Practice Second Midterm

Practice Second Midterm 1. Find the derivatives of the following functions f(x) = (x3−x)4  x+1  5 g(x) =    x−1  ____ √2x+1 h(x) = x2−1  1  k(x) = (x3+2x+1)  2+  x2   2 + l(x) = 2x − x √x 2 4 2. For the function f(x) = x x2+1 (a) Find an equation of the tangent line at the point (0,0). (b) Find the point(s) on the graph where the tangent line is horizontal, and write an equation for the tangent line at those points. Solution. Using the quotient rule, the derivative f′(x) is f′(x) = (x2+1)−x(2x) 2 2 (x +1) = 1−x2 (x2+1)2 (a) The slope of the tangent line to the graph of f(x) at (1,1/2) is f′(0) = 1, so that an equation is y=x. (b) The tangent line is horizontal when its slope is 0. Thus we solve f′(x)=0, or 1−x2 (x2+1)2 =0 The solutions are x=±1. The corresponding points on the graph are (−1,−1/2) and (1,1/2). Equations are y=−1/2 and y=1/2, respectively. 3. The marketing department of Telecom corporation has determined that the demand for their cordless phones is p = −0.02 x+600 (0 ≤ x ≤ 30,000) where p is the phone's unit price in dollars and x is the quantity demanded. (a) Find the revenue function R(x). If the cost of producing x=10,000 phones is C(10,000)=1,000,000 dollars, what is the corresponding profit P(10,000)? (b) Find the marginal revenue function R′(x). Compute R′(10,000) and interpret your result. (c) Determine whether the demand is elastic, inelastic or unitary when x=10,000 and x=20,000. Solution. (a) The revenue function R(x)=xp(x) = x(−0.02 x+600) = −0.02 x2 +600 x The profit function P(x)=R(x)−C(x). Note that we do not have a formula for the cost function C(x), only its value C(10000) = 1000000. But P(10,000)=R(10,000)−C(10,000) = −.02(10000)2 +(600)(10000)−1000000 = 3000000 (b) R′(x) = (2)(−0.02 x)+600 = −.04 x +600 and R′(10000)=200. The revenue corresponding to the sale of the 10,001st phone is about $200. (c) The elasticity of the demand is given by E(p) = −p f′(p) f(p) We have x=f(p) = (600−p)/(.02), thus f′(p) = −1/.02. Plugging into E(p) we obtain E(p) = −p (−1/0.02) (600−p)/(0.02) = p 600−p Moreover, when x=10,000, unit price p=400, and when x=20,000, unit price p=200. Thus 400 E(400)= =2>1 600−400 and the demand is elastic when p=400 E(200)= 200 600−200 =0.5 < 1 so the demand is inelastic when p=200. 4. Find the absolute maximum value and absolute minimum value, if any, of the function f(x)=8x− 1 x2 on the interval [1,3]. Solution. The derivative f′(x) = 8+ 1 x3 Setting f′(x)=0 yields x=(−1/4)1/3. This is a negative number, and hence is not in the interval [1,3], thus we ignore it. From the table x 1 3 f(x) 7 215/9 5. For the function f(x)= x 1+x3 (a) Find the intervals where it is increasing and the intervals where it is decreasing. (b) Find the relative extrema, if any. Solution. (a) The derivative f′(x)= 1−2x3 (1+x3)2 Setting f′(x)=0 gives 1−2x3=0, or x=(1/2)1/3. The derivative does not exists when x=−1 (which is not in the domain of f. Next we see that f′(x) > 0 when x < −1 and when −1 < x < (1/2)1/3, and that f′(x) < 0 when x > (1/2)1/3. Therefore, f(x) is increasing on (−∞, −1)∪(−1,(1/2)1/3), and decreasing on ((1/2)1/3,∞). (b) As we go across x=(1/2)1/3, we see that f(x) changes from increasing to decreasing. Therefore f(x) has a relative maximum at x=(1/2)1/3. 6. For the function f(x) = 2x3 −6x2 +6x +1 (a) Find the intervals where f is concave upward and where it is concave downward. (b) Find the inflection points, if any. 7. The average worker at Wakefield Avionics, Inc., can assemble N(t)=−2t3 +12 t2 +2t (0 ≤ t ≤ 4) ready-to-fly radio-controlled model airplanes t hours into the 8 am to 12 noon morning shift. At what time during this shift is the average worker performing at peak efficiency? Solution. We have to find the inflection points of N(t). Now N′(t) = −6t2 +24 t+2 and N′′(t) = −12t+24. Setting N′′(t)=0 gives t=2. Furthermore, N′′(t) > 0 when t < 2 and N′′(t) < 0 when t > 2. Therefore, t=2 gives an inflection point for N. Thus, the average worker is performing at peak efficiency at 10 am. 8. A man wishes to have an enclosed vegetable garden in his backyard. If the garden is to be a rectangular area of 400 ft2, find the dimensions of the garden that will minimize the amount of fencing material needed. Solution. If x is the length and y is the width of the garden, then the total fencing material needed is f=2x+2y. The area is xy=400, so y=400/x. Substituting into f, we have f(x)=2x+ 800 x The derivative f′(x) = 2 − 800 x2 Setting f′(x)=0 gives x=±20. Only the positive number is meaningful. We further note that f′′(x) = 1600/x3, which is positive for all x > 0. That is f is concave up on (0,∞), and x=20 gives an absolute minimum for f. The dimensions of the graden that will minimize the amount of fencing material are 20 ft.x20 ft. 9. A manufacturer of tennis rackets finds that the total cost C(x) (in dollars) of manufacturing x rackets per day is C(x)=400+4x +0.0001x2 Each racket can be sold at a price of p dollars, where p is related to x by the demand equation p=10−0.0004 x If all the rackets that are manufactured can be sold, find the level of production that will yield a maximum profit for the manufacturer. Solution. The revenue function is R(x)=xp(x) = x(10−0.0004 x)=10x−0.0004x2, and the profit function is P(x) = R(x)−C(x) = 10x−0.0004x2 − (400+4x +0.0001 x2) = −0.0005 x2+6x −400 The derivative P′(x) = −0.001 x +6 and P′(x)=0 gives x=6000. We also observe that P′′(x)=−0.001 < 0 for all x, so that P(x) is concave down on (0,∞). This says that the production level x=6000 will yield a maximum profit for the manufacturer. 10. You wish to construct a closed rectangular box that has a volume of 4ft3. The length of the base of the box will be twice as long as its width. The material for the top and bottom of the box costs 30 cents/square foot. The material for the sides of the box costs 20 cents/square foot. Find the dimensions of the least expensive box that can be constructed. 11. The demand equation for the Roland portable hair dryer is given by ______ p= √225−5x (0 ≤ x ≤ 45) where x (measured in units of a hundred) is the quantity demanded per week and p is the unit price in dollars. (a) Is the demand elastic or inelastic when p=8 and p=10? (b) When is the demand unitary? (c) If the unit price is lowered slightly from $10, will the revenue increase or decrease? (d) If the unit price is lowered slightly from $8, will the revenue increase or decrease? Solution. First we compute the elasticity of demand E(p)=−p f′(p) f(p) where x = f(p) = 45−(p2/5) Then f′(p) = −2p/5 and E(p) = −p −2p/5 = 45−(p2/5) 2p2 225−p2 (a) E(8) = 2(8)2 = 225−(8)2 128 161 ≈ 0.8 so the demand is inelastic when unit price is $8. E(10) = 2(10)2 2 225−(10) = 8 5 ≈ 1.6 so the demand is elastic when the unit price is $1.6. (b) The demand is unitary when E(p)=1. Solving this equation 2p2 225−p2 =1 or 2p2=225 −p2 or 2 p= 225 3 = 75 The demand is unitary when the unit price is p=(75)1/2 ≈ 8.66 dollars. (c) Since the demand is elastic when the unit price is $10, lowering the unit price will cause the revenue to increase. (d) Since the demand is inelastic when the unit price is $8, lowering the unit price will cause the revenue to decrease. 12. Sketch the graph of the function f(x)=−2x3 +3x2 +12 x+2 Solution. (a) Domain. The domain of f is all real numbers, (−∞,∞). (b) Intercepts. The y-intercept is (0,2). (c) Increasing decreasing. The first derivative f′(x) = −6x2 +6x +12. Setting f′(x)=0 yields x=−1 and x=2. The derivative exists for all x. Pick test points −2,0,3 and compute f′(−2)=−24, f′(0)=12, and f′(3)=−24. Thus f(x) is decreasing on (−∞,−1)∪(2,∞) and increasing on (−1,2). (d) Relative extrema. f(x) has a relative minimum at x=−1 and a relative maximum at x=2. (e) Concavity. The second derivative is f′′(x)=−12 x+6. Setting f′′(x)=0 gives x=1/2. Pick test points x=0, 1 and compute f′′(0)=6, f′′(1)=−6. Thus f(x) is concave upward on (−∞, 1/2) and concave downward on (1/2,∞). (f) Inflection Point. There is a change in concavity when going across x=1/2. Since f(1/2) = 17/2, the point in the graph (1/2,17/2) is an inflection point. (g) Asymptotes. The limits limx→ ∞ f(x) and limx→ −∞ f(x) do not exist, thus there are no horizontal asymptotes. Also, polynomials have no vertical asymptotes. (g) Graph. 13. A truck gets 400/x miles per gallon when driven at a constant speed of x mph (between 50 and 40 mph). If the price of fuel is $1/gallon and the driver is paid $8/hour, at what speed between 50 and 70 mph is it most economical to drive? Solution. The fuel cost is x/400 dollars/mile. The labor cost is 8/x dollars/mile. Therefore the total cost is C(x) = fuel cost + labor cost = x 400 + 8 x dollars/mile. The derivative C′(x) = 1 400 + −8 x2 Setting C′(x)=0 gives 1 400 = 8 x2 or x2=3200. That is, x=56,57 is a critical point. Next we find that the second derivative C′′(x) = 16 x3 so that C′′(x) > 0 for x > 0. That is C(x) is concave up and so x=56,57 gives the absolute minimum. 14. The Department of Interior of an African country began to record an index of environmental quality to measure progress or decline in the environmental quality of its wildlife. The index for the years 1984 through 1994 is approximated by the function I(t)= 50 t2 +600 2 t +10 (0 ≤ t ≤ 10) (a) Compute I′(t) and show that I(t) is decreasing on the interval (0,10). (b) Compute I′′(t) and study the concavity of the graph of I on the interval (0,10). (c) Interpret your results. Solution. (a) The first derivative is I′(t) = 2(50)t(t2+10)−2t(50 t2+600) (t2+10)2 = −200 t (t2+10)2 which is negative for all values of t > 0. Thus I(t) is decreasing on the interval (0,10). (b) The second derivative is I′′(t)= 200(3t2−10) (x2+10)3 Setting I′′(t)=0 gives 3t2 −10=0, or x = ±(10/3)1/2. Then I′′(t) < 0 if −(10/3)1/2 < t < (10/3)1/2 and I′′(t) > 0 for t > ((10/3)1/2 and for t < −(10/3)1/2. Thus I(t) is concave upward on ((10/3)1/2,∞) and (−∞,−(10/3)1/2), and concave downward on (−(10/3)1/2,(10/3)1/2). (c) Compute (10/3)1/2 ≈ 1.8. Our calculations show that the rate of decline in the environmental quality of the wildlife was increasing in the first 1.8 years. After that time the rate of decline decreased. 15. Two ships leave the same port at noon with ship A sailing south and ship B sailing west. One-half hour later, ship A is 6 miles from the port and running at 18 mph while ship B is 5 miles from the port and running at 15 mph. How fast is the distance between the two ships changing at that instant of time? Solution. Let A(t) be the position of Ship A at time t, and let B(t) be the position of ship B at time t. The distance between the ships at time t is given by the function f(t)=(A(t)2 +B(t)2)1/2 The question is to find f′(1/2). First find the derivative f′(t) = (1/2) (A(t)2+B(t)2)−1/2(2A(t)A′(t) +2B(t)B′(t)) When t=1/2 (half-hour after leaving port), we know that the position A(1/2)=6 and the velocity A′(1/2)=18, and also B(1/2)=5 and B′(1/2)=15. Plugging in these values into f′(t), we obtain f′(1/2) = ((6)2 + (5)2)−1/2 ((6)(18)+(5)(15)) = 23.43 16. Simplify the following expressions  8  −1/3  81  −1/4      27   256  (4x2)(−3x5)2 Solution. (a) 2 (b) 36 x12