Questions and Answers #1 Analyzing Functions

Questions and Answers Sheet 1 Analyzing Functions Question #1 Find c such that fave = 𝑓(𝑐), 𝑓(𝑥) = (𝑥 − 3)2 , [2,5] Answer: Average value of the function f(x) over the interval [a,b] is given by 𝑓𝑎𝑣𝑔 = 𝑏 1 ∫ 𝑓(𝑥)𝑑𝑥 𝑏−𝑎 𝑎 Average value of 𝑓(𝑥) = (𝑥 − 3)2 ovthe interval [2,5] is given by 𝑓𝑎𝑣𝑔 = 5 1 ∫ (𝑥 − 3)2 𝑑𝑥 5−2 2 Substitute 𝑥 − 3 = 𝑢 and 𝑑𝑥 = 𝑑𝑢 2 Limits of Integration will change from ∫−25 to ∫−2 − 35−3 = ∫−1 1 2 = ∫ 𝑢2 𝑑𝑢 3 −1 2 1  u3  =   ] 3  3  −1 1 8 1  [=  +  = 1 3 3 3 Question #2 Find the length of the curve. 1 𝑟⃗(𝑡) =< 8𝑡, 𝑡 2 , 𝑡 3 > ,0 ≤ 𝑡 ≤ 1 12 Answer: 3 2 𝑡 > 12 1 ⃗⃗⃗⃗ 𝑟 ′ (𝑡) =< 8,2𝑡, 𝑡 2 > 4 ⃗⃗⃗⃗ 𝑟 ′ (𝑡) =< 8,2𝑡, 2 1 ||𝑟⃗⃗⃗⃗′ (𝑡)|| = √(8)2 + (2𝑡)2 + ( 𝑡 2 ) 4 ⇒ ||𝑟⃗⃗⃗⃗′ (𝑡)|| = √64 + 4𝑡 2 + ⇒ ||𝑟⃗⃗⃗⃗′ (𝑡)|| = √ 1 4 𝑡 16 1 ((64 ∗ 16) + 64𝑡 2 + 𝑡 4 ) 16 1 ⇒ ||𝑟⃗⃗⃗⃗′ (𝑡)|| = √1024 + 64𝑡 2 + 𝑡 4 4 1 ⇒ ||𝑟⃗⃗⃗⃗′ (𝑡)|| = √322 + (2 ∗ 32)𝑡 2 + 𝑡 4 4 1 ⇒ ||𝑟⃗⃗⃗⃗′ (𝑡)|| = √(32 + 𝑡 2 )2 4 1 ⇒ ||𝑟⃗⃗⃗⃗′ (𝑡)|| = |32 + 𝑡 2 | 4 Now, 𝑡 2 is always positive. So, (32 + 𝑡 2 ) is always positive. Question #3 1 How to write 𝑥 2 in radical form? Answer: 1 𝑥 2 = √𝑥 I'll explain: 1 𝑚 𝑎 𝑚 = √𝑎 1 2 𝑥 2 = √𝑥 = √𝑥 Question #4 Find the integer a such that 𝑎 ≡ −15 (b mod 27) and −26 ≤ 𝑎 ≤ 0 Answer: Division algorithm Let a be an integer and d a positive integer. Then there are unique q and r with 0 < 𝑟 < 𝑑 such that 𝑎 = 𝑑𝑞 + 𝑟𝑞 is called the quotient and r is called the remainder 𝑞 =𝑎 ÷𝑑 𝑟 = 𝑎 𝑏 𝑚𝑜𝑑 𝑑 𝑎 ≡ −15(𝑏 𝑚𝑜𝑑 27) and −26 ≤ 𝑎 ≤ 0 Since -15 is between -26 and 0: 𝑎 = −15 Question #5 Find the integer a such that 𝑎 ≡ 24(𝑏 𝑚𝑜𝑑 31) and −15 ≤ 𝑎 ≤ 15 Answer: 𝑎 ≡ 24(𝑏 𝑚𝑜𝑑 31) We can find a such that −15 ≤ 𝑎 ≤ 15 by consecutively substracting 31 from 24 until we obtain a value between -15 and 15. 𝑎 ≡ 24(𝑏 𝑚𝑜𝑑 31) ≡ 24 − 31(𝑏 𝑚𝑜𝑑 31) ≡ −7(𝑏 𝑚𝑜𝑑 31) SInce -7 is between -15 and 15: 𝑎 = −7 Question #6 Find the integer a such that 𝑎 ≡ 99(𝑏 𝑚𝑜𝑑 41) and 100 ≤ 𝑎 ≤ 140 Answer: 𝑎 ≡ 99(𝑏 𝑚𝑜𝑑 41) We can find a such that 100 ≤ 𝑎 ≤ 140 by consecutively adding 41 to 99 until we ibtain a vlue between 100 and 140 𝑎 ≡ 99(𝑏 𝑚𝑜𝑑 41) ≡ 99 + 41(𝑏 𝑚𝑜𝑑 41) ≡ 140(𝑏 𝑚𝑜𝑑 41) Since between 100 and 140 (including): 𝑎 = 140 Question #7 Explain the difference between an absolute minimum and a local minimum. a) There is no difference. b) A function f has an absolute minimum at x=c if f(c) is the smallest function value on the entire domain. c) A function f has an absolute minimum at x=c if f(c) is the smallest function value when x is near c, whereas f has a local minimum at c if f(c) is the smallest function value on the entire domain of f. d) A function f has an absolute minimum at x=c if f(c) is the largest function value on the entire domain of f, whereas f has a local minimum at c if f(c) is the largest function value when x is near c. e) A function f has an absolute minimum at x=c if f(c) is the largest function value when x is near c, whereas f has a local minimum at c if f(c) is the largest function value on the entire domain of f. Answer: Option b is correct: A function f has an absolute minimum at x=c if f(c) is the smallest function value on the entire domain of f, whereas f has a local minimum at c if f(c) is the smallest function value when x is near c. Question #8 Determine the end behavior of the following transcendental functions by analyzing appropriate limits. Then provide a simple sketch of the associated graph, showing asymptotes if they exist. 𝑓(𝑥) = 50 𝑒 2𝑥 Answer: Step 1 Here, find function limits at infinity for horizontal asymptotes, and find possible vertical asymptotes at points for which our function is not defined. lim 𝑓(𝑥) = lim 𝑥⇒∞ 50 𝑥⇒∞ 𝑒 2𝑥 substitute for 𝑥 = lim 50 𝑥⇒∞ 𝑒 ∞ compute limit and simplify = 0 lim 𝑓(𝑥) = lim = 50 lim 50 𝑥⇒−∞ 𝑒 2𝑥 𝑥⇒−∞ 1 𝑥⇒−∞ 𝑒 2𝑥 = 50 lim 𝑒 −2𝑥 𝑥⇒−∞ = 50(∞) =∞ We have horizontal asymptote at y=0 Step 2 We have no vertical asymptotes because our function is defined for every x Now, check for slant asymptotes. Using formula lim 𝑓(𝑥) 𝑥⇒±∞ 𝑥 asymptotes) lim 50 𝑒2𝑥 𝑥⇒∞ 𝑥 50 =0 𝑥⇒∞ 𝑥𝑒 2𝑥 = lim As our slope is zero , there is no slant asymptotes. graph: Question #9 Plot the graphs of the given functions = 𝑘 where k represents slope of our line (slant | x | f ( x) =  − x + 1 3 for x  2  m( x ) =  − x for x  2   x −3 Answer: Step 1 | x | f ( x) =  − x + 1 Step 2 for x  2 for x  2 for − 4  x  −1 for − 1  x  3 for x  3 3  m( x ) =  − x   x −3 for − 4  x  −1 for − 1  x  3 for x  3 Question #10 Determine the end behavior of the following transcendental functions by analyzing appropriate limits. Then provide a simple sketch of the associated graph, showing asymptotes if they exist. 𝑓(𝑥) = 2𝑥 Answer: Step 1 To determine the end behavior of the following transcendental functions by analyzing appropriate limits. Given function is 𝑓(𝑥) = 2𝑥 . To find end behavior of function 𝑓(𝑥) = 2𝑥 we have to find asymptotes for 𝑓(𝑥) = 2𝑥 1) To find horizontal asymptotes: The line y=L is a horizontal asymptote of the function y=f(x) if either lim 𝑓(𝑥) = 𝐿𝑜𝑟 lim 𝑓(𝑥) = 𝐿 and L is finite. 𝑥⇒∞ 𝑥⇒−∞ Now here, lim 2𝑥 = 2∞ = ∞ 𝑎𝑛𝑑 lim 2𝑥 = 2−∞ = 0 𝑥⇒∞ 𝑥⇒−∞ Therefore the line y=0 is the horizontal asymptote for 𝑓(𝑥) = 2𝑥 Step 2 2) To find vertical asymptotes: The line x=L is a vertical asymptote of the function 𝑓(𝑥) = 2𝑥 if the limit of the function at this point is infinite. Here the function 𝑓(𝑥) = 2𝑥 is defined for every value of x. Therefore there are no vertical asymptotes for 𝑓(𝑥) = 2𝑥 3) To find slant asymptotes: The line y=mx+b is slant asymptote of a function y=f(x) if either 𝑚 = lim ( 𝑥⇒∞ 𝑓(𝑥) ) = 𝐿 or 𝑚 = lim ( 𝑥 𝑓(𝑥) lim ( 𝑥⇒−∞ 𝑥 ) = lim 2𝑥 𝑥⇒−∞ 𝑥 𝑥⇒−∞ 𝑓(𝑥) 𝑥 ) = 𝐿 and L is finite and zero 𝑚 = lim ( 𝑥⇒∞ 𝑓(𝑥) 𝑥 ) = lim 2𝑥 𝑥⇒∞ 𝑥 = and 𝑚 = =0 Here the value of L is not finite and non zero. Therefore the function 𝑓(𝑥) = 2𝑥 has no slant asymptotes. Step 3 The sketch of the graph of function 𝑓(𝑥) = 2𝑥 is, Question #11 A different interpretation of marginal cost Suppose a large company makes 25,000 gadgets per year in batches of x items at a time. After analyzing setup costs to produce each batch and taking into account storage costs, it has been determined that the total cost C(x) of producing 25,000 gadgets in batches of x items at a time is given by 𝐶(𝑥) = 1250000 + 125000000 𝑥 + 1.5𝑥 a. Determine the marginal cost and average cost functions. Graph and interpret these functions. b. Determine the average cost and marginal cost when x = 5000 . c. The meaning of average cost and marginal cost here is different from earlier examples and exercises. Interpret the meaning of your answer in part (b). Answer: Step 1 𝐶(𝑥) = 1250000 + 125000000 + 1.5𝑥 𝑥 Formula: Marginal Cosr 𝐶̅ (𝑥) = Average Cost ̅̅̅ 𝐶 ′ (𝑥) = 𝑑 𝑑𝑥 𝐶(𝑥) 𝑥 𝐶̅ (𝑥) Step 2 Marginal cost is 𝐶̅ (𝑥) = 1250000 Average cost is: ̅̅̅ 𝐶 ′ (𝑥) = 𝑥 𝑑 𝑑𝑥 ( + 125000000 𝑥2 1250000 𝑥 + + 1.5 125000000 𝑥2 1250000 250000000 + 1.5) ̅̅̅ 𝐶 ′ (𝑥) (− − ) 2 3 𝑥 𝑥 The graph of the marginal(in blue) and average(in black) is As the number of items increases, the marginal function decreases As the number of items increases, the average function increases Step 3 𝐶̅ (5000) = 1250000 5000 + 125000000 50002 + 1.5 250000000 ̅̅̅′ (5000) = (− 1250000 𝐶̅ (5000) = 256.5𝐶 2 − 3 ) 5000 5000 ̅{𝐶}(5000) = −0.052 when the number of items is 5000, the marginal cost is 256.5 and the average cost is -0.052 Question #12 Find the critical points of the following functions.Use the Second Derivative Test to determine (if possible) whether each critical point corresponds to a local maximum, a local minimum,or a saddle point. If the Second Derivative Test is inconclusive,determine the behavior of the function at the critical points 𝑓(𝑥, 𝑦) = (4𝑥 − 1)2 + (2𝑦 + 4)2 + 1 Answer: Given function is 𝑓(𝑥) = (4𝑥 − 1)2 + (2𝑦 + 4)2 + 1 = 16𝑥 2 + 4𝑦 2 − 8𝑥 + 16𝑦 + 18 We find critical points: 𝑓𝑥 = 𝑓𝑦 = 𝜕𝑓 𝑑𝑦 𝜕𝑓 𝜕𝑥 = 32𝑥 − 8 1 = 8𝑦 + 16. Setting these equal to zero, 32𝑥 − 8 = 0, ⇒ 𝑥 = 8𝑦 + 16 = 0, ⇒ 𝑦 = −2. So, the 4 1 critical point is ( , − 2) 4 Now we find the second derivatives, 𝑓𝑥𝑥 = 32 𝑓𝑦𝑦 = 8 𝑓𝑥𝑦 = 0 2 So, 𝐷 = 𝑓𝑥𝑥 × 𝑓𝑦𝑦 − (𝑓𝑥𝑦 ) 1 For ( , − 2) 4 𝐷 = 32 × 8 − 0 = 256 > 0 1 Since D>0 and 𝑓𝑥𝑥 ( , − 2) = 32 > 0, so the obtained critical point is a local minimum. 4 Question #13 Find the critical points of the following functions. 𝑓(𝑥, 𝑦) = 𝑥 4 + 𝑦 4 − 4𝑥 − 32𝑦 + 10 Answer: Step 1 The function given is 𝑓(𝑥, 𝑦) = 𝑥 4 + 𝑦 4 − 4𝑥 − 32𝑦 + 10 Take the first derivative of the function, 𝑓𝑥 (𝑥, 𝑦) = 4𝑥 3 − 4 4𝑥 3 − 4 = 0 4(𝑥 3 − 1) = 0 𝑥=1 𝑓𝑦 (𝑥, 𝑦) = 4𝑦 3 − 32 4𝑦 3 − 32 = 0 4(𝑦 3 − 8) = 0 𝑦3 − 8 = 0 𝑦=2 Step 2 The critical are: (1, 2) Take the second derivative of the 𝑓𝑥 (𝑥, 𝑦) 𝑓𝑥𝑥 (𝑥, 𝑦) = 12𝑥 2 At (1, 2), 𝑓𝑥𝑥 (1,2) = 12 ⋅ 12 = 12 Take the second derivative of the 𝑓𝑦 (𝑥, 𝑦) 𝑓𝑦𝑦 (𝑥, 𝑦) = 12𝑦 2 Step 3 At (1, 2) 𝑓𝑦𝑦 (1,2) = 12 ⋅ 22 = 48 The other second derivative with respect to y, 𝑓𝑥𝑦 (𝑥, 𝑦) = 0 2 Consider the condition, 𝐷 = 𝑓𝑥𝑥 ⋅ 𝑓𝑦𝑦 − (𝑓𝑥𝑦 ) = 12 ⋅ 48 − 0 = 576 Here, the value of discriminant is greater than 0. Therefore the critical point (1, 2) is a local minimum point. Question #14 Find the critical points of the following functions. 𝑓(𝑥, 𝑦) = −4𝑥 2 + 8𝑦 2 − 3 Answer: Step 1 To find critical points, first find the partial derivative of function f with respect to x and y respectively. 𝑓𝑥 = −8𝑥𝑓𝑦 = 16𝑦 W quate it to zero, −8𝑥 = 0 𝑥 = 0 And 16𝑦 = 0 𝑦 = 0 Hence critical point is (0, 0) Step 2 Let's use the second derivative test to classify the critical points. Hence, compute 𝑓𝑥𝑥 = −8𝑓𝑦𝑦 = 16𝑓𝑥𝑦 = 0 2 And 𝐷 = 𝑓𝑥𝑥 𝑓𝑦𝑦 − [𝑓𝑥𝑦 ] . At critical point (0, 0) 𝐷 = (−8)(16) − (0)2 , 𝐷 = −128, 𝐷 < 0. This indicates, point (0,0) is a saddle point. Question #15 Find the critical points of the following functions. 𝑓(𝑥, 𝑦) = 𝑦𝑒 𝑥 − 𝑒 𝑦 Answer: Step 1 To find the critical point, find 𝑓𝑥 , 𝑓𝑦 and set 𝑓𝑥 = 0, 𝑓𝑦 = 0 𝑓𝑥 = 𝑦𝑒 𝑥 𝑓𝑦 = 𝑒 𝑥 − 𝑒 𝑦 Now set 𝑓𝑥 = 0 𝑓𝑥 = 𝑦𝑒 𝑥 = 0 ⇒ 𝑦 = 0 or 𝑒 𝑥 = 0 For 𝑒 𝑥 = 0 have no solution for 𝑥 ∈ 𝑅. Therefore y=0 Set 𝑓𝑦 = 0 𝑓𝑦 = 𝑒 𝑥 − 𝑒 𝑦 = 0 Substitute y=0 𝑓𝑦 = 𝑒 𝑥 − 𝑒 0 = 0 ⇒ 𝑒 𝑥 − 1 = 0 ⇒ 𝑒 𝑥 = 1 Taking 'ln on both sides, ln(𝑒 𝑥 ) = ln(1) x=0 Thus (0, 0) is the crirtical point. Step 2 𝐷(𝑥, 𝑦) = 𝑓𝑥𝑥 (𝑥, 𝑦)𝑓𝑦𝑦 (𝑥, 𝑦) − (𝑓𝑥𝑦 (𝑥, 𝑦)) 2 1) If 𝐷(𝑎, 𝑏) > 0 amd 𝑓𝑥𝑥 (𝑎, 𝑏) > 0 then (𝑎, 𝑏) is local minimum of f 2) If 𝐷(𝑎, 𝑏) > 0 and 𝑓𝑥𝑥 (𝑎, 𝑏) < 0 then (a, b) is local maximum of f 3) If 𝐷(𝑎, 𝑏) < 0 then f(a, b) is saddle point. 4) If 𝐷(𝑎, 𝑏) = 0 then this test is inconclusive. Here 𝑓𝑥𝑥 = 𝑦𝑒 𝑥 , 𝑓𝑦𝑦 = −𝑒 𝑦 and 𝑓𝑥𝑦 = 𝑒 𝑥 𝐷(𝑥, 𝑦) = 𝑦𝑒 𝑥 (−𝑒 𝑦 ) − (𝑒 𝑥 )2 = −𝑦𝑒 𝑥+𝑦 − 𝑒 2𝑥 . The critical point is (0, 0) 𝐷(0,0) = −0𝑒 0 − 𝑒 2⋅0 = −𝑒 {0) = −1. Thus 𝐷(0,0) = −1 < 0 then f(0, 0) is saddle point, 𝑓(0,0) = (0)𝑒 0 − 𝑒 0 = −𝑒 0 = −1 Therefore -1 is the saddle point. Question #16 Find the critical points of the following functions.Use the Second Derivative Test to determine (if possible) whether each critical point corresponds to a local maximum, a local minimum,or a saddle point. If the Second Derivative Test is inconclusive,determine the behavior of the function at the critical points. 1 1 𝑓(𝑥, 𝑦) = sin(2𝜋𝑥) cos(𝜋𝑦) , for |𝑥| ≤ and |𝑦| ≤ 2 2 Answer: Step 1 To find the critical point we need to find the partial derivatives of the given function with respect to x and with respect to y 𝑓𝑥 (𝑥, 𝑦) = 2𝜋(cos 2 𝜋𝑥) cos(𝜋𝑦) 𝑓𝑦 (𝑥, 𝑦) = 𝜋(− sin 𝜋 𝑦) sin(2𝜋𝑥) Step 2 Equating the derivatives to 0, to find the critical points 2𝜋(cos 2 𝜋𝑥) cos(𝜋𝑦) = 0 ⇒ cos 2 𝑥 = 0 ⇒ 2𝜋𝑥 = cos −1 (0) ⇒ 2𝜋𝑥 = 𝜋 1 𝜋 1 ⇒ 𝑥 = cos 𝜋 𝑦 = 0 ⇒ 𝜋𝑦 = cos −1 (0) ⇒ 𝜋𝑦 = ⇒ 𝑦 = 2 4 2 2 1 1 4 2 Therefore the critical points from this is ( , 𝑎) and (𝑏, ) 𝜋(− sin 𝜋 𝑦) sin(2𝜋𝑥) = 0 ⇒ − sin 𝜋 𝑦 = 0 ⇒ sin 𝜋 𝑦 = 0 ⇒ 𝜋𝑦 = sin−1 (0) ⇒ 𝑦 = 0 sin(2𝜋𝑥) = 0 ⇒ 2𝜋𝑥 = 𝑠𝑖𝑛−1 (0) ⇒ 𝑥 = 0 The critical points from this are (0,c) and (d,0) Therefore the critical points of the given function are 1 1 4 2 ( , 𝑎) , (𝑏, ) , (0, 𝑥) and (d,0) Step 3 Let us use the second derivative test to determine whether the critical points correspond to a local maximum, a local minimum,or a saddle point. 𝐷(𝑥, 𝑦) = 𝑓𝑥𝑥 (𝑥, 𝑦)𝑓𝑦𝑦 (𝑥, 𝑦) − (𝑓𝑥𝑦 (𝑥, 𝑦)) 2 1. If 𝐷(𝑥, 𝑦) > 0 and 𝑓𝑥𝑥 (𝑥, 𝑦) < 0, then f has a local maximum value at (x,y) 2. If 𝐷(𝑥, 𝑦) > 0 and 𝑓𝑥𝑥 (𝑥, 𝑦) > 0, then f has a local minimum value at (x,y) 3. If 𝐷(𝑥, 𝑦) < 0 then f has a saddle point at (x,y) 4. If 𝐷(𝑥, 𝑦) = 0 then the test is inconclusive. Let us find 𝑓𝑥𝑥 (𝑥, 𝑦), 𝑓𝑦𝑦 (𝑥, 𝑦), 𝑓𝑥𝑦 (𝑥, 𝑦) 𝑓𝑥𝑥 (𝑥, 𝑦) = 4𝜋 2 (−𝑠𝑖𝑛2𝜋𝑥) cos(𝜋𝑦) 𝑓𝑦𝑦 (𝑥, 𝑦) = 𝜋 2 (−𝑐𝑜𝑠𝜋𝑦)(sin 2 𝜋𝑥) 𝑓𝑥𝑦 (𝑥, 𝑦) = 2𝜋 2 (cos 2 𝜋𝑥)(− sin 𝜋 𝑦) Step 4 1 At ( , 𝑎) 4 2 1 1 1 1 1 1 𝐷 ( , 𝑎) = 𝑓𝑥𝑥 ( , 𝑎) 𝑓𝑦𝑦 ( , 𝑎) = (𝑓𝑥𝑦 ( , 𝑎)) 𝑓𝑥𝑥 ( , 𝑎) = 4𝜋 2 (− sin 2 𝜋 ) cos(𝜋𝑎) = −4𝜋 2 cos(𝜋𝑎) 4 4 4 4 4 4 1 𝑓𝑦𝑦 (𝑥, 𝑦) = 𝜋 2 (− cos 𝜋 𝑎) (sin 2 𝜋 ) = −𝜋 2 cos 𝜋 𝑎 4 1 𝑓𝑥𝑦 = 2𝜋 2 (cos 2 𝜋 ) (− sin 𝜋 𝑎) = 0 4 1 1 D ( , a ) = (−4 2 cos( a )( − 2 cos( a )) − 0  D( , a) = 4 4 cos 2 ( a)  0 4 4 1 1 4 4 and 𝑓𝑥𝑥 ( , 𝑎) < 0. Which implies f has a local maximum at ( , 𝑎) Step 5 1 1 1 1 2 2 2 2 2 1 1 2 3 At (𝑏, ) 𝐷 (𝑏, ) = 𝑓𝑥𝑥 (𝑏, ) 𝑓𝑦𝑦 (𝑏, ) − (𝑓𝑥𝑦 (𝑏, 𝑓𝑟𝑎𝑐12)) 𝑓𝑥𝑥 (𝑏, ) = 4𝜋 2 (− sin 2 𝜋𝑏) cos (𝜋 ) = 0 1 𝑓𝑦𝑦 (𝑥, 𝑦) = 𝜋 2 (− cos 𝜋 ) (sin 2 𝜋𝑏) = 0 2 1 𝑓𝑥𝑦 = 2𝜋 2 (cos 2 𝜋𝑎) (− sin 𝜋 ) = −2𝜋 2 (cos 2 𝜋𝑎) 2 1 𝐷 (𝑏, ) = 0 − 4𝜋 2 (cos 2 (2𝜋𝑎)) 2 1 Which implies f has a saddle point at (𝑏, ) 2 Step 6 At (0, c) 𝐷(0, 𝑐) = 𝑓𝑥𝑥 (0, 𝑐)𝑓𝑦𝑦 (0, 𝑐) − (𝑓𝑥𝑦 (0, 𝑐)) 2 𝑓𝑥𝑥 (0, 𝑐) = 4𝜋 2 (− sin 2 𝜋(0)) cos(𝜋𝑐) = 0 𝑓𝑦𝑦 (𝑥, 𝑦) = 𝜋 2 (− cos 𝜋 𝑐)(sin 2 𝜋(0)) = 0 𝑓𝑥𝑦 (𝑥, 𝑦) = 2𝜋 2 (cos 2 𝜋(0))(− sin 𝜋 𝑐) = −2𝜋 2 (sin 𝜋 𝑐) 𝐷(0, 𝑐) = 0 − 4𝜋 2 (sin 𝜋 𝑐) < 0 2 Which implies f has a saddle point at (0, c) At (𝑑, 0)𝐷(𝑑, 0) = 𝑓𝑥𝑥 (𝑑, 0)𝑓𝑦𝑦 (𝑑, 0) − (𝑓𝑥𝑦 (𝑑, 0)) 𝑓𝑥𝑥 (𝑑, 0) = 4𝜋 2 (− sin 2 𝜋𝑑) cos(𝜋0) = −4𝜋 2 sin 2 𝜋𝑑 𝑓𝑦𝑦 (𝑑, 0) = 𝜋 2 (− cos 𝜋 0)(sin 2 𝜋𝑑) = −𝜋 2 𝑠𝑖𝑛2𝜋𝑑 𝑓𝑥𝑦 (𝑑, 0) = 2𝜋 2 (cos 2 𝜋𝑑)(− sin 𝜋 0) = 0 𝐷(𝑑, 0) = (−4𝜋 2 sin 2 𝜋𝑑)(−𝜋 2 sin 2 𝜋𝑑) − 0 = 4𝜋 4 sin2 (2𝜋𝑑) ⇒ 𝐷(𝑑, 0) > 0 and 𝑓𝑥𝑥 (𝑑, 0) < 0 Which implies f has a local maximum at (d, 0) Question #17 Find the critical points of the following functions.Use the Second Derivative Test to determine (if possible) whether each critical point corresponds to a local maximum, a local minimum,or a saddle point. If the Second Derivative Test is inconclusive,determine the behavior of the function at the critical points. 𝑓(𝑥, 𝑦) = √𝑥 2 + 𝑦 2 − 4𝑥 + 5 Answer: Step 1 The given function is, 𝑓(𝑥, 𝑦) = √𝑥 2 + 𝑦 2 − 4𝑥 + 5 Apply the procedure for Maxima-Minima, Solve for critical points, ⇒𝑥=2 𝜕𝑓 −1 (2𝑦) = 9 =0⇒ 𝜕𝑥 2𝑓 ⇒𝑦=0 Thus, (2, 0) is a critical point. Step 2 Solve for the second derivative as, 𝜕2𝑓 𝜕 𝜕𝑓 = ( ) 𝜕𝑥 2 𝜕𝑥 𝜕𝑥 = 𝜕 −1 ( (2𝑥 − 4)) 𝜕𝑥 2𝑓 = 𝜕 1 ( (2 − 𝑥)) 𝜕𝑥 𝑓 𝑓(−1) − (2 − 𝑥) (− = 1 ) (2𝑥 − 4) 2𝑓 𝑓2 = 1 1 [−𝑓 + (2𝑥 − 𝑥) (2𝑥 − 4)] 2 𝑓 2𝑓 = 1 [−𝑓 2 + (2 − 𝑥)(𝑥 − 2)] 𝑓3 = 1 [−(𝑥 2 + 𝑦 2 − 4𝑥 + 5) − (𝑥 2 + 4 − 4𝑥)] 𝑓3 𝜕𝑓 𝜕𝑥 =0⇒ −1 2𝑓 (2𝑥 − 4) = 0 = 1 [−𝑥 2 − 𝑦 2 + 4𝑥 − 5 − 𝑥 2 − 4 + 4𝑥] 𝑓3 = 1 [−2𝑥 2 − 𝑥 2 + 8𝑥 − 9] 𝑓3 Step 3 Substitute the values, 𝑟= 𝜕2𝑓 |(2,0) 𝜕𝑥 2 1 = (√(2)2 + (0)2 1 = 3 − 4(2) + 5) 3 [−2(2)2 − (0)2 + 8(2) − 9] [−8 + 16 − 9] (√4 − 8 + 5) = 1 ⋅ [−1] Now, Solve for s as, = = 𝜕 −1 ( (2𝑦)) 𝜕𝑥 2𝑓 = 𝜕 −𝑦 ( ) 𝜕𝑥 𝑓 𝜕2 𝑓 𝜕𝑥𝜕𝑦 = 𝜕 𝜕𝑓 ( ) 𝜕𝑥 𝜕𝑦 −1 = (−𝑦) ( 2 ) 𝑓 = 𝑦 𝑓2 Step 4 Substitute the values, 𝑠 = 𝜕2 𝑓 | 𝜕𝑥𝜕𝑦 (2,0) (0)2 (2)2 + (0)2 − 4(2) + 5 =0 Similarly, the second derivative as, = 𝜕 −𝑦 ( ) 𝜕𝑦 𝑓 𝑓(1) − 𝑦 ( =− −1 ) (2𝑦) 2𝑓 𝑓2 =− 1 𝑦2 (𝑓 + ) 𝑓2 𝑓 =− 1 2 (𝑓 + 𝑦 2 ) 𝑓3 Step 5 𝜕2 𝑓 𝜕𝑦 2 = 𝜕 𝜕𝑓 ( ) 𝜕𝑦 𝜕𝑦 Substitute the values, 𝑡 = 𝜕2 𝑓 | 𝜕𝑦 2 (2,0) 1 =− (√(2)2 + (0)2 1 =− 3 − 4(2) + 5) 3 [(2)2 + (0)2 − 4(2) + 5 + (0)2 ] [4 − 8 + 5] (√4 − 8 + 5) 1 = − (1) = −1 1 Apply theSecond Derivative Test, 𝑟𝑡 − 𝑠 2 = (−1)(−1) − (0)2 = 1 Step 6 Since 𝑟𝑡 − 𝑠 2 > 0 ⇒ 𝑓(𝑥, 𝑦) is minimum at (2, 0) Solve for the minimum value, 𝑓𝑚𝑖𝑛 (𝑥, 𝑦) = 𝑓(2,0) = √(2)2 + (0)2 − 4(2) + 5 = √4 + 0 − 8 + 5 = √1 = 1 Thus, the minimum value of the given function is 1, which occurs at the point (2,0). Question #18 2 Without integrating, explain why ∫−2 𝑥(𝑥 2 + 1)2 𝑑𝑥 = 0 Answer: Step 1 Odd Function A function f(x) is a odd function if and only if 𝑓(−𝑥) = −𝑓(𝑥) Even Function A function f(x) is an even function if and only if 𝑓(−𝑥) = 𝑓(𝑥) 𝑎 Now, if we have 𝐼 = ∫−𝑎 𝑓(𝑥)𝑑𝑥 , then 0 if f(x) is an odd function   I = 2 2 f ( x)dx if f(x) is an even function   0 With the help of this result we will solve our given integration. Step 2 2 Given integration is ∫−2 𝑥(𝑥 2 + 1)2 𝑑𝑥 Here, 𝑓(𝑥) = 𝑥(𝑥 2 + 1)2 and a=2 Now, to check the function, that is function is odd or even, consider f(-x), replace x by -x, we get 𝑓(−𝑥) = −𝑥((−𝑥)2 + 1)2 = −𝑥(𝑥 2 + 1)2 = −(𝑥(𝑥 2 + 1)2 ) = −𝑓(𝑥) ⇒ 𝑓(−𝑥) = −𝑓(𝑥) 2 hence function 𝑥(𝑥 2 + 1)2 is an odd function. So, by the result of integration, ∫−2 𝑥(𝑥 2 = 1)2 𝑑𝑥 = 0 Hence proved Question #19 Consider 𝑓(𝑥) = 𝑥𝑒 −𝑘𝑥 for k>0 . Find the relative extrema and the points of inflection of the function. Answer: 𝑓(𝑥) = 𝑥𝑒 −𝑘𝑥 For finding relative extrema we need to us first derivative test Now for that 𝑓 ′ (𝑥) = 0 here 𝑓 ′ (𝑥) = −(𝑘𝑥 − 1)𝑒 𝑘𝑥 0 = −(𝑘𝑥 − 1)𝑒 −𝑘𝑥 let 𝑘 = 20 = −(2𝑥 − 1)𝑒 −2𝑥 𝑥 = 0.5 is a critical point f(x) in domain P (−∞, 0.5), (0.5, ∞) 𝑚𝑖𝑛 = (0.5, 1 ) 𝑒 0.5 Question #20 Analizing the polynomial 𝑦 = (𝑥 − 1)(3𝑥 − 2)(𝑥 + 3) a) Determine the roots of the polinomial b) Determine the end behaviors c) Determine how many turning points Answer: Step 1 Explanation: Given that, 𝑦 = (𝑥 − 1)(3𝑥 − 2)(𝑥 + 3) a) Determine the roots of the polynomial. To find the roots, put y = 0 and solve for x (𝑥 − 1)(3𝑥 − 2)(𝑥 + 3) = 0 𝑥−1=0 𝑥=1 3𝑥 − 2 = 0 𝑥= 3 2 𝑥 + 3 = 0 𝑥 = −3 Roots are: −3,1, 3 2 Step 2 b) Determine end behaviour. Write the polynomial in standard form. 𝑦 = (𝑥 − 1)(3𝑥 − 2)(𝑥 + 3) 𝑦 = 3𝑥 3 + 4𝑥 2 − 13𝑥 + 6 Here we see that, leading coefficient is +3 that is positive. and degree of the polynomial is 3. Rule says, When degree is odd and leading coefficient is positive end behaviour of the polynomial is: 𝑓(𝑥) ⇒ −, as 𝑥 ⇒ −∞ f(x) \Rightarrow \infty, as x ⇒ ∞ c) Determine the number of turning points. A polynomial of degree n will have at most n - 1 turning points. Here degree of given polynomial is 3 Number of turning points = 3 - 1 = 2