f(x) = lim 2) = 2 2 = 0 (c) Provide a rough sketch of f(x). Be sure to include your scale, intercepts and label your axis.


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1 Math 16  Final Exam Solutions  Fall 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. { 6 x 2 x < 2 1. (15 pts) Consider the piecewise defined function f(x) = x 2 x 2 (a) Find lim x 2 f(x), lim x 2 + f(x) and lim x 2 f(x). lim f(x) = lim x 2 x 2 (6 x2 ) = 6 4 = 2 lim x 2 f(x) = lim + +(x 2) = 2 2 = x 2 lim f(x) = DNE because above are not equal. x 2 (b) Is f(x) continuous at x = 2 and explain why. f(x) is not continuous at x = 2 because lim x 2 f(x) DNE (there is a break at this point). (c) Provide a rough sketch of f(x). Be sure to include your scale, intercepts and label your axis. 2. (1 pts) Consider the cost function C(x) =.1x x + 35 Find the average and marginal costs for producing x = 35 units. C (x) =.2x + 55 Marginal Costs: C (35) =.2(35) + 55 = $48 per unit Average Costs: C(35) = C(35) 35 = = $61.5 per unit
2 Math 16  Final Exam Solutions  Fall Jaimos F Skriletz 2 3. (15 pts) Find the following limits: (a) lim x 5 25 x 2 x 2 + 3x 1 = lim (5 + x)(5 x) x 5 (x + 5)(x 2) = lim 5 x x 5 x 2 = 1 7 = 1 7 (b) lim x 25 x 2 x 2 + 3x 1 = lim x (25 x 2 ) 1 x 2 (x 2 + 3x 1) 1 x 2 = lim x 25 x x 1 x 2 = = 1 4. (1 pts) The market research team for a particular model of HDTVs has calculated that at the current price of $12 per unit, the monthly demand is 8 TVs. While if the price were dropped to $1 per unit, the monthly demand increased to 12 TVs. Assuming the demand relation is linear, find the monthly demand and revenue from selling x HDTVs per month. Let x be monthly demand and p be unit price, then the line that passes though (x 1, p 1 ) = (8, 12) and (x 2, p 2 ) = (12, 1) has slope m = p 1 12 = = 2 = 5 x Thus the equation of the line is And the monthly revenue would be p p 1 = m(x x 1 ) p 12 = 5(x 8) p = 5x + 16 R(x) = xp = x( 5x + 16) = 5x x
3 Math 16  Final Exam Solutions  Fall Jaimos F Skriletz 3 5. (1 pts) Use the Limit Definition (4step process) to find the derivative of the function f(x) = 3 x 2. f(x + h) = 3 (x + h) 2 = 3 (x 2 + 2xh + h 2 ) = 3 x 2 2xh h 2 f(x + h) f(x) = (3 x 2 2xh h 2 ) (3 x 2 ) = 2xh h 2 f(x + h) f(x) h = 2xh h2 h = h( 2x h) h = 2x h f f(x + h) f(x) (x) = lim = lim ( 2x h) = 2x h h h 6. (5 pts) Give one definition for the number e. In this course we saw three limit equivalent limit definitions for e. ( e = lim n n e = lim(1 + s) 1/s s e h 1 lim = 1 h h ) n
4 Math 16  Final Exam Solutions  Fall Jaimos F Skriletz 4 7. (25 pts) Find the derivatives of the following functions: (a) f(x) = 6 x 1 x 2 = 6x1/2 x 2 f (x) = 3x 1/2 + 2x 3 = 3 x + 2 x 3 (b) g(x) = x 4 (4 3x) 3 g (x) = [4x 3 ](4 3x) 3 + [3(4 3x) 2 ( 3)](x 4 ) = x 3 (4 3x) 2 [4(4 3x) 9x] = x 3 (4 3x) 2 (16 21x) (c) f(x) = x2 x + x 2 f (x) = (1 + 2x)(x2 ) (2x)(x + x 2 ) (x + x 2 ) 2 = x2 (x + x 2 ) 2 (d) h(t) = ln(1 + 4e t ) h (t) = e t (4e t ( 1)) = 4e t 1 + 4e t
5 Math 16  Final Exam Solutions  Fall Jaimos F Skriletz 5 8. (2 pts) Find the following indefinite or definite integrals (a) (x 3 1x ) 3 dx = (x 3 x 3 )dx = 1 4 x2 1 2 x 2 + C = 1 4 x x 2 + C (b) xe x2 dx Let u = x 2 then du = 2xdx or 1 2du = xdx. Thus this integral becomes: 1 xe x2 dx = 2 eu du = 1 2 eu + C = ex + C (c) e 1 5 x dx = 5 ln x e = 5 ln(e) 5 ln(1) = 5 1
6 Math 16  Final Exam Solutions  Fall Jaimos F Skriletz 6 9. (15 pts) Consider the function f(x) = x 4 4x 3. (a) Find all critical points of f(x) and use the first or second derivative to determine their nature. f (x) = 4x 3 12x 2 = 4x 2 (x 3) If f (x) = then x = or x = 3 so these are the two critical points. The sign table for f 3 (x) is ( ) ( ) (+) This shows that (, f()) = (, ) is a flat point while (3, f(3)) = (3, 27) is a local minimum. (b) Provide a rough sketch of f(x). Be sure to include all critical points, intercepts, show your scale and label your axis. 1. (1 pts) Consider the graph of the function y = f(x). (a) What interval(s) is f(x) increasing? (a, b) (b) What interval(s) is f(x) decreasing? (, a) and (b, ) (c) What interval(s) is f(x) concave up? (, b) and (b, c) (d) What interval(s) is f(x) concave down? (c, ) (e) List all local extrema of f(x). Local Min at x = a and Local Max at x = b (f) List all inflection points of f(x). Inflection point at x = c
7 Math 16  Final Exam Solutions  Fall Jaimos F Skriletz (1 pts) The demand for and cost to produce a certain product is given by the functions p(x) =.3x + 4 C(x) = 22x + 75 (a) Find the revenue and profit for producing and selling x units. R(x) = xp(x) = x(.3x + 4) =.3x 2 + 4x P (x) = R(x) C(x) = [.3x 2 + 4x] [22x + 75] =.3x x 75 (b) What is the price per unit that maximizes profit? Possible maximum when P (x) =.6x + 18 = x = 18.6 = 3 Since P (x) =.6 < for all x, the maximum occurs with a demand of x = 3 and a price of p(3) =.3(3) + 4 = $31 per item 12. (15 pts) Consider the function of two variables f(x, y) = 4x 3 y 2 3xy 2 + 2x 3. Find all four second order partial derivatives: f xx (x, y), f xy (x, y), f yx (x, y) and f yy (x, y). f x (x, y) = 12x 2 y 2 3y 2 + 6x 2 f xx (x, y) = 24xy x f xy (x, y) = 24x 2 y 6y f y (x, y) = 8x 3 y 6xy f yx (x, y) = 24x 2 y 6y f yy (x, y) = 8x 3 6x
8 Math 16  Final Exam Solutions  Fall Jaimos F Skriletz (1 pts) The velocity, v in feet per second, of a skydiver t seconds after jumping from a perfectly good plane is given by the function v(t) = 18(1 e.2t ) (a) What is the initial velocity of this skydiver? v() = 18(1 e ) = feet per second (b) What is the terminal velocity of this skydiver? lim v(t) = 18(1 ) = 18 feet per second t 14. (1 pts) (a) Use a Riemann Sum to approximate the area under the curve y = x from x = to x = 4 using a partition of n = 4 rectangles and the midpoints of each subinterval. x = b a n = 4 = 1 4 Midpoints:.5, 1.5, 2.5, f(c i ) x = x[f(.5)+f(1.5)+f(2.5)+f(3.5)] = 25 i=1 (b) Use a definite integral to find the exact area of the above region. 4 [ 1 (x 2 + 1)dx = 3 x3 + x 4 = = 76 3
9 Math 16  Final Exam Solutions  Fall Jaimos F Skriletz (1 pts) Solve the initial valued differential equation f (x) = 12x 3 6x + 5 and f() = 11 The family of solutions is f(x) = f (x)dx = (12x 3 6x + 5)dx = 3x 4 3x 2 + 5x + C Use the initial value f() = 11 to get Thus the function is f(x) = 3x 4 3x 2 + 5x 11 f() = C = (1 pts) The demand function for a certain product is given by the function p(x) =.3x Find the Consumers Surplus for this demand relation at the current market price of p = 23. Include a rough sketch of the area the Consumers Surplus represents. CS = = 2 2 p(x) = 23.3x = 23.3x 2 = 12 x 2 = 4 x = ±2 (p(x) p)dx = 2 (.3x )dx (.3x )dx = [.1x x =.1(2) (2) = 16 2
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