1. Done in class. Because answers will vary, and it is hard to type any solution, I will not display a solution here. 2. (a) We should start by simplifying: f (x) = 3x2 (x + 2)(x − 1) 3x4 + 3x3 − 6x2 = = x3 − 8x2 + 16x x(x − 4)2

3x(x + 2)(x − 1) . Now, vertical asymptotes can potentially occur at zeroes of the de(x − 4)2 nominator of a rational function. So we set the denominator equal to 0: (x − 4)2 = 0 =⇒ x = 4. So x = 4 is a potential vertical asymptote. Because of our simplification, this should be a vertical asymptote, but we should look at lim f (x) to be sure. But x→4 216 . Already this is enough to say that x = 4 is a vertical asymptote. If lim f (x) → x→4 0 you want to be extra careful, though, you can look at the left and right limits. It should be noted that x = 0 is not a vertical asymptote, even though as we originally wrote the function, x = 0 is a zero of the denominator. (b) The line y = b is a horizontal asymptote of a function f (x) if lim f (x) = b or x→∞ x→−∞

lim f (x) = b. Technically we should consider both limits, but because it is a well-

known fact that horizontal asymptotes of rational functions are completely determined 3x4 + 3x3 − 6x2 by only one of these limits. So we need only look at lim f (x) = lim 3 = x→∞ x→∞ x − 8x2 + 16x 3 6 1 + 3x − 3x2 3x4 3x4 lim = lim Remember that the later (rather nasty 8 16 x→∞ x→∞ x3 x3 1 − x + x2 looking, and even more nasty to encode) fraction disappears because its limit is 1. But 3x4 = lim 3x = ∞. Therefore, f does not have any horizontal asympthen lim x→∞ x→∞ x3 totes. (c) lim f (x) → 0 which is indeterminate. So we simplify f as we did in part a and get x→0 0 3x(x + 2)(x − 1) 0 lim f (x) = lim = = 0. x→0 x→0 (x − 4)2 16

(d) Remember that a function can potentially change sign at its points of discontinuity or its zeroes. We can get a rational functions points of discontinuity by setting the denominator equal to 0 (Yielding x = 0, 4) and can get the function’s zeroes by setting its numerator equal to 0 (Yielding x = −2, 1). Construct a sign chart for f with the points −2, 0, 1, 4. Picking test points in each subinterval, we’ll get: f (−3) <

1 0, f (−1) > 0, f ( ) < 0, f (3) > 0, f (5) > 0. So, we find that f (x) < 0 on the interval 2 (−∞, −2) ∪ (0, 1). However, the question asks for where f (x) ≤ 0, so we should include points where f (x) = 0. So the solution is the interval (−∞, −2] ∪ (0, 1]. Note that we do not include x = 0 because 0 is not in the domain of f (in the original way we wrote f , we would be dividing by 0 if we attempted to plug 0 into f ). 3. (a) The ball’s initial height is the height the ball was at at time t = 0. In other words the ball’s initial height is y(0) = 80 ft. So determining when the ball will return to its initial height is solving y(t) = 80. That is −16t2 + 64t + 80 = 80 =⇒ −16t2 + 64t = 0 =⇒ −16t(t − 4) = 0 =⇒ t = 0, t = 4. Of course t = 0 is the instant that the ball is tossed, so the ball will return to its initial height after 4 seconds. (b) The ball will reach its maximum height when it has neither upward velocity (else it is still going up) nor downard velocity (as it has already reached its maximum height and is falling). So the maximum height will occur when the ball’s velocity is 0. As velocity is the derivative of position, v(t) = y (t) = −32t + 64. Solving v(t) = 0 yields t = 2 seconds. So the ball reaches its maximum height 2 seconds after it is thrown. (c) The ball will hit the ground when its height is 0 ft. That is, we want to solve y(t) = 0. That is −16t2 + 64t + 80 = 0 =⇒ −16(t2 − 4t − 5) = 0 =⇒ −16(t − 5)(t + 1) = 0 so t = 5, −1. Of course we cannot have negative time, so the ball will hit the ground 5 seconds after its thrown. To determine its velocity when it hits the ground, we need to find v(5) = −32(5) + 64 = −96. So the ball has a velocity of 96 ft/s (downward) when it hits the ground. f (x + h) − f (x) = 4. Computing f (x) from the definition means computing lim h→0 h 1 lim (f (x + h) − f (x)) (I’m writing it this way for this specific problem, because it makes h→0 h 1 1 something later on easier to see). So if f (x) = then lim (f (x + h) − f (x)) = h→0 h x 1 1 1 lim − . So we can combine these 2 fractions. To do, we need a common h→0 h x+h x x denominator. To get a common denominator, we multiply the first term by and multix

ply the second by

1 1 1 x x+h − = lim − . h→0 h x+h x x(x + h) x(x + h) 1 x − (x + h) So we combine fractions and combine the into the fraction to get lim = h→0 hx(x + h) h −1 −1 −h = lim = 2 lim h→0 x(x + h) h→0 hx(x + h) x

x+h 1 . We get lim h→0 h x+h

5. To get the equation of the tangent line, we need both a point on the line and a slope. In either case, we should start by simplifying h(u). We need to simplify h to find its derivative, and (u − 3)2 − (u + 4)(u − 3)(u − 2) = simplifying will make finding h(1) easier. So, h(u) = u2 (u − 3) (u − 3)((u − 3) − (u + 4)(u − 2)) (u − 3) − (u + 4)(u − 2) (u − 3) − (u2 + 2u − 8) = = = u2 (u − 3) u2 u2 −u2 − u + 5 u2 u 5 1 5 = − 2 − 2 + 2 = −1− + 2 = −1−u−1 +5u−2 . So, h(1)−1−1+5 = 3. u2 u u u u u h (u) = u−2 − 10u−3 . So h (1) = 1 − 10 = −9. Thus, the line tangent to h at u = 1 is given by y = −9(u − 1) + 3, or y = −9u + 12 6. (a) The 101st computer will cost exactly C(101) − C(100) = 18160 − 18000 = $160. If we use marginal analysis to approximate the cost of the 101st computer, we find C (x) = 160 so C (100) = $160. Note this is the same value. (Think about why) (b) To determine the maximum profit, we need to get our profit function, P (x). To get that, we need our revenue function, R(x) = xp = x(400 − 0.4x) = 400x − 0.4x2 . So P (x) = R(x) − C(x) = 400x − 0.4x2 − (2000 + 160x) = −0.4x2 + 240x − 2000. The maximum profit will occur at a production level when the marginal profit is 0, so we look at P (x) = −0.8x + 240. Solving P (x) = 0, we get x = 300. Thus the maximum profit will be P (300) = $34000. (c) We determined in b that the maximum profit will occur at a production level of 300 computers. We can get the price we should sell the computers at from our price demand equation: p = 400 − 0.4x. In particular, we should set our price to be p = 400 − 0.4(300) = $280. (d) The average profit, P (x) = P (x) −0.4x2 + 240x − 2000 2000 = = −0.4x + 240 − . x x x 2000 So P (300) =≈ $113.33. The marginal average profit, P (x) = −0.4 + 2 . So x

P (300) ≈ −$0.38