Math Practice Final - solutions

Transcription

1 Math Practice Final - solutions Problem 1 Indicate the following from looking at the graph of f(x) above. All answers are small integers, ±, or DNE for does not exist. a) lim x 1 f(x) = 1 b) lim x 1 + f(x) = c) lim x 1 f(x) = d) lim x 1 + f(x) = e) lim x 2 f(x) = 1 g) lim x 3 f(x) = 1.5 h) lim x 3 f(x) = DNE i) All points where f(x) is not continuous are x = 1, 1, 2, 3. Sorry the answer to part g) is not an integer! Other comments: for the limit of a function at a point, it does not matter what the value of the function is at that point itself - only what the function does as you approach the point. At x = 3, the limit does not exist because the left-handed limit, 1.5, is different from the right-handed limit, 0. Problem 2 Find the equation of the tangent line to the graph of f at the point (1, 0), where f(x) = x ln x. First, we find the derivative using the product rule, f (x) = x 1 + ln x = 1 + ln x. x

2 So f (1) = 1. This is the slope of our the tangent line at (1, 0). The equation is then y = 1 (x 1) + 0 = x 1. Problem 3 Find f (x) where f(x) = x + 1. We need to use the quotient rule, ex f (x) = ex (x + 1) e x (x + 1) 2 = xex (x + 1) 2 Problem 4 Find dy dx where y = 2x3 (4x 3) 5. We use the product rule again, dy dx = 6x2 (4x 3) x 3 (4x 3) 4 I want you to make simplifications like combining constant factors into one factor and putting it in front, also writing fractions in lowest terms like 6/4 = 3/2, but not more unless you need this result for later. Problem 5 In a certain production facility, the cost function is and the revenue function is C(x) = 2x + 5 R(x) = 8x x 2 where x is the number of units produced and sold, and R and C are measured in millions of dollars. Find the following. a) the marginal revenue b) the marginal cost c) the break-even point(s) [the number(s) x for which R(x) = C(x)] d) the number x for which the marginal revenue equals the marginal cost.

3 e) For which value of x does the facility make maximal profit? a) R (x) = 8 2x. b) C (x) = 2. c) We need to solve 8x x 2 = 2x + 5, which has solutions x = 1 and x = 5. d) We need to solve 8 2x = 2, which gives x = 3. e) We need to maximize the profit P (x) = R(x) C(x) = x 2 + 6x 5 which has derivative P (x) = 2x + 6, equaling zero for x = 3. Or we could just look at item d) and find the same answer! This is why there were all these questions phrased as marginal revenue = marginal cost in the book (why is it not a coincidence that the profit is maximal there?). Problem 6 The cost of making x units of a product is C(x) = x ln(x + 5). a) Find the marginal cost of making 100 units. b) How much will the cost increase approximately, compared to part a), if 4 extra units are made? Use approximation and the marginal cost the exact value will not get any credit. a) First, from the product rule and chain rule, C (x) = x x+5 + ln(x + 5). So C (100) = ln(105) = ln(105). b) By 4C (100) = 80/ ln 105. Note the exact value that does not get any credit means C(104) C(100) = 104 ln(109) 100 ln 105. Sure it s fun to compare that to our approximation, but it is not part of the problem. Another meaning of exact values that is used in the preamble of this exam is not a rounded decimal fraction, leave ln(105) and such things as they are. This is something we do want, unless there are instructions for rounding (because it makes your work easier to follow, easier to give partial credit, and accuracy of rounding can be left to any potential reader in the future). Problem 7 If a rock falls from a height of 40 meters on the planet Jupiter, then its height H after t seconds is approximately H(t) = 40 10t 2

4 a) What is the average velocity of the rock from t = 0 to t = 1? b) What is the instantaneous velocity at time t = 1? c) What is the acceleration of the rock? d) When does the rock hit the ground? a) The average velocity is H 0,1 = H(1) H(0) 1 0 = 10 (in meters per second). b) H (t) = 20t, so H (1) = 20. c) H (t) = 20 for all times t (in meters per second squared). d) We solve H(t) = 0 and get t = 2 (discard the negative answer). Problem 8 The marginal profit from selling x units of Hokum is given by two formulas: for 0 x 5, it is and for 5 < x 7, it goes down to P (x) = 12 P (x) = x 2 Given that P (0) = 0, find the profit from selling 7 units of Hokum. We need to compute P (7) = P (0) P (x) dx = = [12x] [ 20x 0.4x3 3 = ] dx x 2 dx = 12(5 0) + 20(7 5) ( ) (In the real exam, I ll ask for a rounded answer in such a word problem). Problem 9 Consider the graph of the function f(x) = x 3 6x 2 36x + 1.

5 a) Find the intervals on which f is increasing. b) Find the intervals on which f is decreasing. c) Find all local maxima of f. d) Find all local minima of f. First, we need the derivative f (x) = 3(x 2 4x 12) = 3(x 6)(x + 2). This is zero for x = 2 and x = 6. From f (5) < 0 and f ( 3) > 0, f (7) > 0, the sign pattern is + +. So this function is increasing in (, 2) and in (6, ). b) From the same sign pattern, f(x) is decreasing in ( 2, 6). Closed intervals are also OK in a), b). c) Only one, at x = 2 (sign of f (x) changes from + to ). d) Only one, at x = 6 (sign of f (x) changes from to +). Problem 10 What is the slope of the tangent line to the curve given by xy 2 x 2 y = 2 at the point (1, 2)? We use implicit differentiation. Differentiating both sides gives y 2 + 2xyy 2xy x 2 y = 0.

6 Then we solve this for y (factor out y to get (2xy x 2 )y = 2xy y 2, then divide by 2xy x 2 ), y 2xy y2 = 2xy x. 2 Finally, substitute x = 1 and y = 2 to get y = 0. So that tangent line is horizontal, with equation y = 0(x 1) + 2, simplified to y = 2. Problem 11 Find the absolute maximum and absolute minimum values of the function f(x) = x 16 x 2 on the closed interval [1, 3]. Indicate at which x-values each of these extrema occur. f (x) = x x 2 2x2 = 16 x 2 16 x 2 This equals zero exactly for x = ± 8. Other critical points are x = ±4 because the function f(x) is not differentiable there. But we discard 8 and ±4 because they are outside of [1, 3]. The only points to look at are x = 1, 8, 3. A table of values of f(x) is x f(x) We can compare these numbers using a calculator. But even without a calculator, we know 3 7 = 63 and 8 = 64. So 15 < 3 7 < 8. Therefore the maximum is 8, located at x = 8. The minimum is 15, located at x = 1. Problem 12 Consider the function y = 5x x a) Find the equation of the tangent line to the graph of this function at x = 6. b) Approximate y for x = Use the equation from part a) (amounts to the same as using differentials). a) First, we need to compute the derivative y = dy dx 5(x 1) 5x 5 = = (x 1) 2 (x 1). 2

7 Second, for x = 6, we get y = 7 and y = 1/5. So the tangent line is described by y = 1 (x 6) b) Just substitute x = 6.02 in the tangent line equation, so y = Alternative: If you liked differentials, you could have written dy = 1 dx = and then y 7 + dy, with the same answer. Not part of the problem: The exact value to 8 digits is Problem 13 A stock has price p(t) at time t and the demand for it is x(t). Suppose p and x are differentiable and p (t) = 0.3, x (t) = at a time when p = 400 and x = 100. a) Find the rate of change of revenue at this time. b) Is the revenue increasing or decreasing at this time? a) Start with R = px. Each of these letters stands for a function of time. So we have to use the product rule to differentiate R, and we get R = p x + px = = = 20 (in units of money per units of time). b) Revenue is increasing at this time because R is positive. 3 dx Problem 14 Evaluate the indefinite integral 2x 5. We use the substitution u = 2x 5, so du = 2 dx. This gives 3 dx 3 2x 5 = 2u du = 3 2 ln( u ) + K = 3 ln( 2x 5 ) + K. 2 Problem 15 Evaluate the indefinite integral x x dx. We use the substitution u = x 2 + 5, so du = 2x dx. This gives

8 x x dx = 1 2 u du = 1 3 u3/2 + K = 1 3 (x2 + 5) 3/2 + K. Problem 16 Find the area enclosed by the graph of f(x) = 20 x 2 and the line g(x) = 7x First, we solve f(x) = g(x) to find x values where the graphs of these two functions intersect. We find the solutions x = 8 and x = 1. In the interval [ 8, 1], f(x) is always bigger than g(x) (you can either plot them to see this, or make a test evaluation at eg x = 0). So all we need is to compute A = 1 8 f(x) g(x) dx = = [8x 7x2 2 x3 3 = = ] x x 2 dx = 8(1 + 8) 7 2 (1 64) 1 3 ( ) Problem 17 Approximate the area between the x-axis, the graph of f(x) = 1/ x 3 + 1, and the lines x = 1 and x = 3 using four rectangles. a) Use the left endpoints of your subintervals of [1, 3]. b) Use the right endpoints of your subintervals of [1, 3]. For a) and b) combined, we divide the interval [1, 3] into four subintervals. We will need the x-values and their associated values of f(x) from this table. x f(x) f(x) approx a) For the left endpoints, we don t need x = 3. Our approximation is R L = 0.5( ) = 0.88

9 b) For the right endpoints, we don t need x = 1. Our approximation is R L = 0.5( ) = 0.62 Note. I have rounded all results to two decimals because it becomes clearer how this is set up than when writing the exact values 1/ 28 etc. Note that we still use the much more accurate results from the calculator for all intermediate results to compute the final answer - only after that do we round it. This is important, because the rounding errors from intermediate results could accumulate. On the real exam, I will ask for rounded results as shown here.