Math 211 Business Calculus TEST 3. Question 1. Section 2.2. Second Derivative Test.

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1 Math 211 Business Calculus TEST 3 Question 1. Section 2.2. Second Derivative Test. p. 1/??

2 Math 211 Business Calculus TEST 3 Question 1. Section 2.2. Second Derivative Test. Question 2. Section 2.3. Graph with asymptotes. Exercise 50. p. 1/??

3 Math 211 Business Calculus TEST 3 Question 1. Section 2.2. Second Derivative Test. Question 2. Section 2.3. Graph with asymptotes. Exercise 50. Question 3. Section 2.6. Marginal Profit. p. 1/??

4 Math 211 Business Calculus TEST 3 Question 1. Section 2.2. Second Derivative Test. Question 2. Section 2.3. Graph with asymptotes. Exercise 50. Question 3. Section 2.6. Marginal Profit. Question 4. Section 2.5. Optimization Problem. p. 1/??

5 Math 211 Business Calculus TEST 3 Question 1. Section 2.2. Second Derivative Test. Question 2. Section 2.3. Graph with asymptotes. Exercise 50. Question 3. Section 2.6. Marginal Profit. Question 4. Section 2.5. Optimization Problem. Question 5. Section 3.1. Derivative with e x p. 1/??

6 Question 1 Prototype Second Derivative Test f(x) = x 3 12x p. 2/??

7 Question 1 Prototype Second Derivative Test f(x) = x 3 12x f (x) = 3x 2 12 p. 2/??

8 Question 1 Prototype Second Derivative Test f(x) = x 3 12x f (x) = 3x 2 12 = 3(x 2)(x + 2) p. 2/??

9 Question 1 Prototype Second Derivative Test f(x) = x 3 12x f (x) = 3x 2 12 = 3(x 2)(x + 2) Critical points: 2 and 2 p. 2/??

10 Question 1 Prototype Second Derivative Test f(x) = x 3 12x f (x) = 3x 2 12 = 3(x 2)(x + 2) Critical points: 2 and 2 f (x) = 6x p. 2/??

11 Question 1 Prototype Second Derivative Test f(x) = x 3 12x f (x) = 3x 2 12 = 3(x 2)(x + 2) Critical points: 2 and 2 f (x) = 6x p. 2/??

12 Question 1 Prototype Second Derivative Test f(x) = x 3 12x f (x) = 3x 2 12 = 3(x 2)(x + 2) Critical points: 2 and 2 f (x) = 6x x f(x) = x 3 12x f (x) = 6x Sign Concl p. 2/??

13 Question 1 Prototype Second Derivative Test f(x) = x 3 12x f (x) = 3x 2 12 = 3(x 2)(x + 2) Critical points: 2 and 2 f (x) = 6x x f(x) = x 3 12x f (x) = 6x Sign Concl = neg Max p. 2/??

14 Question 1 Prototype Second Derivative Test f(x) = x 3 12x f (x) = 3x 2 12 = 3(x 2)(x + 2) Critical points: 2 and 2 f (x) = 6x x f(x) = x 3 12x f (x) = 6x Sign Concl = neg Max = pos Min p. 2/??

15 Question 2 Prototype Graph with Asymptotes Sketch the graph of y = f(x) = x 2 x + 1 p. 3/??

16 Question 2 Prototype Graph with Asymptotes Sketch the graph of y = f(x) = x 2 x + 1 We need the derivatives p. 3/??

17 Question 2 Prototype Graph with Asymptotes Sketch the graph of y = f(x) = x 2 x + 1 We need the derivatives By the quotient rule f (1)(x + 1) (x 2)(1) (x) = (x + 1) 2 = x + 1 x + 2 = (x + 1) 2 3 (x + 1) 2 p. 3/??

18 Question 2 Prototype Graph with Asymptotes Sketch the graph of y = f(x) = x 2 x + 1 We need the derivatives By the quotient rule f (1)(x + 1) (x 2)(1) (x) = (x + 1) 2 = x + 1 x + 2 (x + 1) 2 3 = (x + 1) 2 By the power rule f (x) = 6(x + 1) 3 p. 3/??

19 Graphs continued Since the first derivative f 3 (x) = is always (x + 1) 2 positive, the function is always increasing. p. 4/??

20 Graphs continued Since the first derivative f 3 (x) = is always (x + 1) 2 positive, the function is always increasing. Since the second derivative f (x) = 6 (x + 1) is { 3 negative if x > 1 positive if x < 1, { concave down if x > 1 the graph is concave up if x < 1, p. 4/??

21 Graphs continued Since the first derivative f 3 (x) = is always (x + 1) 2 positive, the function is always increasing. Since the second derivative f (x) = 6 (x + 1) is { 3 negative if x > 1 positive if x < 1, { concave down if x > 1 the graph is concave up if x < 1, There are no critical points and no inflection points. p. 4/??

22 Find the Asymptotes For y = f(x) = x 2 x + 1 p. 5/??

23 Find the Asymptotes For y = f(x) = x 2 x + 1 Vertical asymptotes occur when the denominator x + 1 = 0 or when x = 1 p. 5/??

24 Find the Asymptotes For y = f(x) = x 2 x + 1 Vertical asymptotes occur when the denominator x + 1 = 0 or when x = 1 Note that lim f(x) = x 1 + lim x 2 x 1 + x + 1 = 3 0 = + p. 5/??

25 Find the Asymptotes For y = f(x) = x 2 x + 1 Vertical asymptotes occur when the denominator x + 1 = 0 or when x = 1 Note that lim f(x) = lim x 2 x 1 + x 1 + x + 1 = 3 0 = + To find the horizontal asymptote(s), compute the limit x 2 1/x = lim x x + 1 1/x = lim 1 2/x x 1 + 1/x = = 1 p. 5/??

26 Find the Asymptotes For y = f(x) = x 2 x + 1 Vertical asymptotes occur when the denominator x + 1 = 0 or when x = 1 Note that lim f(x) = lim x 2 x 1 + x 1 + x + 1 = 3 0 = + To find the horizontal asymptote(s), compute the limit x 2 1/x = lim x x + 1 1/x = lim x Vertical asymptote: x = 1 Horizontal asymptote: y = 1 1 2/x 1 + 1/x = = 1 p. 5/??

27 Graph of (x 2)/(x + 1) x = 1 vertical asymptote y = 1 horizontal asymptote p. 6/??

28 Graph of (x 2)/(x + 1) x = 1 vertical asymptote y = 1 horizontal asymptote p. 6/??

29 Question 3 Prototype Maximize Profit The revenue for selling x thousand units of a product is given by R(x) = 0.05x 2 + 2x + 60 and the cost for producing x thousand units is given by C(x) = 1.5x + 20 What level of sales maximizes profit? p. 7/??

30 Question 3 Prototype Maximize Profit The revenue for selling x thousand units of a product is given by R(x) = 0.05x 2 + 2x + 60 and the cost for producing x thousand units is given by C(x) = 1.5x + 20 What level of sales maximizes profit? Solution: R (x) = 0.1x + 2 and C (x) = 1.5 p. 7/??

31 Question 3 Prototype Maximize Profit The revenue for selling x thousand units of a product is given by R(x) = 0.05x 2 + 2x + 60 and the cost for producing x thousand units is given by C(x) = 1.5x + 20 What level of sales maximizes profit? Solution: R (x) = 0.1x + 2 and C (x) = 1.5 Setting R (x) = C (x) gives 0.1x + 2 = x =.5 x = 5 p. 7/??

32 Question 4 Prototype Optimization Word Problem The owner of a long building wishes to attach a rectangular fence as shown p. 8/??

33 Question 4 Prototype Optimization Word Problem The owner of a long building wishes to attach a rectangular fence as shown Building $4 y y $4 x $5 p. 8/??

34 Question 4 Prototype Optimization Word Problem The owner of a long building wishes to attach a rectangular fence as shown Building $4 y y $4 x $5 The sides that make up the width cost $4 per foot The side that makes up the length costs $5 per foot The area must be 810 square feet p. 8/??

35 Question 4 Prototype Optimization Word Problem The owner of a long building wishes to attach a rectangular fence as shown Building $4 y y $4 x $5 The sides that make up the width cost $4 per foot The side that makes up the length costs $5 per foot The area must be 810 square feet What dimensions minimize the cost? p. 8/??

36 Fencing Problem Setup Let x = length of the pen and y = width The two side pieces cost 4y each and the one length piece costs 5x The total cost is The area is A = xy C = 2 4y + 5x = 8y + 5x p. 9/??

37 Fencing Problem Setup Let x = length of the pen and y = width The two side pieces cost 4y each and the one length piece costs 5x The total cost is C = 2 4y + 5x = 8y + 5x The area is A = xy The problem is to minimize the cost function C subject to the constraining equation x y = 810 p. 9/??

38 Fence Solution Solve xy = 810 for y: y = 810/x p. 10/??

39 Fence Solution Solve xy = 810 for y: y = 810/x Substitute y = 810/x into the cost equation C = 8y + 5x p. 10/??

40 Fence Solution Solve xy = 810 for y: y = 810/x Substitute y = 810/x into the cost equation C = 8y + 5x = x + 5x = 6480 x + 5x p. 10/??

41 Fence Solution Solve xy = 810 for y: y = 810/x Substitute y = 810/x into the cost equation C = 8y + 5x = x = x x + 5x Differentiate (with respect to x) and set to 0: dc dx = = 0 x 2 p. 10/??

42 Fence Solution Solve xy = 810 for y: y = 810/x Substitute y = 810/x into the cost equation C = 8y + 5x = x = x x + 5x Differentiate (with respect to x) and set to 0: dc dx = = 0 x 2 if and only if 6480 = 5 x 2 if and only if 5x 2 = 6480 if and only if x 2 = 1296 p. 10/??

43 Fence Solution Solve xy = 810 for y: y = 810/x Substitute y = 810/x into the cost equation C = 8y + 5x = x = x x + 5x Differentiate (with respect to x) and set to 0: dc dx = = 0 x 2 if and only if 6480 = 5 x 2 if and only if 5x 2 = 6480 if and only if x 2 = 1296 Taking the square root x = 1296 = 36 p. 10/??

44 Question 5 Prototype Derivatives with e x Differentiate y = x 2 e x p. 11/??

45 Question 5 Prototype Derivatives with e x Differentiate y = x 2 e x By the product rule, y = d ( ) x 2 e x + x 2 d dx dx (ex ) = 2xe x + x 2 e x p. 11/??