Solutions to Study Guide for Test 3. Part 1 No Study Guide, No Calculator

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1 Solutions to Study Guide for Test 3 Part 1 No Study Guide, No Calculator 1. State the definition of the derivative of a function. Solution: The derivative of a function f with respect to x is the function f f (x) = lim h 0 f(x + h) f(x). h The domain of f is the set of all x where the limit exists. 2. State the definition of a critical point. bf Solution: A critical point (or number) of a function f is any number x in the domain of f such that f (x) = 0 or f (x) does not exist. 3. State the definition of an inflection point. bf Solution: A point on the graph of a continuous function f where the tangent line exists and where the concavity changes is called an inflection point. 4. State the definition of an anti-derivative of a function on an interval I. bf Solution: A function F is an antiderivative of f on an interval I if F (x) = f(x) for all x in I. 5. State the fundamental limit that defines the number e. bf Solution: The number e is defined to be the number that satisfies e h 1 lim h 0 h = Find the derivatives of the following functions. (a) f(x) = e x. bf Solution: f (x) = e x. 1

2 (b) f(x) = ln(x) for x > 0. f (x) = 1 x (c) f(x) = ln( x) for x < 0. bf Solution: By the chain rule, if y = ln( x) = ln(u) and u = x, then dy/du = 1/u and du/dx = 1. Thus the derivative is 1/( x) ( 1) = 1 x. So f (x) = 1 x. (d) g(x) = x 2 e x. bf Solution: By the product rule with u = x 2 and v = e x so u = 2x and v = e x, we have g (x) = u v + uv = 2xe x + x 2 e x = (2x + x 2 )e x. (e) g(x) = e ex2 +3x. bf Solution: Write y = e u u = e v v = x 2 + 3x dy du = eu du Dv = ev dv = 2x + 3. dx Putting this together by the chain rule, we have dy dx = dy du dv du dv dx = e u e v (2x + 3) = e ev e x2 +3x (2x + 3) = e ex2 +3xe x2 +3x (2x + 3). (f) h(t) = ln(t 3 + e t ). bf Solution: By the chain rule: Let y = h(t) = ln(u) and u = t 3 + e t, so that dy/du = 1/u and du/dt = 3t 2 + e t. By the chain 2

3 rule, we then have dy dt = dy du dudt = 1 u (3t2 + e t ) = 1 t 3 + e t (3t2 + e t ) = 3t2 + e t t 3 + e t. 7. Find f (3) if f(x) = ( x 2 + 8) 4 (e x + 1) 4. bf Solution: This is a product and chain rule. u = ( x 2 + 8) 4 and v = (e x + 1) 4, so that u = 4( x 2 + 8) 3 ( 2x) and v = 4(e x + 1) 3 e x. The result is then f (x) = u v + uv = 4( x 2 + 8) 3 ( 2x)(e x + 1) 4 + 4( x 2 + 8) 4 (e x + 1) 3 e x For f (3) = 4( 9 + 8) 3 ( 6)(e 3 + 1) 4 + 4( 9 + 8) 4 (e 3 + 1) 3 e 3 = 24(e 3 + 1) 4 + 4(e 3 + 1) 3 e Find the indefinite integrals 1 (a) x dx. bf Solution: = ln x + C. (b) 4x 3 + 6e 3x dx. bf Solution: The integral is x 4 + 2e 3x + C. (c) e 2 dt. bf Solution: The indefinite integral is = e 2 x + C. (d) 5 xdx. bf Solution: 10/3 x 3/2 + C. 9. Find all functions satisfying (a) f (x) = 8x 1/3. bf Solution: The integral is f(x) = 6x 4 3 3

4 10. Which one of the following is ln(x)dx? (a) x ln(x) + C (b) x ln(x) x + C (c) e ln(x) (d) None of the above bf Solution: The solution is (b). This can be found by taking the derivatives of each of the responses. Part II Calculator and Study Sheet Allowed 11. True or false (a) ln(a b) = ln(a) ln(b). bf Solution: This is false. The correct rule is that ln(a b) = ln(x) + ln(b). (b) The derivative of ln( x) is 1 for x < 0. x bf Solution: False, The derivative of ln( x) is f(x)=1/x. (c) If a is a critical point of the function f(x), then the graph of y = f(x) has either a local maximum or a local minimum at x = a. bf Solution: False, it is possible that a critical point could be an inflection point. 12. Suppose we have the functions f(x) and g(x), both of which are differentiable. Suppose further that we have the following chart: x f(x) g(x) f (x) g (x)

5 (a) If h(x) = e f(x), find h (6). bf Solution: This is a chain rule with y = e u and u = f(x). By the chain rule, we have Substituting t = 6, we have f (x) = e u + f (x) = e f(x) + f (x). f (6) = e f(6) f (6) = e 3 1 = e 3. (b) If h(x) = ln(g(x)), find h (3). bf Solution: Again as a chain rule with y = h(x) = ln(u) and u = g(x), we have h (x) = 1 g(x) g (x). This says that h (3) = 5/ Draw the graph of a function f(x) that has the following characteristics: (a) f(1) = 2, f(3) = 0 (b) f(x) has a horizontal asymptote at y = 1. (c) f (x) > 0 on the interval (3, ]. (d) f (x) > 0 on the interval [0, 3) (e) f (x) < 0 on the interval[, 0]. Solution:

6 14. An apple orchard has an average yield of 36 bushels of apples/tree if tree density is 22 trees/acre. For each unit increase in tree density, the yield decreases by 2 bushels/tree. How many trees should be planted in order to maximize yield? Solution: Let x denote the unit density of trees. We note that since for every tree above 22/acre, the yield/tree is 2 less, so that the yield per tree, y t (x) = 36 2(x 22) (where y t is the per tree yield. To find the total yield, we multiply the yield per tree by the number of tees. Consequently the function for the yield of the orchard (per acre) is given by y(x) = x(36 2(x 22)) = 2x x. We now find the critical points, which are where y (x) = 0 or it does not exist. In this case, we have y (x) = 4x + 80 and the only critical point is x = 20. Now, there are two possible interpretations of the problem for end points. If you assume that the tree density can be less than 22, so that if there are only 10 trees, then the production per tree is 60 bushels per tree, we have that 0 x 40 (at which point there are no apples produced by any tree. In this case, to maximize the answer, we find y(0) = 0, y(20) = 800, and y(40) = 0. Clearly the greatest of these is 800, so we should plant 20 trees per acre. The second interpretation would require that 22 x 40 which interprets the statement about density to have a minimum of 22 trees per acre. In this case, no critical point is within the closed interval [22, 40], so we can just evaluate the yield at the endpoints.. Checking y(22) = 792 bushels per tree and we should plant 22 trees. 15. Graph the function Solution: f(x) = x 3 7x 2 + 7x + 15 = (x 5)(x 3)(x + 1). 6

7 If you invest $15,000 at 5% interest per year (compounded monthly), how many months will it take for you to have $20,000 in the account. Solution: This is like the problem we did in class. If the original principle is $15000, then the principle after t months is: P (t) = 15000( )t, and in particular we want the value of t in months for which we first cross the $20,000 threshold, so we want to solve or = 15000( )t 4/3 = ( )t Taking the log of both sides, we have But using the log rules we have Solving for t we obtain ln(4/3) = ln(( )t ) ln(4/3) = t ln(( )) t = ln(4/3) ln( ) =

8 Since the interest is paid at the end of the month, we have to wait 70 months. 8