MATH A - FINAL EXAM DELUXE - SOLUTIONS PEYAM RYAN TABRIZIAN. ( points, 5 points each) Find the following limits (a) lim x x2 + x ( ) x lim x2 + x x2 + x 2 + + x x x x2 + + x x 2 + x 2 x x2 + + x x x2 + + x = = (b) lim x + xx2 ) Let y = x x2 2) Then ln(y) = x 2 ln(x) 3) ln(x) lim ln(y) x2 ln(x) x + x + x + x 2 4) Hence lim y = x e = H x + x 2 x 3 x 2 x + 2 = Date: Friday, August 2th, 2.
2 PEYAM RYAN TABRIZIAN 2. ( points) Use the definition of the derivative to calculate f (x), where: f(x) = x 2 f f(x) f(a) (a) x a x a x 2 a 2 ) x a x a (x a)(x + a) x a x a x + a x a =2a Hence, f (x) = 2x
MATH A - FINAL EXAM DELUXE - SOLUTIONS 3 3. ( points, 5 points each) Find the derivatives of the following functions (a) y, where x y = y x Hint: Take lns first! Taking lns: y ln(x) = x ln(y) Differentiating and solving for y : y ln(x) + y xy = ln(y) + x y ( y ln(x) x ) = ln(y) y y x y = ln(y) y x ln(x) x y (b) y at (, ), where sin(y) = x 2 y 2 Differentiating: cos(y)y = 2x 2yy Now plug in x = and y = : ()y = y =
4 PEYAM RYAN TABRIZIAN 4. (5 points) Assume Peyam s happiness function is given by: Where: H = M 2 L + 2G M is the happiness due to holding office hours L is the happiness due to lecturing G is the happiness due to grading exams Assume that at the end of the summer: Peyam s happiness due to holding office hours is 5 utils/week, and is decreasing by 2 utils/week Peyam s happiness due to lecturing is utils/week, and is decreasing by util/week Peyam s happiness due to grading exams, is 2 utils/week, and is decreasing by utils/week. Question: By how much is Peyam s happiness increasing/decreasing at the end of the summer? ) No picture needed 2) WTF dh dt 3) H = M 2 L + 2G 4) Differentiating, we get: dh dt = 2M dm dt L + M 2 dl dt + 2dG dt 5) Plug in M = 5, dm dt = 2, L =, dl dt dh dt = 2(5)( 2)() + (5)2 ( ) + 2( ) =, dg dt = :
MATH A - FINAL EXAM DELUXE - SOLUTIONS 5 6) Fortu- dh = 2 25 2 = 227 dt Peyam s happiness is decreasing by 227 utils/week. nately, this doesn t correspond to reality :) 5. (2 points) What is the area of the largest rectangle that can be put inside the parabola y = 4 x 2? ) Picture: A/Math A Summer/Exams/FDrectangle.png 2) In the picture above, the length of the rectangle is 2x and the width is y, so the area is: A = 2xy Now (x, y) is on the parabola, so y = 4 x 2, whence: A(x) = 2x(4 x 2 ) = 8x 2x 3
6 PEYAM RYAN TABRIZIAN 3) Constraint: The constraint is x 2. (you find the 2 by solving 4 x 2 = ) 4) A (x) = 8 6x 2 = 6x 2 = 8 x 2 = 8 6 = 4 4 3 x = 3 = 2 3 Now ( A() ) = and A(2) =, so by the closed interval method, 2 A 3 is the biggest area, and: A ( 2 3 ) = 2 ( 2 3 ) ( 4 ( 2 3 ) 2 ) = 4 3 ( 4 4 3 ) = 4 3 ( 8 3 ) = 32 3 3
MATH A - FINAL EXAM DELUXE - SOLUTIONS 7 6. (5 points) (a) (3 points) Show that the following equation has exactly one solution: cos(x) = 2x Let f(x) = cos(x) 2x At least one solution: f() = = >, f(π) = 2π <, f is continuous, so by the IVT, f has at least one zero. At most one solution: Suppose f has two zeros a and b. Then f(a) = f(b) =, so by Rolle s theorem, there is some c with f (c) =. But = f (c) = sin(c) 2 <, so <, contradiction! Therefore, f has exactly one zero, and hence cos(x) = 2x has exactly one solution. (b) (2 points) Use part (a) to show that the following function has exactly one critical point: g(x) = sin(x) x 2 g (x) = cos(x) 2x = f(x). But we ve shown in (a) that f has exactly one zero, hence g (x) has exactly one zero, that is g has exactly one critical point.
8 PEYAM RYAN TABRIZIAN 7. (2 points) Use the definition of the integral to evaluate: ( x 3 2 ) dx You may use the following formulas: = n i= i = i= n(n + ) 2 Note: 2 for not writing lim i = i= n(n + )(2n + ) 6 i= i 3 = n2 (n + ) 2 4 Preliminary work: f(x) = x 3 a =, b =, x = n x i = i n = n 2 x 3 2dx xf(x i ) i= ( ) ( ( ) ) 3 i n n ( ) ( ) i 3 n n 3 i= i= i= n 4 n 4 i 3 n 4 ( ) i 3 i= (n + ) 2 4 = 4 ( ) n 2 (n + ) 2 4
MATH A - FINAL EXAM DELUXE - SOLUTIONS 9 Check: (not required, but useful) [ ] x x 3 4 dx = = 4 4 = 4 8. (3 points, 5 points each) Find the following: (a) x2 dx Note: Don t spend too much time on this one, either you know it or you don t! The integral represents the area of a semicircle of radius, hence: x2 dx = 2 π()2 = π 2 (b) The antiderivative F of f(x) = 3e x + 4 sec 2 (x) which satisfies F () =. The MGAD of f is F (x) = 3e x + 4 tan(x) + C. To find C, use the fact that F () =, so 3 + + C =, so C = 2, hence: F (x) = 3e x + 4 tan(x) 2 (c) g (x), where g(x) = e x sin(t 3 )dt x 2 Let f(t) = sin(t 3 ), then: g(x) = F (e x ) F (x 2 ), so: g (x) = F (e x )(e x ) F (x 2 )(2x) = f(e x )e x f(x 2 )(2x) = sin ( e 3x) e x sin ( x 6) (2x) (d) (cos(x)) 3 sin(x)dx Let u = cos(x), then du = sin(x)dx, so: (cos(x)) 3 sin(x)dx = u 3 ( du) = u4 4 + C = (cos(x))4 + C 4
PEYAM RYAN TABRIZIAN ( (ln(x)) 3 (e) e 2 e x ) dx Let u = ln(x), then du = dx, and u(e) = ln(e) = and x u(e 2 ) = 2, so: e 2 e ( (ln(x)) 3 x ) dx = 2 [ u u 3 4 du = 4 ] 2 = 5 4 ) (f) The average value of f(x) = sin (x 5 ) (cos (x 2 ) + e x2 + x 4 on [ π, π] π π sin (x5 ) (cos (x 2 ) + e x2 + x 4 ) dx π because f is an odd function. = 2π =
MATH A - FINAL EXAM DELUXE - SOLUTIONS 9. ( points) Find the area of the region enclosed by the curves: y = cos(x) and y = cos(x) from to π. Hint: It might help to notice a certain symmetry in your picture! Picture: A/Math A Summer/Exams/Finalarea.png Then determine the points of intersection between the two curves: cos(x) = cos(x) 2 cos(x) = cos(x) = x = π 2 On [, π 2 ], cos(x) is above cos(x), and on [ π 2, π], cos(x) is above cos(x), so we ll have to figure out A + B as in the picture. However, notice the symmetry! Namely, A = B, so all we really need to calculate is A + B = 2A, that is:
2 PEYAM RYAN TABRIZIAN 2 π 2 (cos(x) ( cos(x))) dx π =2 2 2 cos(x)dx =4 π 2 =4 [sin(x)] π 2 =4( ) =4 cos(x)dx
MATH A - FINAL EXAM DELUXE - SOLUTIONS 3. ( points) If f(x) = x3 3 x2 2, find: (a) Intervals of increase and decrease, and local max/min f (x) = x 2 x = x(x ) = if x =,. Now drawing a sign table, you should see that: f is increasing on (, ), decreasing on (, ), and increasing on (, ). And hence f has a local max f() =, and a local min f() = 3 2 = 6 (b) Intervals of concavity and inflection points (just give me the x coordinate of the IP) f (x) = 2x = if x = 2. Hence f is concave down on (, 2) and concave up on ( 2, ). Moreover, f has an inflection point at x = 2.
4 PEYAM RYAN TABRIZIAN Bonus (5 points) Fill in the gaps in the following proof that the function f is not integrable on [, ]: { if x is rational f(x) = if x is irrational Step : Pick x i such that x i is rational. Then: f(x)dx = xf(x i ) i= i= n () Step 2: Pick x i such that x i is irrational. Then: f(x)dx xf(x i ) i= n () i= n i= n (n) = Since we get two different answers for the integral, we have a contradiction.. And hence f is not integrable on [, ]. Note: See the handout Integration sucks!!! for a nice discussion of this problem!
MATH A - FINAL EXAM DELUXE - SOLUTIONS 5 Bonus 2 (5 points) Another way to define ln(x) is: ln(x) = x t dt Show using this definition only that ln(e x ) = x. Hint: Let g(x) = ln(e x ) = e x dt. Differentiate g, simplify, t and antidifferentiate. Make sure you face the issue of the constant! Let f(t) = t, then g(x) = F (ex ) F (), so: g (x) = F (e x )(e x ) = f(e x )(e x ) = e x ex = Since g (x) =, we get g(x) = x + C. To figure out what C is, plug in x =, and we get: e g() = + C dt =C t dt =C t =C C = Hence g(x) = x, so ln(e x ) = x
6 PEYAM RYAN TABRIZIAN Bonus 3 (5 points) Define the Product integral b a f(x)dx as follows: If we define x, x i, and x i as usual, then: b a f(x)dx (f(x )) x (f(x 2)) x (f(x n)) x (that is, instead of summing up the f(x i ), we just multiply them!) Show that this is nothing new, that is, express b a f(x)dx in terms of b f(x)dx a Hint: How do you turn a product into a sum? Let P = b a f(x)dx. Then: ( ln(p ) = ln lim (f(x )) x (f(x 2)) x (f(x n)) x) ln ((f(x )) x (f(x 2)) x (f(x n)) x) ln ( f(x ) x) + ln ( f(x 2) x) + + ln ( f(x n) x) x ln (f(x )) + x ln (f(x 2)) + + x ln (f(x n)) = b a x ln (f(x i )) i= ln(f(x))dx So ln(p ) = b a ln(f(x))dx, so P = e b a ln(f(x))dx, hence: b a f(x)dx = e b a ln(f(x))dx