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Tutorial 6 - April 04, 2014 Questions related to SECTION 4.3 1. Sketch the graph of f (x) = x4 − 6x2 (a) Domain: (−∞, ∞) (b) x− and y− intercepts: x−intercept: y = 0 ⇔ x4 − 6x2 = 0 So the x−intercepts are (0, 0), y−intercept: x = 0 ⇔ √ (− 6, 0), x = 0, √ x = − 6, x= √ 6 √ ( 6, 0) ⇔y=0 So the y−intercept is (0, 0). (c) Critical Points and Local Extreme Values: i. Find f 0 (x) f 0 (x) = 4x3 − 12x = 4x(x2 − 3) ii. Find all critical points: f 0 (x) = 0 : f 0 (x) = 4x(x2 − 3) = 0 ⇔ x = 0, √ x = − 3, x= √ 3 f 0 (x) is not defined: f 0 (x) is defined for all real x. iii. Increasing and Decreasing Intervals: √ √ x x< − 3 <x< 0 <x< 3 <x 0 f (x) − 0 + 0 − 0 + f (x) & loc. min % loc. max & loc. min % iv. Evaluate f (x) at each critical point: √ √ local minimums at (− 3, −9) , ( 3, −9) and local maximum at (0, 0) (d) Determine Concavity and Points of Inflection: 00 i. Find f (x) 00 f (x) = 12x2 − 12 = 12(x2 − 1) = 12(x − 1)(x + 1) c 2012 Department of Mathematics, Eastern Mediterranean University 1 Tutorial 6 - April 04, 2014 ii. Find all potential points of inflection: 00 f (x) = 0 : 00 f (x) = 12(x − 1)(x + 1) = 0 00 ⇔ x = 1, x = −1 00 f (x) is not defined: f (x) is defined for all real x. iii. Intervals of Concavity: x x< 00 f (x) + f (x) ∪ −1 <x< 1 <x 0 − 0 − infl. ∩ infl. ∪ point point iv. The points of inflection are : (−1, −5) and (1, −5) (e) Asymptotes: Horizontal Asymptote: None Vertical Asymptote: None Oblique Asymptote: None (f) Sketch the graph: 10 5 -4 -2 2 4 -5 -10 (g) The range of f (x) is [−9, ∞). c 2012 Department of Mathematics, Eastern Mediterranean University 2 Tutorial 6 - April 04, 2014 2. Sketch the graph of y = (x + 1)2 1 + x2 (a) Domain: x ∈ R (b) x− and y− intercepts: x−intercept: y = 0 ⇔ (x + 1)2 =0 1 + x2 ⇔ (x + 1)2 = 0 ⇔ x = −1 So the x−intercept is (−1, 0). y−intercept: x = 0 ⇔y= (0 + 1)2 1 + 02 ⇔y=1 So the y−intercept is (0, 1). (c) Critical Points and Local Extreme Values: i. Find f 0 (x) f 0 (x) = (1 + x2 ) · 2(x + 1) − (x + 1)2 · 2x 2(1 − x2 ) = (1 + x2 )2 (1 + x2 )2 ii. Find all critical points: f 0 (x) = 0 : f 0 (x) = 2(1 − x2 ) =0 (1 + x2 )2 ⇔ x = −1 and x = 1 f 0 (x) is not defined: f 0 (x) is defined for all real x. iii. Increasing and Decreasing Intervals: x x< −1 <x< 1 <x 0 f (x) − 0 + 0 − f (x) & loc. min. % loc. max. & iv. Evaluate f (x) at each critical point: local minimum at (−1, 0) and local maximum at (1, 2) (d) Determine Concavity and Points of Inflection: 00 i. Find f (x) 00 f (x) = (1 + x2 )2 · 2(−2x) − 2(1 − x2 )[2(1 + x2 ) · 2x] 4x(x2 − 3) = (1 + x2 )4 (1 + x2 )3 c 2012 Department of Mathematics, Eastern Mediterranean University 3 Tutorial 6 - April 04, 2014 ii. Find all potential points of inflection: 00 f (x) = 0 : 00 f (x) = 00 4x(x2 − 3) =0 (1 + x2 )3 √ ⇔ x = 0, x = ± 3 00 f (x) is not defined: f (x) is defined for all real x. iii. Intervals of Concavity: √ x x< − 3 <x< 0 <x< 00 f (x) − 0 + 0 − f (x) ∩ infl. ∪ infl. ∩ point point √ 3 <x 0 + infl. ∪ point √ √ iv. The points of inflection are : (− 3, 0.1340) , (0, 1) and ( 3, 1.8660) (e) Asymptotes: Horizontal Asymptote: x2 1 + x2 + x12 x2 + 2x + 1 (x + 1)2 =1 = lim = lim lim x→±∞ x→±∞ x→±∞ 1 + x2 1 + x2 x2 1 + x12 So we have a horizontal asymptote at y = 1. Vertical Asymptote: None Oblique Asymptote: None (f) Sketch the graph: c 2012 Department of Mathematics, Eastern Mediterranean University 4 Tutorial 6 - April 04, 2014 2 1 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 -1 (g) The range of f (x) is [0, 2]. Questions related to SECTION 4.5 1. (a) Write the equation of the line that represents the linear approximation to the following functions at the given point a. (b) Use the linear approximation to estimate the given function value. (c) Compute the percent error in your approximation by the formula: 100 · i. f (x) = 12 − x2 ; a = 2; |approx − exact| |exact| f (2.1) (a) f (a) = f (2) = 8 and f 0 (a) = −2a = −4 so the linear approximation has the equation: y = L(x) = f (a) + f 0 (a)(x − a) = 8 − 4(x − 2) = −4x + 16 (b) f (2.1) ≈ L(2.1) = 7.6 (c) The percent error is : 100 · |7.6 − 7.59| ≈ 0.13% |7.59| c 2012 Department of Mathematics, Eastern Mediterranean University 5 Tutorial 6 - April 04, 2014 ii. f (x) = sin x; a = π/4; f (0.75) √ (a) f (a) = f (π/4) = sin(π/4) = 2 2 and f 0 (a) = cos a = cos(π/4) = √ 2 2 so the linear approximation has the equation: π π π y = L(x) = f (a) + f 0 (a)(x − a) = sin + cos x− = 4 4 4 √ √ √ π π 2 2 2 = + x− = x+1− 2 2 4 2 4 (b) f (0.75) ≈ L(0.75) ≈ 0.68 (c) The percent error is : iii. f (x) = ln (1 + x); 100 · a = 0; |0.68 − sin 0.75| ≈ 0.064% | sin 0.75| f (0.9) (a) f (a) = f (0) = ln 1 = 0 and f 0 (a) = 1 1 = =1 1+a 1+0 so the linear approximation has the equation: y = L(x) = f (a) + f 0 (a)(x − a) = 0 + 1(x − 0) = x (b) f (0.9) ≈ L(0.9) = 0.9 (c) The percent error is : iv. f (x) = ex ; a = 0; 100 · |0.9 − ln 1.9| ≈ 40% | ln 1.9| f (0.05) (a) f (a) = f (0) = e0 = 1 and f 0 (a) = ea = e0 = 1 so the linear approximation has the equation: y = L(x) = f (a) + f 0 (a)(x − a) = 1 + 1(x − 0) = 1 + x (b) f (0.05) ≈ L(0.05) = 1.05 (c) The percent error is : 100 · |1.05 − e0.05 | ≈ 0.12% |e0.05 | c 2012 Department of Mathematics, Eastern Mediterranean University 6 Tutorial 6 - April 04, 2014 2. Use linear approximations to estimate the following quantities. Choose a value of ’a’ to produce a small error. √ (a) 146 Let f (x) = f 0 (x) = √ x 1 √ 2 x and ⇒ a = 144. Then f 0 (a) = √1 2 144 = f (a) = √ 144 = 12 1 24 So the linear approximation near a = 144 is L(x) = f (a) + f 0 (a)(x − a) = 12 + 1 (x 24 − 144) Therefore √ (b) √ 3 146 = f (146) ≈ L(146) = 12 + 1 1 (146 − 144) = 12 24 12 65 1 and Let f (x) = x 3 f 0 (x) = 13 x −2 3 f 0 (a) = 13 64 ⇒ 1 a = 64. Then −2 3 f (a) = 64 3 = 4 1 48 = So the linear approximation near a = 64 is L(x) = f (a) + f 0 (a)(x − a) = 4 + 1 (x 48 − 64) Therefore √ 3 65 = f (65) ≈ L(65) = 4 + 1 1 (65 − 64) = 4 48 48 (c) e0.06 Let f (x) = ex f 0 (x) = ex ⇒ and a = 0. Then f (a) = e0 = 1 f 0 (a) = e0 = 1 So the linear approximation near a = 0 is L(x) = f (a) + f 0 (a)(x − a) = 1 + 1(x − 0) Therefore e0.06 = f (0.06) ≈ L(0.06) = 1 + 1(0.06 − 0) = 1.06 c 2012 Department of Mathematics, Eastern Mediterranean University 7 Tutorial 6 - April 04, 2014 (d) cos 31◦ Let f (x) = cos x f 0 (x) = − sin x a = π6 (= 30). Then f (a) = and ⇒ f 0 (a) = − sin 30 = √ 3 2 −1 2 So the linear approximation near a = 30◦ is : L(x) = f (a) + f 0 (a)(x − a) = √ 3 2 − 12 (x − π6 ) Therefore ◦ cos 31 = cos 31π 180 =f 31π 180 ≈L 31π 180 √ 3 1 = − 2 2 31π π − 180 6 ≈ 0.857 Note that we must convert 31◦ to radians before applying the linear approximation formula. c 2012 Department of Mathematics, Eastern Mediterranean University 8