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

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

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/??

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/??

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/??

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

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

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

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/??

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/??

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/??

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 2 8 + 24 = 16 12 neg Max p. 2/??

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 2 8 + 24 = 16 12 neg Max 2 8 24 = 16 12 pos Min p. 2/??

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

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

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/??

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/??

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

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/??

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

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/??

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/??

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 0 1 + 0 = 1 p. 5/??

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

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/??

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

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/??

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/??

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/??

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/??

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

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

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

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

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

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/??