Math 160 - Final Solutions - Spring 2011 - Jaimos F Skriletz 1 Answer each of the following questions to the best of your ability. To receive full credit, answers must be supported by a sufficient amount of work using the methods presented in class. 1. (10 pts) Find the implicit domain of the real valued function f(x) = ln(x+4) The logarithm can only accept positive inputs thus x+4 > 0 x > 4 Thus the domain is ( 4, ). 2. (10 pts) Evaluate the following limits: 3x 2 27 (a) lim x 3 x 2 x 6 = lim 3(x+3)(x 3) x 3 (x+2)(x 3) = lim 3(x+3) = 3(6) = 18 x 3 x+2 5 5 (b) lim x 3x 2 27 x 2 x 6 = lim x x 2 (3 27 x 2 ) x 2 (1 1 x 6 x 2 = 3 0 1 0 0 = 3
Math 160 - Final Solutions - Spring 2011 - Jaimos F Skriletz 2 3. (15 pts) Consider the piecewise defined function f(x) = (a) Find f(3) = (3) 2 5 = 9 5 = 4 { x+6 x > 3 x 2 5 x 3 (b) Find lim x 3 f(x) = 32 5 = 4 (c) Find lim x 3 + f(x) = 3+6 = 3 (d) Find lim x 3 f(x) D.N.E. This limit does not exist since the right limit is not equal to the left limit above. (e) Determine if f(x) continuous at x = 3 by using the three criteria for continuity. f(x) is not continuous at x = 3 because it breaks the second property that lim x 3 f(x) exists. In this case since the directional limits exist but are not equal, there is a break in the function. (f) Provide a rough sketch of f(x).
Math 160 - Final Solutions - Spring 2011 - Jaimos F Skriletz 3 4. (15 pts) Use the limit definition of the derivative to find the derivative of f(x) = 9 4x 2. f (x) = lim h 0 f(x+h) f(x) h [9 4(x+h) 2 ] [9 4x 2 ] = lim h 0 h [9 4x 2 8xh 4h 2 ] [9 4x 2 ] = lim h 0 h 8xh 4h 2 = lim h 0 h h( 8x 4h) = lim h 0 h = lim( 8x 4h) = 8x h 0 5. (10 pts) Find the equation of the tangent line to the curve y = 16 x at the point (2,8). If y = 16x 1 then y = 16x 2 = 16 x. 2 The slope of the tangent line when x = 2 is m = y (2) = 16 2 = 4. 2 Thus the equation of this tangent line is y y 1 = m(x x 1 ) y 8 = 4(x 2) y 8 = 4x+8 y = 4x+16
Math 160 - Final Solutions - Spring 2011 - Jaimos F Skriletz 4 6. (25 pts) Find the derivatives of the following functions, simplification is not required: (a) f(x) = x 1+x 2 f (x) = (1)(1+x2 ) (2x)(x) (1+x 2 ) 2 = 1 x2 (1+x 2 ) 2 (b) g(x) = 15e x2 +2x 3 g (x) = 15e x2 +2x 3 (2x+2) = 30(x+1)e x2 +2x 3 (c) h(x) = (1+x 2 ) 120 x 2 = (1+x 2 )(120 x 2 ) 1/2 h (x) = [(2x)](120 x 2 ) 1/2 +[(1/2)(120 x 2 ) 1/2 ( 2x)](1+x 2 ) = x(120 x 2 ) 1/2 [2(120 x 2 ) (1+x 2 )] = x(239 3x2 ) 120 x 2
Math 160 - Final Solutions - Spring 2011 - Jaimos F Skriletz 5 7. (15 pts) Consider the function f(x) = 3x 2 x 3 +9x 5. Use the methods of calculus to find the following: (a) Find the intervals of increasing/decreasing. (b) Find all local extrema. (c) Find the intervals of concavity. (d) Find all inflection point(s). f (x) = 6x 3x 2 +9 = 3(x 2 2x 3) = 3(x 3)(x+1) If f (x) = 0 then x = 1 or x = 3 are the critical points. A sign table of f (x) shows 1 3 ( ) (+) ( ) Thus f(x) is increasing on ( 1,3). f(x) is decreasing on (, 1) and (3, ). ( 1,f( 1)) = ( 1, 10) is a local minimum. (3,f(3)) = (3,22) is a local maximum. y 20 10 f (x) = 6 6x = 6(x 1) If f (x) = 0 then x = 1. Thus the sign table for f (x) is 1 ( ) (+) Thus f(x) is concave down on (,1). f(x) is concave up on (1, ). (1,f(1)) = (1,6) is an inflection point. K4 K2 0 2 4 x K10 8. (10 pts) The marketing research team for a particular model of HDTVs has calculated that at the current price of $700 per HDTV, the monthly demand is 150 HDTVs. Furthermore for every $50 increase in the price, the monthly demand drops by 5. What is the price for HDTVs that maximizes monthly revenue? Assume the demand is linear, we have a line that passes though the point (x 1,p 1 ) = (150,700) and has slope Thus the equation of the line is m = p x = +50 5 = 10 p p 1 = m(x x 1 ) p 700 = 10(x 150) p 700 = 10x+1500 p = 10x+2200 Thus the revenue is R(x) = xp = x( 10x+2200) = 10x 2 +2200x is a parabola facing down with a maximum when R (x) = 20x+2200 = 0 20x = 2200 x = 110 Thus the revenue is maximal when x = 110 or p = 10(110)+2200 = $1100 per HDTV.
Math 160 - Final Solutions - Spring 2011 - Jaimos F Skriletz 6 9. (15 pts) Newtons Law of Cooling states that if a pot of boiling water (100 C) is placed in a room and left to cool, its temperature after t minuets is given by the function T(t) = 20+80e 0.075t. (a) Show that T(t) solves the differential equation: T (t) = 0.075(T(t) 20) First the derivative of T(t) is T (t) = 0+80e 0.075t ( 0.075) = 6e 0.075t Second the RHS of the equation is 0.075(T(t) 20) = 0.075(20+80e 0.075t 20) = 6e 0.075t Since the above two are equal, T(t) = 20+80e 0.075t is a solution to the above differential equation. (b) What is the initial temperature of this pot of water? The initial temperature happens when t = 0, thus T(0) = 20+80e 0.075(0) = 20+80e 0 = 100 (c) What is the eventual temperature of the water after a long time (t )? lim T(t) = lim t t (20+80e 0.075t ) = 20+80(0) = 20 (d) Make a rough sketch of T(t). 100 80 60 40 20 0 0 10 20 30 40 50 t
Math 160 - Final Solutions - Spring 2011 - Jaimos F Skriletz 7 10. (25 pts) Find the following indefinite integrals: (a) (x 2 1x ) 2 dx = (x 2 x 2 )dx = 1 3 x3 +x 1 +C = 1 3 x3 + 1 x +C (b) 5 1 x dx Let u = 1 x then du = dx and this integral becomes 5 1 ( du) = 5 u du = 5ln u +C = 5ln(1 x)+c u (c) 2x(x 2 +4) 3 dx Let u = x 2 +4 then du = 2xdx thus this integral becomes u 3 du = 1 4 u4 +C = 1 4 (x2 +4) 4 +C
Math 160 - Final Solutions - Spring 2011 - Jaimos F Skriletz 8 11. (10 pts) The monthly demand, x, for a certain product is x = 100 1 4p where p is the price per unit. The marginal cost to produce x units of the product is C (x) = 2x+140 and the fixed costs are C(0) = 3500. Find the cost, revenue and profit of this product as a function of units sold. First the demand relation says that x = 100 1 4 p 4x = 400 p p = 400 4x Thus the revenue is R(x) = xp = x(400 4x) = 400x 4x 2. Second if C (x) = 2x+140 then C(x) = C (x)dx = (2x+140)dx = x 2 +140x+C Since C(0) = C = 3500 the cost function is C(x) = x 2 +140x+3500. Last the profit is P(x) = R(x) C(x) = (400x 4x 2 ) (x 2 +140x+3500) = 5x 2 +260x 3500 12. (10 pts) Find the area bounded by the two curves y = x 2 and y = x+2. Include a graph of the enclosed region. First if x 2 = x+2 then x 2 = x+2 x 2 x 2 = 0 (x+1)(x 2) = 0 x = 1 or x = 2 Thus the area is A = 2 1 [(x+2) x 2 ]dx [ 1 = 2 x2 +2x 1 2 3 x3 1 = 10/3 ( 7/6) = 9/2
Math 160 - Final Solutions - Spring 2011 - Jaimos F Skriletz 9 13. (10 pts) The Gini Index for a Lorenz curve y = f(x) is defined to be G = 2 Find the Gini Index for the Lorenz curve y = x 2.3. 1 0 (x f(x))dx G = 2 1 0 (x x 2.3 )dx [ 1 = 2 2 1 [ 1 = 2 2 1 3.3 = 13 33 0.394 1 ] [0] 3.3 x3.3 0 14. (10 pts) Given f(x,y) = 5xy 2 +x 3 y answer the following: (a) Find f x (x,y) and f y (x,y). f x (x,y) = 5y 2 +3x 2 y f y (x,y) = 10xy +x 3 (b) Find f yy (x,y) and f xy (x,y). f yy (x,y) = y (10xy +x3 ) = 10x f xy (x,y) = y (5y2 +3x 2 y) = 10y +3x 2
Math 160 - Final Solutions - Spring 2011 - Jaimos F Skriletz 10 15. (15 pts) The demand curves for two types of beans are x = 18 3 2 p+ 1 3 q y = 6+ 2 3 p 1 2 q Where x is the weekly demand, in hundreds of pounds, for type 1; y is the weekly demand, in hundreds of pounds, for type 2; p is the price per pound, in cents, for type 1; and q is the price per pound, in cents, for type 2. The weekly revenue is then R(p,q) = xp+yq = 3 2 p2 1 2 q2 +18p+6q +pq Find the prices (p, q) that maximize weekly revenue. (Hint: Find the local maximum for the 2-variable function R(p, q). For the D-test, D(p,q) = [R pp (p,q)][r qq (p,q)] [R pq (p,q)] 2 ) The first order partial derivatives are Solve equation (2) for q to get q = 6+p. Substitute this back into equation (1) to get R p (p,q) = 3p+18+q = 0 (1) R q (p,q) = q +6+p = 0 (2) 3p+18+(6+p) = 0 2p+24 = 0 2p = 24 p = 12 If p = 12 then q = 6+p = 18. Thus the critical point is (p,q) = (12,18). To show this point is a maximum notice that R pp = 3 Thus D(p,q) = [ 3][ 1] [1] 2 = 3 1 = 2. R qq = 1 R pq = 1 R qp = 1 Since D(12,18) = 2 > 0 and R pp (12,18) = 3 < 0, the point (p,q) = (12,18) is a local maximum.