAN TABRIZIAN x 0 + x 0 y0 + y 0 = ( x 0 ) + x 0 y0 + ( y 0 ) = ( x 0 + y 0 ) But since (x 0, y 0 ) is on the curve x + y = c, we get x 0 + y 0 = c. And so, finally we get that the sum of the x and y intercepts is: x 0 + x 0 y0 + y 0 = ( x 0 + y 0 ) = ( c) = c (or, either answer is fine) (x+) (x +x) G (x) = x x arccos(x) x +x +(x = +x ) (+x ) Note: This is a ridiculous simplification, and don t worry about this too much, but here are the steps: () First put everything under a common denominator () Then expand out the + (x + x ) in the denominator, and simplify and factor out the (3) Then multiply the numerator and the denominator by + x + x (conjugate form) (4) Then expand out the ( + x + x)( + x x + x )-part of the denominator to get + x Look up the handout Proof of the derivative of arccos on my website! (a) f(f (x)) = x, let y = f (x), then f(y) = x, so f (y)y =, so y = (b) 3 f(y) = f(f (x)) Let s denote the point of intersection between the ellipse and the tangent line by (a, b). Then, using implicit differentiation, we can show that the slope of the tangent line is a 4b Now, let K be the altitude of the lamp, our goal is to find K. Notice that the same tangent line goes through the points ( 5, 0) and (3, K), so by the slope formula, we have: Slope = K 0 3 ( 5) = K 8 In particular, since the slope is also equal to a 4b, we have:

21 SOLUTION BANK So K 8 = a 4b K = 8 a 4b = a b So all we really need to do to solve this problem is to find a b! Now we also know that the tangent line goes through the points ( 5, 0) and b 0 (a, b), so its slope is a ( 5) = b a+5, but again we know that its slope is also a 4b, and so we get: b a + 5 = a 4b So cross-multiplying, we have 4b = (a)(a + 5), that is a + 4b = 5a. HOWEVER, We also know that (a, b) is on the ellipse, so it satisfies the equation of the ellipse, and so a + 4b = 5, whence we get 5a = 5, and so a =. And plugging a = into a + 4(b) = 5 and assuming b > 0, we get b =, and so K = a b = =, and we re done! Section 3.6: Derivatives of logarithmic functions ( x ) g (x) = x + x x x y = log 0 ( x) + x ln(0) x x = log 0( x) + ln(0) ( ) y = x x 4 + (x ) + x3 x 4 + ( ) y = x cos(x) sin(x) ln(x) + cos(x) x Section 3.7: Rates of change in the natural and social sciences (a) f (t) = e t t e t = e t (b) f (3) = e 3 (c) t = (d) When t < ( ) ( ) t (e) f() f(0) + f() f(8) = e 0 + e 8e 4 = 4e 8e 4 (f) The particle is moving to the right between t = 0 and t =, and then to the left from t = to t = 8.

22 PEYAM RYAN TABRIZIAN ( ) (g) f (t) = e t t + e t ( ) f (3) = e 3 4 ( ) ( = e t + t 4 ) ( = e t t 4 ) ; (h) Use a calculator (i) Speeding up when f (t) > 0 and f (t) > 0 or when f (t) < 0 and f (t) < 0. But solving those equations reveals that none of the two situations can happen! Hence the particle is constantly slowing down! (a) First solve for v(t) = 0, where v(t) = ds 80 dt = 80 3t, you get t = 3 = 5. So the maximum height s is s = s( 5 ) = = 00 (b) To find the time t when the ball is 96ft above the ground, we need to solve the equation s(t) = 96, and you get t =, 3, whence v() = 80 3 = 6 ft s and v(3) = = 6 ft s f (x) = 6x = linear density at x. f () = 6, f () =, f (3) = 8. The density is highest at 3 and lowest at First of all, we know two things, namely f(0) = 0 and f (0) =. But by the chain rule: f (t) = 0.7be 0.7t a ( + be 0.7t ) = 0.7abe 0.7t ( + be 0.7t ) So from f(0) = 0, we get: And from f (0) =, we get: a + b = 0 0.7ab ( + b) = From a +b = 0, we get a = 0(+b), and plugging this into the second equation, we get: (0.7)(0)( + b)b ( + b) = 4b + b = 4b =( + b) 4b = + b b = b = 6 And so a = 0( + 6) = 0(7) = 40. Therefore, we have a = 40 and b = 6.

23 SOLUTION BANK 3 Finally, to find out what happens in the long run, we need to calculate lim t f(t). But notice that lim t e 0.7t = 0, and so lim t f(t) = a +0 = a = (a) C (x) = 0.x x (b) C (00) = 3; The cost of producing one more yard of a fabric once 00 yards have been produced (c) C(0) C(00) = 3.005, which is pretty close to C (00) Section 3.8: Exponential growth and decay (a) Hint: Solve the differential equation y = ky with y(0) = 36 and y(0) = 439 and find y(50) (b) Hint: Solve the differential equation y = ky with y(0) = 439 and y(0) = 683 and find y(39) and y(49) (a) y(t) = 00e ln( ) t 30 = 00 ( (b) y(00) = 00 ( ) 00 (c) t = 30 ln( 00 ) ln( ) ) t The problem asks about radioactive decay, so as usual, we have y = ky, so y(t) = Ce kt. Now we re given two things: First of all, the half-life is t = 5730 years, so y(5730) = y(0) = C. Moreover, we know that at a certain time t (we want to find t ), y(t ) = 0.74y(0) = 0.74C. Now even though we don t know what C is, we can still solve for t. The following calculation helps us find k: y(5730) = C Ce 5730k = C e 5730k = 5730k = ln( ) k = ln( ) 5730 Whence y(t) = Ce ln( ) 5730 t = C( ) t 5730 Now we re given that y(t ) = 0.74C, and the following calculation helps us solve for t :

24 4 PEYAM RYAN TABRIZIAN y(t ) =0.74C C( ) t 5730 =0.74C ( ) t 5730 =0.74 t 5730 ln( ) = ln(0.74) t =5730 ln(0.74) ln( ) t 489 So t 489 years (notice how we didn t even need info about C to figure this out!) (a) (i) 3000 ( ) ()(5) 388 (ii) 3000 ( ) ()(5) 3840 (iii) 3000 ( ) ()(5) 3850 (iv) 3000 ( ) (5)(5) (v) 3000 ( ) (365)(5) (vi) 3000e 0.05(5) (b) A = 0.05A, A(0) = dh dt = 3 5π (Use V = πr h) Section 3.9: Related rates ) First of all, let s draw a picture of the situation, and remember to only label things that are constant! Here, x is the distance between the street light and the man, and y is the distance between the man and the shadow. Also, let z = x + y, the total length of the shadow. ) We want to figure out z when x = 40. 3) Looking at the picture, we can use the law of similar triangles to conclude: That is: y x + y = 6 5 y = (x + y) y = 5 x y = 3 x

25 SOLUTION BANK 5 A/Math A Summer/Solution Bank/Street Light.png It follows that: z = x + y = x + 3 x = 5 3 x 4) Hence z = 5 3 x 5) However, we know that x = 5 (because the man is walking with a speed of 5 ft/s). Hence we get z = (5) = 3 So z = 5 3. Note: We did not need the fact that x = 40! dd dt = 65mph (use the pythagorean theorem to conclude D = x + y ) dh dt = 6 5π (use the fact that V = π 3 r h = π h3 because h = r ) ) Again, draw a picture of the situation: Here, x is the distance between P and the beam of light.

26 6 PEYAM RYAN TABRIZIAN A/Math A Summer/Solution Bank/Lighthouse.png ) We want to figure out dx dt when x = 3) Looking at the picture, because we have info about the derivative of θ (see below), we use the definition of tan(θ): tan(θ) = x 3 So x = 3 tan(θ) 4) Hence dx dt = 3 sec (θ) dθ dt 5) First of all, we actually know what dθ dt is! Since the lighthouse makes 4 revolutions per minute and one revolution corresponds to π, we know that dθ dt = 8π (think of speed = distance/time, and here time = minute, and distance = 8π, also you put a minus-sign since x is decreasing!). Moreover, by drawing the exact same picture as above, except with x =, we can calculate sec (θ), namely: sec(θ) = hypothenuse adjacent = 0 And the 0 we get from the Pythagorean theorem! 3

27 SOLUTION BANK 7 ( 0 ) It follows that sec (θ) = 3 = 0 9. Now we got all of the info we need to conclude the problem: Whence dx dt = 80π 3 rad/min dx dt =3 sec (θ) dθ dt dx dt =3(0 9 )( 8π) dx dt = 40π 9 dx dt = 80π 3 ) As usual, let s draw a picture of the situation: A/Math A Summer/Solution Bank/Runners.png Here, x is the distance between the runner and the friend, and s is the length of the arc corresponding to θ

28 8 PEYAM RYAN TABRIZIAN ) We want to figure out dx dt when x = 00 3) Looking at the picture, it looks like we should use the law of cosines (because we have info about θ and about of the 3 sides of the triangle) In other words: 4) Hence x dx dt x = (00)(00) cos(θ) x = cos(θ) = sin(θ) dθ dt 5) First of all x = 00 in this case. Also, by definition of a radian, we know that s = 00θ, whence ds dθ ds dt = 00 dt. But we re given that dt = 7m/s, so dθ dt = 7 00 = So all we got to figure out is sin(θ) For this, draw the same picture as above, except you let x = 00. And in this case, we use the law of cosines again: 00 = (00)(00) cos(θ) 0000 = cos(θ) cos(θ) = 4 Now you can use either the triangle method to figure out what sin(θ) is (all you gotta do is calculate sin(cos ( 4 ), or, even easier, notice that sin(θ) = cos (θ) (this works because sin(θ) > 0 because we assume that θ is between 0 and π. Hence sin(θ) = 6 = 5 6 = 5 4. PHEW!!! Now we have all the info we need to solve the problem: Hence dx dt = x dx =40000 sin(θ)dθ dt dt (00) dx 5 dt =40000( 4 )(0.07) 400 dx dt =700 5 dx dt = As is usual for related rates problems, let s draw a picture: Let θ be the angle between the hour hand and the minute hand. Now, what we want to calculate is D (θ), where the indicates differentiation with respect to the time variable.

29 SOLUTION BANK 9 A/Math A Summer/Solution Bank/Clock.png How can we relate D(θ) with what we know? This is easy! We know an angle θ and the lengths of AB and AC in the picture, so let s just use the law of cosines. We get: That is: BC = AC + AB AC AB cos(θ) Which you can write as: D(θ) = cos(θ) D(θ) = cos(θ) Now differentiate with respect to time! We get: D(θ)D (θ) = 64 sin(θ) dθ dt And now, all we need to do is to plug in everything we know!

30 30 PEYAM RYAN TABRIZIAN First of all θ = π = π 6 (basically, the whole circle corresponds to π, and so of the circle corresponds to π ). In particular, sin(θ) = sin( π 6 ) =. Now, for d(θ), we use the law of cosines again, this time with the value θ = π 6 : D( π 6 ) = cos( pi 3 ) = = So, taking square roots, we get d( π 6 ) = = mm. Finally, we need to compute Dθ dt. But think about it! In hours, θ = π, so the speed of θ should be π = π 6 = π 6 rad/h (notice that θ is decreasing, so we wanted a negative answer!) Finally, we have all our information to get our final answer: D(θ)D (θ) = 64 sin(θ) Dθ dt (4 5 ( 3) D π ) = 64 6 π 6 (4 5 ( 3) D π ) = 3 π ( 6 6 D π ) π 6 = ( D π ) π 6 = ( D π ) π = ( 3 D π ) 8.55mm/h 6 So our final answer is D ( ) π 6 = π mm/h 8.55mm/h Section 3.0: Linear approximations and differentials L(x) = + ( ) 3 x π (a) dy = (x sin(x) + x cos(x))dx ( ) (b) dy = t +t +t dt = (a) dy = 0 e x 0 dx (b) dy = 0 (0.) = 0.0 t +t dt (y) = y(5) y(4) = 5 4 = 0 = 0. dy = 4 () = 8 = 0.5

31 SOLUTION BANK l = πr = 84, so r = 84 π = 4 π. We know dl = 0.5, so πdr = 0.5, so dr = 0.5 π = 4π (a) S = 4πr, so ds = 8πrdr = 8π 4 π 4π = 84 π. Also the relative error is ds S = 8πrdr 4πr = dr r = π π 4 = (b) V = 4 3 πr3, so dv = 4πr dr = 4π 4 π 4π = 764 π 79. Also the relative error is dv V = 4πr dr 4 = 3dr 3 πr3 r = 3 4π 4 = 3 68 = π df = 4kR 3 dr, so: df F = 4kR3 dr = 4kR3 dr F F And when dr R = 0.05, df F = 4(0.05) = 0. = 4kR3 dr kr (a) L(x) = f() + f ()(x ) = 5 (x ) = 6 x f(0.9) L(0.9) = 5 (0.9 ) = 5. f(.) L(.) = 5 (. ) = 4.9 = 4 dr R (b) Notice that (f ) (x) < 0, hence f (x) < 0, so f is concave down (Think for example x, we ll discuss that more in section 4.3), which means that the linear approximations will be overestimates (in other words, the tangent line to f at will be above the graph of f; again, think of the case x) cosh(x) + sinh(x) = ex + e x Section 3.: Hyperbolic functions + ex e x = ex + e x + e x e x = ex = ex 3... Just expand out the right-hand-side and use the fact that sinh(x) = e x e x, cosh(y) = ey e y, cosh(x) = ex +e x and sinh(y) = ey e y sinh(x) cosh(x) = ex e x e x + e x = 4 (ex e x )(e x +e x ) = (ex e x ) = ex e x 3... sinh(x) = 4 3 (use the fact that cosh (x) sinh (x) = and the fact that sinh(x) > 0 when x > 0). Then you get tanh(x) = sinh(x) cosh(x) = = 4 5, sech(x) = cosh(x) = 3 5, etc. = sinh(x) 3..3(a). (use tanh(x) = ex e x e x +e x the denominator) and factor out e x from the numerator and This is very similar to example 3 on page 57. However, there is a subtle point involved, check out the document Subtle point in 3..6 for more info! 3..9(a)(b). This is also very similar to example 4 on page 57. For (a), use the fact that cosh(cosh (x)) sinh(cosh (x)) = and cosh(cosh (x)) = x. Also, you ll need to fact that sinh(cosh (x)) 0 (and this is because cosh (x) 0 by definition, and sinh(x) 0 if x 0). (b) is even easier, use the fact that: tanh(tanh (x)) = sech(tanh (x)) and tanh(tanh (x)) = x.

32 3 PEYAM RYAN TABRIZIAN f (x) = sinh(x) + x cosh(x) sinh(x) = x cosh(x) y = +tanh (x) (sech (x)) = sech (x) +tanh (x) Chapter 3 Review TRUE FALSE. () TRUE () FALSE (3) TRUE (4) TRUE (5) FALSE ( f ( x) x ) (6) FALSE (e is a constant, so 0) (7) FALSE (y = ln(0)0 x, exponential rule) (8) FALSE (ln 0 is a constant, so 0) (9) FALSE ( tan(x) sec (x)) (0) FALSE ((x + ) x +x x +x ; Write x + x = (x + x) and use the chain rule) () TRUE () TRUE (f is a polynomial of degree 30, so its 3st derivative is 0) (3) TRUE (By the quotient rule and ()) (4) FALSE (y 4 = 4(x + ); it s not even the equation of a line!) (5) TRUE (= g () = 5() 4 = 80) 3.R.68. (a) sin(x) = sin(x) cos(x) (b) cos(x + a) = cos(x) cos(a) sin(x) sin(a) (The important thing here is that you differentiate with respect to x, leaving a constant) 3.R.87. v(t) = s (t) = Ace ct cos(ωt+δ) Aωe ct sin(ωt+δ) = Ae ct (c cos(ωt + δ) + ω sin(ωt + δ)) a(t) = v (t) = Ace ct (c cos(ωt + δ) + ω sin(ωt + δ)) Ae ct ( cω sin(ωt + δ) + ω cos(ωt + δ) ) 3.R.94. y(t) = 00 t 5.4 (a) y(0) = mg (b) t = 34.8 years 5.4 ln(00) ln() 3.R.99. () D = x +y (Typical Pythagorean theorem problem; Draw a right triangle in the shape of an L, and let x be the bottom side, y be the left-hand-side, and D be the hypothenus) () D dd dx dy dt = x dt + y dt (3) x = 3 5 = 45 (velocity time), y = = 60 (initial height + velocity time), dx dy dt = 5, dt = 5, and D = x + y = = 565 = 75 which gives: dd dt = = 3

33 SOLUTION BANK 33 3.R.0. () tan(θ) = 400 x (Typical trigonometry-problem. Draw another triangle in the shape of an L, let 400 be the left-hand-side, x be the bottom, and the angle on the right be θ) () sec (θ) dθ dt = 400 dx x dt (3) x = (redraw the same triangle, but this time with θ = π 6 ), dθ dt = 0.5, and θ = π 6, which gives: dx dt = R.05. The area of the window is given by y = x + π Then: dy = ( x ) ( ) = + π 8 x. ( + π ) ( xdx = + π ) (0)(0.) = + 3π R.. f (x) = x, so f (x) = ( x ) = x Section 4.: Maximum and Minimum Values - Absolute maximum: Does not exist (NOT 5) - Absolute minimum: f(4) = - Local minimum: f() =, f(4) = - Local maximum: f(3) = 4, f(6) = (g does not exist), (makes g (c) = 0) (F does not exist), 4, 8 7 (makes F (c) = 0) f (θ) = sin(θ) + sin(θ) cos(θ), which gives θ = πm or θ = πm, which can be just written as θ = πm (m is an integer) f (x) = xe 3x 3x e 3x, which gives x = 0 and x = f (x) = x 3 x 4x = x(x x ) = x(x )(x + ), which gives x =, 0,. Candidates: f( ) = 33, f(3) = 8 (endpoints), f(0) =, f( ) = 4, f() = 3. Absolute maximum: f( ) = 33, Absolute minimum: f() = ) Evaluate f at the endpoints 0 and π f(0) = + 0 =, f( π ) = = 0

34 34 PEYAM RYAN TABRIZIAN ) Find the critical numbers of f f (t) = sin(t) + cos(t) f (t) = 0 sin(t) + cos(t) = 0 sin(t) = cos(t) Here comes the tricky part! This seems impossible to solve, but ideally we d like to write the right-hand-side just in terms of sin(x) in order to have a shot at solving this! Start with cos(t) = cos (t) sin (t) (the double-angle formula for cos). Moreover cos (t) = sin (t) (because cos (t) + sin (t) = ) So we get cos(t) = sin (t) sin (t) = sin (t). So our original equation becomes: sin(t) = sin (t) which you can rewrite as sin (t) + sin(t) = 0. This again looks impossible to solve, but notice that this is just a quadratic equation in sin(t)! So let X = sin(t), then we get: X + X = 0 And using the quadratic formula (or your factoring skills), we get: (X )(X + ) = 0 So X = or X =. That is, sin(t) = or sin(t) =. HOWEVER, remember that we re only focusing on [0, π ], so in particular sin(t) = has only one solution in [0, π ], namely t = π 6, and sin(t) = has NO solution in [0, π ]. It follows that the only critical number of f in [0, π ] is t = π 6 (there are no numbers where f is not differentiable, so in fact those are all the critical numbers). And we get f( π 6 ) = = 3 3 3) Compare all the candidates you have: Our candidates are f(0) =, f( π ) = 0 and f( π 6 ) = Hence, the absolute minimum of f on [0, π ] is f( π ) = 0 (the smallest candidate), and the absolute maximum of f on [0, π ] is f( π 6 ) = 3 3 (the largest candidate)

35 SOLUTION BANK 35 Section 4.: The Mean Value Theorem f(0) = f(π) = 0, but f (x) = sec (x) > 0. This does not contradict Rolle s Theorem because f is not continuous on [0, π] (it is discontinuous at π ln(x) is continuous on [, 4], differentiable on (, 4); c = 3 ln(4) Let f(x) = x sin(x) At least one root: f(0) = < 0 and f(π) = π > 0 and f is continuous, so by the Intermediate Value Theorem (IVT) the equation has at least one root. At most one root: Suppose there are two roots a and b. Then f(a) = f(b) = 0, so by Rolle s Theorem there is at least one c (a, b) such that f (c) = 0. But f (c) = cos(c) 0, which is a contradiction, and hence the equation has at most one root Let f(x) = x 3 + e x At least one root: f( ) = + e < 0 and f(0) = 0 + e 0 = > 0 and f is continuous, so by the Intermediate Value Theorem (IVT) the equation has at least one root. At most one root: Suppose there are two roots a and b. Then f(a) = f(b) = 0, so by Rolle s Theorem there is at least one c (a, b) such that f (c) = 0. But f (c) = 3c + e c e c > 0, so f (c) 0, which is a contradiction, and hence the equation has at most one root By the MVT, f(4) f() 4 = f (c) for some c in (, 4). Solving for f(4) and using f() = 0, we get f(4) = 3f (c) = This is equivalent to showing: sin(a) sin(b) a b Which is the same as: sin(b) sin(a) b a Which is the same as: sin(b) sin(a) b a But by the MVT applied to f(x) = sin(x), we get: sin(b) sin(a) = cos(c) b a for some c in (a, b). However, cos(c), and so we re done!

36 36 PEYAM RYAN TABRIZIAN Let f(x) = sin (x), g(x) = cos ( x ). Then f (x) = x and: g 4x (x) = ( x ) = 4x + 4x 4x 4 = 4x 4x 4x 4 = (you get this by factoring out 4x out of the square root. This works because x 0) Hence f (x) = g (x), so f(x) = g(x) + C To get C, plug in x = 0, so f(0) = g(0) + C. But f(0) = g(0) = 0, so C = 0, whence f(x) = g(x) Let f(t) be the speed at time t. By the MVT with a = : 00 and b = : 0, we get: f( : 0) f( : 00) = f (c) : 0 : 00 But : 0 : 00 = 0 minutes = 6 h, so: = f (c) Whence: f (c) = 0 for some c between : 00 pm and : 0 pm. But f (c) is the acceleration at time c, and so we re done! This is again a proof by contradiction! Suppose f has (at least) two fixed points a and b. Then, by definition of a fixed point, f(a) = a, and f(b) = b. However, by the Mean Value Theorem, there is a c in (a, b) such that: f(b) f(a) b a = f (c) Now using the fact that f(b) = b and f(a) = a, we get: 4x x x = = f (x) x So b a b a = f (c) = f (c) That is, f (c) =. However, by assumption, f (x) for all x, so in particular setting x = c gives f (c). but since f (c) =, we get, which is a contradiction! Hence f has at most one fixed point!

37 SOLUTION BANK Section 4.3: How derivatives affect the shape of a graph (a) (0, ) (3, 7) (b) (, 3) (c) (, 4) (5, 7) (d) (0, ) (4, 5) (e) (, ), (4,.5), (5, 4) (a) f (x) = 6x + 6x 36 = 6(x )(x + 3); on (, 3) (, ), on ( 3, ) (b) Local max: f( 3) = 8; Local min: f() = 44 (c) f (x) = x + 6; CU on (, ), CD on (, ), IP (, f( 0.5) = 37 ) (a) f (x) = cos(x) sin(x); on (0, π 4 ) ( 5π 4, ), on ( π 4, 5π 4 ) (b) Local max: f( π 4 ) = ; Local min: f( 5π 4 ) = (c) f (x) = sin(x) + cos(x); CU on ( 3π 4, 7π 4 ), CD on (0, 3π 4 ) ( 7π 4 ( 3π 4, 0), ( 7π 4, 0), π), IP A possible graph looks like this:

38 38 PEYAM RYAN TABRIZIAN A/Math A Summer/Solution Bank/Concave-Kink.png (a) f (x) = 3x, on (, ) (, ), on (, ) (b) Local min: f() = 4, Local max: f( ) = 8 (c) f (x) = 6x; CD on (, 0), CU on (0, ); Inflection point at (0, ) (d) Draw the graph! (a) f (θ) = sin(θ) cos(θ) sin(θ) = sin(θ)( + cos(θ)); on (π, π), on (0, π) (b) Local min: f(π) =, no local max. (c) f (x) = cos(θ) + sin (θ) cos (θ) = cos(θ) + 4 cos (θ) = 4(cos (θ) cos(θ) ) = 4(cos(θ) + )(cos(θ) ); CU on ( π 3, 5π 3 on (0, π 3 ) ( 5π 3, π), IP ( π 3, 5 4 ), ( 5π 3, 5 4 ) (d) Draw the graph! (a) VA: x = 0, HA: y = (at ± ) (b) f (x) = x + x = x 3 x ; on (, 0) (, ), on (0, ) 3 (c) Local maximum at (, 5 4 ), No local minimum ), CD

39 (d) f (x) = 6x +x 3 x 6 IP at (3, 9 ) (e) Draw the graph! = 6+x x (a) No VA; HA: y = 0 (at ± ) (b) f (x) = ( x)e x, on (, 0), on (0, ) (c) Local maximum at (0, ), no local minimum SOLUTION BANK 39 (d) f (x) = ( + 4x )e x = (x )e x ; CU on ( ) IP at ±, e (e) Draw the graph! = x 6 x ; CD on (, 0) (0, 3), CU on (3, ); 4 ) ( ) (,, ; Let f(x) = tan(x) x, then f (x) = sec (x) = + tan (x) + = tan (x) > 0 on ( ) ( ) 0, π, hence f is increasing on 0, π. In particular, f(x) > f(0) = 0, so tan(x) x > 0, so tan(x) > x on ( ) 0, π (a) No, (b) Yes, (c) No, (a) Yes, 0 0 (b) No, 0 (c) Yes, (d) Yes, 0 (e) No, (f) Yes, Section 4.4: L Hopital s Rule cos(x) lim x ( π ) + sin(x) = x ( lim sin(x) π ) + cos(x) = 0 = 0 + = lim θ π sin(θ) csc(θ) = lim θ π cos(θ) (l Hopital s rule) cos(θ) sin (θ) = lim sin (θ)(cancelling out) θ π = ln(x) lim = lim x x x x x = lim = 0 x x tanh(x) lim x 0 tan(x) = lim sech (x) x 0 sec (x) =

40 40 PEYAM RYAN TABRIZIAN sin (x) lim = lim x 0 x x 0 x = x 3 lim x x3 e x = lim x e x = lim x 3x xe x = 3 lim x x e x = 3 lim = 3 x xe x 0 = lim x x x ln(x) = lim x ln(x) (x ) ln(x) + = lim x (x ) ln(x) x ln(x) + x = lim x x x + = x ln(x) lim x ln(x) = lim x( ) = ( 0) = x x x ) bx ) Let y = ( + a x ) ln(y) = bx ln( + a x ) 3) lim ln(y) = lim bx ln(+ a x x x ) = lim ln( + a x ) ( + )( a x = lim a x x ) x bx ( x )( b ) = lim x ab + a x = ab ( ) 4) So lim x + a bx x = e ab ) Let y = x x ) Then ln(y) = ln(x) x 3) So lim x ln(y) = lim x ln(x) x = 0 4) Hence lim x x x = lim x y = e 0 = (a) Here, you want to calculate lim t, so treat t as our x, and leave everything else as a constant!!!!. In particular, we get: So, we get: ct lim e m = 0 because c > 0 and m > 0 t lim v = mg t c ( 0) = mg c (b) Here, we let c 0 +, so we treat c as our x, and leave everything else as a constant!. Then, we have:

41 SOLUTION BANK 4 lim v = lim mg c 0 + c 0 + = lim c 0 + mg ( e ct m c ( ( t m (( t ) ) m e 0 =mg (( t ) ) m =mg = mgt m =gt ) ) e ct m ) (by l Hopital s rule) All you gotta do is show that: ) nt ( lim + r ) nt = e rt n n ) Let y = ( + r n ) ln(y) = nt ln( + r n ) 3) The important thing to realize is that you re taking the limit as n goes to, which means that r and t are constants! lim ln(y) = lim nt ln(+ r n n n ) = lim ln( + r n ) n = lim n nt r n + r n n t rn t n = lim n + r n = lim n rt + r n = rt + 0 = rt ( ) 4) So lim n + r nt ( ) n = e rt, and hence lim n A 0 + r nt n = A0 e rt Section 4.5: Summary of curve sketching D : R I : x intercepts: 0, 4, y intercept: 0 S : None A : None, but lim x ± f(x) = I : f (x) = (x 4) 3 + 3x(x 4) = (x 4) (x 4 + 3x) = (x 4) (4x 4) = 4(x 4) (x ); f is decreasing on (, ) and increasing on (, ); Local minimum f() = 7 C : f (x) = 8(x 4)(x ) + 4(x 4) = 4(x 4)(x + x 4) = 4(x 4)(3x 6) = (x 4)(x ); f is concave up on (, ), concave down on (, 4), and concave up on (4, ). Inflection points: (, 6), (4, 0)

42 4 PEYAM RYAN TABRIZIAN A/Math A - Fall 03/Homeworks/Quartic.png D : R {±3} I : No x intercepts, y intercept: y = 9 S : f is even A : Horizontal Asymptote y = 0 (at ± ), Vertical Asymptotes x = ±3 I : f (x) = x (x 9) ; f is increasing on (, 3) ( 3, 0) and decreasing at 0. on (0, 3) (3 ). Local maximum of 9 C : f (x) = 6 x +3 (x 9) 3 ; f is concave up on (, 3) (3, ) and concave down on ( 3, 3); No inflection points

43 SOLUTION BANK 43 A/Archive/Homeworks - Spring 0/hw0graph.png D : R I : No x intercepts, y intercept: y = S : No symmetries A : Horizontal Asymptotes: y = 0 (at ), y = (at ) I : f (x) = e x (+e x ) > 0, so f is decreasing on R C : f (x) = ex e x e x + 3 (multiply numerator and denominator by (e x ) 3 after simplifying), so f is concave down on (, 0) and concave up on (0, ). Inflection point at (0, )

44 44 PEYAM RYAN TABRIZIAN A/Math A Summer/Solution Bank/hw0graph3.png D : x > 0 I : No x intercept because f(x) > 0 for all x (see Increasing/Decreasing section). No y intercept (not defined at 0) S : No symmetries A : Vertical asymptote x = 0, No Horizontal Asymptote, because: ( lim x ln(x) = lim x ln(x) ) = ( 0) = x x x Also no slant asymptote, because if there were such a slant asymptote y = ax + b, then: And then: ln(x) x a = lim = x x b = lim (ln(x) x) ( )x = lim ln(x) = x x which is a contradiction! I : f (x) = x = x x, so decreasing on (0, ) and increasing on (, ); local minimum f(x) =. In particular f(x) for all x, and so f(x) > 0 (hence no x intercept) C : f (x) = x, concave up on (0, ); No inflection points.

45 SOLUTION BANK 45 A/Math A - Fall 03/Homeworks/x - ln(x).png Note: : First of all, f is periodic of period π, so from now on we may assume that x [0, π] D : We want sin(x) > 0, so the domain is (0, π) I : No y intercepts, x intercepts: Want ln(sin(x)) = 0, so sin(x) =, so x = π S : Again, f is periodic of period π A : No horizontal/slant asymptotes, but lim x 0 + ln(sin(x)) = ln(0 + ) =, so x = 0 is a vertical asymptote. Also lim x π ln(sin(x)) =, so x = π is also a vertical asymptote. I : f (x) = cos(x) sin(x) = cot(x), then f (x) = 0 x = π, and using a sign table, we can see that f is increasing on (0, π ) and decreasing on ( π, π). Moreover, f( π ) = ln() = 0 is a local maximum of f. C : f (x) = csc (x) < 0, so f is concave down on (0, π).

46 46 PEYAM RYAN TABRIZIAN A/Math A Summer/Solution Bank/hw0graph.png At : Suppose the slant asymptote is y = ax + b, then: x tan (x) a = lim = lim x x tan π (x) = x x = b = lim x x tan (x) x = lim x tan (x) = π Hence x tan (x) has a slant asymptote of y = x π at At : Suppose the slant asymptote is y = ax + b, then: a = b = x tan (x) lim = lim x x tan π (x) = x x = lim x x tan (x) x = lim x tan (x) = Hence x tan (x) has a slant asymptote of y = x + π ( π at ) = π D Dom = R I y intercept: f(0) = 0, x intercept: 0 (there are no others, because f is increasing; see Increasing/Decreasing section) S No symmetries A No vertical asymptotes (f is defined everywhere), Slant Asymptotes y = x π at, y = x + π at ; No H.A. because there are already two S.A.

47 SOLUTION BANK 47 I f (x) = +x = x +x 0, so f is increasing everywhere; No local max/min C f (x) = x(+x ) x (x) = x, so f is concave down on (, 0) and (+x ) (+x ) concave up on (0, ). Inflection point (0, f(0)) = (0, 0) A/Math A - Fall 03/Homeworks/x - arctan(x).png Section 4.7: Optimization Problems - Want to minimize x + y - But xy = 00, so y = 00 x - Let f(x) = x + 00 x - x > 0 (x is positive), so x + y = x + 00 x - f (x) = 0 00 x = 0 x = 00 x = 0 - By FDTAEV, x = 0 is the absolute minimizer of f - Answer: x = 0, y = 00 0 = The picture is as follows:

48 48 PEYAM RYAN TABRIZIAN A/Math A Summer/Solution Bank/Fence.png Want to minimize 3w + 4l - But lw =.5, so l = 0.75 w, so 3w + 4l = 3w + 3 w - Let f(w) = 3w + 3 w - w > 0 - f (x) = w = 0 w = w = - By FDTAEV, w = is the absolute minimum of f - Answer: w =, l =.5 - Want to minimize S = x + 4xh (where x is the length of the base-side and h is the height) - However, V = x h = 3000, so h = 3000 x, so x + 4xh = x + 4x 3000 x = x x - Let f(x) = x x - x > 0 - f (x) = 0 x 8000 x = 0 x 3 = 8000 x = = 40 - By FDTAEV, x = 40 is the absolute minimizer of f - Answer: x = 40, h = 3000 (40) = = 0 - We have D = (x ) + y, so D = (x ) + y - But y = 4 4x, so D = (x ) + 4 4x - Let f(x) = (x ) + 4 4x - No constraints - f (x) = (x ) 8x = 6x = 0 x = 3 - By the FDTAEV, x = 3 is the maximizer of f. - Since y = 4 4x, we get y = = 3 9, so y = ± 3 9 = ± 4 3

49 SOLUTION BANK 49 - Answer: ( 3, 4 3 ) ( and 3, 4 3 ) Picture: A/Math A Summer/Solution Bank/hw0opt.png We have A = xy, but the trick here again is to maximize A = x y (thanks for Huiling Pan for this suggestion!) - But x + y = r, so y = r x, so A = x (r x ) = x r x 4 - Let f(x) = x r x 4 - Constraint 0 x r (look at the picture) - f (x) = xr 4x 3 = 0 x = 0 or x = r - By the closed interval method, x = r f(0) = f(r) = 0 - Answer: x = r, y = r r = r is a maximizer of f (basically - A = rh + πr - But P = πr + r + h = 30, so h = 5 r π r - Let f(r) = r ( 5 r π r) + πr = 30r r πr + πr = 30r r π r - Constraint r > 0 - f (r) = 30 4r πr = 0 r = 30 π+4 - By FDTAEV, r = 30 π+4 is the minimizer of f - r = π+4, h = 5 π+4 5π π+4 = 30 π+4 = r

50 50 PEYAM RYAN TABRIZIAN Let t AB be the time spent rowing from A to B and t BC be the time spent walking from B to C - By the formula time = distance velocity, we have: t AB = AB = cos(θ)ac = 4 cos(θ) = cos(θ) t BC = BC 4 = BOC 4 = θ 4 = θ (here O is the origin; it is a geometric fact that BOC = BAC) - Let f(θ) = cos(θ) + θ - Constraint: 0 θ π (see the picture!) - f (θ) = sin(θ) + = 0 sin(θ) = θ = π 3 - f(0) =, f ( ) π = π and f ( ) π 3 = 3 + π 3 = 3 + π 3. By the closed interval method, θ = π is an absolute minimizer. - Therefore, she should just walk! (which makes sense because she walks much faster than she rows!) (a) c (x) = C (x)x C(x) x. When c is at its minimum, c (x) = 0, so C (x)x C(x) = 0, so C (x) = C(x) x = c(x), so C (x) = c(x), i.e. marginal cost equals the average cost! (a) p(x) = 550 x 0 (Basically imitate Example 6 on page 33) (b) The revenue function is R(x) = xp(x) = 550x x 0. R (x) = = x 5 x = 750, and the corresponding price is p(750) = = 75 and the rebate is = 75 dollars (c) Here the profit function is P (x) = R(x) C(x) = 550x x x. P (x) = x 5 50 = 0 x = 000, so the corresponding price is p(000) = = 350, so the corresponding rebate is = 00 dollars (thank you Brianna Grado-White for the solution to this problem!) The picture is as follows:

51 SOLUTION BANK 5 A/Math A Summer/Solution Bank/hw0opt.png Here, h and h and L are fixed, but x varies. Now the total time taken is t = t + t = d v + d v. Now, by the Pythagorean theorem: d = x + h and d = (L x) + h, so we get: And t(x) = x + h v + (L x) + h v t x x L (x) = + v x + h v (L x) + h = x v d + x L v d Setting t (x) = 0 and cross-multiplying, we get: v d (L x) = v d x So, by definition of sin(θ ) and sin(θ )), we get: v v = d x (L x)d = x d = sin(θ ) L x sin(θ d ) The picture is as follows (Note that the two θ s are indeed the same!)

52 5 PEYAM RYAN TABRIZIAN A/Math A - Fall 03/Homeworks/Pipe.png - We want to minimize L + L - cos(θ) = L 9, so L = 9 cos(θ), sin(θ) = L 6, so L = 6 sin(θ) - Let f(θ) = 9 cos(θ) + 6 sin(θ) - Constraint: 0 < θ < π (Notice that at 0 and π, we can t carry the pipe horizontally around the corner; it would break at that corner) - f (θ) = 9 sin(θ) 6 cos(θ) cos (θ) + = 9 sin3 (θ) 6 cos 3 (θ) = 0 sin (θ) cos (θ) sin (θ) ( ) 3 9 sin 3 (θ) 6 cos 3 (θ) = 0 sin(θ) cos(θ) = 6 9 = 3 tan3 (θ) = 3 ( ) θ = tan 3 3 ( ) - By FDTAEV, θ = tan 3 3 is the absolute minimizer of f ( ) 9 - Answer: cos(θ) + 6 sin(θ), where θ = tan 3 3 (if you want to, you can simplify this using the triangle method: = + x and cos(tan (x)) sin(tan (x)) = +x x, but I think this is enough torture for now :) Section 4.8: Newton s method As usual, a good picture is the key to solving the problem:

53 SOLUTION BANK 53 A/Math A Summer/Solution Bank/Newton.png Note: This picture is very complete. It is meant to illustrate all the points I am making below. Here, we are given a lot of information, so let s try to tackle this problem one step at a time! First, let s calculate the equation of any tangent line to the graph of f(x) = sin(x) that goes through (0, 0). Later, we will be worrying about finding the one that has the largest slope. By definition of the derivative, the tangent line to the graph of f at c has slope f (c) = cos(c), so any such tangent line that also goes through (0, 0) has equation: y 0 = cos(c)(x 0), i.e. y = cos(c)x. Finally, we know that the tangent line goes through (c, sin(c)) (i.e. goes through the graph of f at c), so we get: sin(c) = cos(c) c, i.e. tan(c) = c. So any tangent line at c with the above properties must solve tan(c) = c, i.e. tan(c) c = 0. So what we really need to do is to approximate the zero of the function g(x) = tan(x) x. Now this looks like a Newton s method problem! But remember, that for Newton s method, we need to find a good initial guess, and here is where we use the information that the tangent line must have largest slope! First of all, notice that f(x) = sin(x) is odd, so the graph is symmetric about the origin! If you look at the picture above, you ll see that the tangent line to the graph at c is the same as the tangent line at c. This means we can restrict ourselves to the right-hand-side of the picture, i.e. c 0! (if you don t understand this argument, don t worry, it s just a simplification) And if you look at the picture again, you ll notice that if your initial guess is between 0 and π, your successive approximations will got to 0. And you don t want that because the slope of the tangent line at 0 is (which is not the greatest slope). The same problem arises with the initial guess between 3π and π (the

54 54 PEYAM RYAN TABRIZIAN approximations go to π) Finally, notice that when c gets larger and larger, the tangent line at c has smaller and smaller slope (see picture: The brown line has a smaller slope than the blue line, which has a smaller slope than the green line), so you d like your initial guess not to be too large. In particular, we don t want the initial guess to be larger than 3π! From this analysis, we conclude that any initial guess between π and 3π is good enough! (also look at the picture above: the green tangent line seems to be the winner here) For example, with x 0 = 4.5 (or you could try x 0 = π), we get the successive approximations (remember that you are applying Newton s method to g(x) = tan(x) x, NOT f(x)!): x 0 =4.5 x = x = x 3 = x 4 = And so, our approximation is: c And hence the largest slope is approximately equal to cos( ) (because f (c) = cos(c)). To summarize: Draw a picture Derive the function that you want to apply Newton s method to (i.e. g(x) = tan(x) x) Argue that your intial approximation must be between π and 3π (YOU NEED TO JUSTIFY THIS PART!, maybe not as precise as I did, but there needs to be some justification Apply Newton s method to g with initial approximation between π and 3π (4 would work) R(θ) = sec(θ) e θ Section 4.9: Antiderivatives f(x) = sin(t) + tan(t) + C, but 4 = f( π 3 ) = C = C, so f(x) = sin(t) + tan(t) If f (θ) = sin(θ) + cos(θ), then f (θ) = cos(θ) + sin(θ) + C. f (0) = 4, so C = 4, so C = 5. Hence f (θ) = cos(θ) + sin(θ) + 5. Hence f(θ) = sin(θ) cos(θ) + 5θ + C.

55 SOLUTION BANK 55 f(0) = 3, so C = 3, so C = 4. Hence f(θ) = sin(θ) cos(θ) + 5θ a(t) = 0 sin(t) + 3 cos(t), so v(t) = 0 cos(t) + 3 sin(t) + A, so s(t) = 0 sin(t) 3 cos(t) + At + B Now, s(0) = 0, but s(0) = 0(0) 3() + A(0) + B, so 3 + B = 0, so B = 3 So s(t) = 0 sin(t) 3 cos(t) + At + 3 Moreover, s(π) =, but s(π) = 0(0) 3() + A(π) + 3 = A(π), so A(π) =, so A = π = 6 π So altogether, you get: s(t) = 0 sin(t) 3 cos(t) + 6 π t Suppose the acceleration of the car is a(t) = A. Then v(t) = At + B and s(t) = A t + Bt + C. However, at t = 0, the car is moving at 00 km/h, so v(0) = 00, so B = 00, hence v(t) = At + 00 and s(t) = A t + 00t + C. Moreover, at t = 0, the car is at its initial position 0, so s(0) = 0, so C = 0, hence s(t) = A t + 00t Now let t be the time needed to real the pile-up. We want the car to have 0 velocity at t, hence v(t ) = 0, hence At + 00 = 0, so At = 00 Moreover, we want s(t ) = 80m = 0.08 km, so A (t ) + 00t = 0.08, but using the fact that At = 00, this just becomes: 00t + 00t = 0.08, so 50t = 0.08, so t = 65. Therefore A = 00 t = = 6500 km/h, so the answer is 6500 km/h (a) (i) x =, so Section 5.: Areas and Distances L 6 = f(0)()+f()()+f(4)()+f(6)()+f(8)()+f(0)() = (ii) ++8 = R 6 = f()()+f(4)()+f(6)()+f(8)()+f(0)()+f()() = = (iii) M 6 = f()()+f(3)()+f(5)()+f(7)()+f(9)()+f()() = =

56 56 PEYAM RYAN TABRIZIAN (b) Overestimate (c) Underestimate (d) M 6 (just right, does not overshoot, like L 6, but not undershoot either, like R 6 ) (a) If n = 3, then x =, and if n = 6, x =, so: R 3 = f(0)() + f()() + f()() = = 8 R 6 =f( 0.5)(0.5) + f(0)(0.5) + f(0.5)(0.5) + f()(0.5) + f(.5)(0.5) + f()(0.5) =.5(0.5) + (0.5) +.5(0.5) + (0.5) + 3.5(0.5) + 5(0.5) =6.875 (b) L 3 = f( )() + f(0)() + f()() = + + = 5 L 6 =f( )(0.5) + f( 0.5)(0.5) + f(0)(0.5) + f(0.5)(0.5) + f()(0.5) + f(.5)(0.5) =(0.5) +.5(0.5) + (0.5) +.5(0.5) + (0.5) + 3.5(0.5) =5.375 (c) M 3 = f( 0.5)() + f(0.5)() + f(.5)() = 5.75 M 6 = f( 0.75)(0.5)+f( 0.5)(0.5)+f(0.5)(0.5)+f(0.75)(0.5)+f(.5)(0.5)+f(.75)(0.5) = (d) M Here n = 6 and x = 0.5 L 6 =v(0)(0.5) + v(0.5)(0.5) + v()(0.5) + v(.5)(0.5) + v()(0.5) + v(.5)(0.5) =0(0.5) + 6.(0.5) + 0.8(0.5) + 4.9(0.5) + 8.(0.5) + 9.4(0.5) =34.7 R 6 =v(0.5)(0.5) + v()(0.5) + v(.5)(0.5) + v()(0.5) + v(.5)(0.5) + v(3)(0.5) =6.(0.5) + 0.8(0.5) + 4.9(0.5) + 8.(0.5) + 9.4(0.5) + 0.(0.5) = The midpoint sum seems to best approximate the area: M 6 = v(0.5)()+v(.5)()+v(.5)()+v(3.5)()+v(4.5)()+v(5.5)() = = 53ft

57 SOLUTION BANK Here x = n and x i = + i n, so: lim n n i= Here x = π n and x i = πi n, so: lim n i= ( ) + i n ( ) + i n + π n sin ( ) πi n 5... The area under the curve of f(x) = x 0 from 5 to 7 (or, if you want, the area under the curve of f(x) = (x + 5) 0 from 0 to ) The area under the curve of f(x) = tan(x) from 0 to π (a) We are given that the polygon is made out of n congruent triangles, so A n = n T, where T is the area of each triangle. So all we need to find is T. Here again, a picture tells a thousand words, so by drawing the picture of such a triangle, we can figure out its area: A/Math A Summer/Solution Bank/Polygon.png Using the picture, you ll notice that: T = A B = A B And we can divide the triangle into two right triangles, and hence use trigonometry to calculate A and B! Here, θ = π n, the central angle!

58 58 PEYAM RYAN TABRIZIAN We get: cos ( ) θ = B r B = r cos ( ) θ sin ( ) θ = A r A = r sin ( ) θ And so, we get: T = A B = r sin ( θ ) r cos ( ) θ = r sin ( ) θ cos ( ) θ = r ( sin θ ) = r sin(θ) = r sin ( ) π n Here, we used the fact that, in general, sin(x) cos(x) = sin(x), so sin(x) cos(x) = sin(x). And so, we have: A n = n T = n r sin ( ) π = n nr sin ( ) π n (b) Actually, the hint tells us that we don t even have to use l Hopital s rule, but rather the rule that: sin(x) lim = x 0 x We have that:

59 SOLUTION BANK 59 ( ) lim A π n = lim n n r n sin n = ( ) π r lim n sin n n = sin ( ) π r n lim n = r lim n n =π r lim n =πr lim x 0 sin(x) x =πr () =πr π sin ( ) π π n n sin ( ) π π n n ( x = π ) n The basic idea is that, if we have an indeterminate form of the form 0, we rewrite as 0, or we rewrite 0 as. Here, for example, we wrote n = in order to apply the hint in the problem! n And hooray, you just proved that the formula for the area of a circle of radius r is πr. But actually, you didn t, because trigonometry, which you used in (a), relies heavily on this formula! π π cos(x) x dx Section 5.: The definite integral 5... First of all, a =, b = 5, x = 3 n and x i = + 3i n

60 60 PEYAM RYAN TABRIZIAN 5 4 xdx = lim n 3 n n n i= 3 = lim n n 3 = lim n n i= 8 = lim n n 8 = lim n n = lim n = lim n = lim n = 9 ( n ( ( 4 + 3i n ( 4 4 6i n i= ) 6 n i ) ( n i i= ( ) n(n + ) 9n n + n 9n n ( ) + n 9 ( ) + n ) )) First of all, a =, b = 0, x = n and x i = + i n

61 SOLUTION BANK 6 0 n ( x + xdx = lim + i ) ( + + i ) n n n n i= n ( = lim 4 8i n n n + 4i n + i ) n i= n ( 4i = lim n n n 6i ) n + i= n ( ) 4i = lim n n n n ( ) 6i + n () n n n i= i= i= ( n ) ( 8 = lim n n 3 i n ) ( n i + 4 n ) n i= i= i= ( ) 8 n(n + )(n + ) = lim n n 3 ( ) n(n + ) 6 n + 4 n n ( 8 = lim + ) ( + ) ( 6 + ) + 4 n 6 n n n = = 8 3 = x = 9 n, so x i = + 9i n, and so: x 4 ln(x)dx = lim n i= n (x)(x i 4 ln(x i )) = lim n n i= (( 9 + 9i ) ( 4 ln + 9i )) n n n (a) 4 (the area of the large triangle) (b) π (minus the area of the semicircle) (c) 4 π + = 9 π (the area of the large triangle minus the area of the semicircle plus the area of the small triangle) π 4 (the area of a rectangle plus the area of a quarter of a circle of radius 3) x dx = 5 0 dx 6 0 x dx = = 5 = ex dx = 3 ex dx 3 dx = (e3 e)

62 6 PEYAM RYAN TABRIZIAN f(x)dx+ 5 f(x)dx f(x)dx = 5 f(x)dx f(x)dx = 5 f(x)dx m f(x) M, so integrating from 0 to, we get m f(x)dx M 0 f(x)dx = 5 f(x)dx On [0, ], x x, so + x + x, so + x + x, so integrating from 0 to, we get 0 + x dx 0 + xdx As usual, let f(x) be as in the problem, a = 0, b =, x i = i n, and (x) = n. First, let s draw a picture of what s going on: A/Math A Summer/Solution Bank/Nonintegrable.png In the picture above, the green dots represent where f(x) = 0 and the red lines represent where f(x) =. This problem is unlike the problem above! In this case, the function does not blow up to infinity, but it can t make up its mind! We need to somehow use this fact in order to show that f is not integrable! But even though this problem is different, the general strategy is almost the same. f integrable means that no matter how we choose the x i, we get the same answer! So to show that something is NOT integrable, we have to pick two different sets of points x i and y i that give us two different answers!