\f\f\f\f\f\f\fSolutions to questions on 1.pdf file 8. Given, g ( x) x3 3x 2 9 x 7 For , g '( x ) 3 x 2 6 x 9 0 x {1,3} Now, g ''( x ) 6 x 6 Testing ; g ''(1) 6(1) 6 12 0 Thus, point (-1, 12) is a local maximum Similarly, g ''(3) 6(3) 6 12 0 Thus, point (3, -20) is a local minimum 10. Given, f ( x) x ln( x) For , 1 f '( x) x ln( x) 0 x f '( x) 1 ln( x) 0 x 1 e Now, 1 f ''( x) x Testing ; f ''( 1 ) e 2.7183 0 e Thus, point (1/e, -e) is a local minimum Page 1 of 10 Considering the profile of the above graph, and appreciating that at as point of inflection the slope after the local extremum is same as that before; the points of inflection in the above graph are points b, d and h. Solutions to questions on 2.pdf file 16. Given, f ( x) 7 x 2 5 x 3 Now, f '( x) 14 x 5 Thus, f ''( x) 14 18. Given, f ( x) (6 x x 2 )10 By chain rule of differentiation: f '( x) 10(6 x x 2 )9 (6 2 x) 20(6 x x 2 )9 (3 x) 20 x 9 ( x 6) 9 ( x 3) Applying chain rule of differentiation and the product rule: Page 2 of 10 f ''( x ) 20{(3 x )(9)(6 x x 2 )8 (6 2 x) (6 x x 2 )9 (1)} f ''( x ) 20{18(3 x ) 2 (6 x x 2 )8 (6 x x 2 )9 } f ''( x ) 20(6 x x 2 )8{18(3 x) 2 (6 x x 2 )} 20(6 x x 2 )8{18(9 6 x x 2 ) 6 x x 2 } f ''( x ) 20(6 x x 2 )8 (162 114 x 19 x 2 ) 20 x 8 (6 x)8 (19 x 2 114 x 162) Thus, f ''( x ) 20 x 8 ( x 6)8 (19 x 2 114 x 162) 20. Given, f ( x ) (2 x3 3)6 By chain rule of differentiation: f '( x) 6(2 x 3 3)5 (6 x 2 ) 36 x 2 (2 x3 3)5 Applying chain rule of differentiation and the product rule: f ''( x ) 36{(2 x 3 3)5 (2 x) x 2 (5)(2 x 3 3) 4 (6 x 2 )} f ''( x ) 36(2 x 3 3) 4{(2 x)(2 x 3 3) 30 x 4 } f ''( x ) 36(2 x 3 3) 4{4 x 4 6 x 30 x 4 } 36(2 x 3 3) 4 (34 x 4 6 x) Thus, f ''( x ) 72 x (2 x 3 3) 4 (17 x 3 3) 22. Given, f ( x ) ln( x 2 4) By chain rule of differentiation: f '( x ) 2x x 4 2 Applying quotient rule of differentiation: Page 3 of 10 ( x 2 4)(2) (2 x)(2 x) 2 x 2 8 4 x 2 8 2 x2 2( x 2 4) f ''( x) 2 2 ( x 2 4) 2 ( x 2 4)2 ( x 4)2 ( x 4) 2 Thus, f ''( x) 2( x 2 4) ( x 2 4) 2 Solutions to questions on 3.pdf file 8. Given, f ( x) 2 x 3 96 x 42 For , f '( x) 6 x 2 96 0 x 2 16 0 Thus, x 4 x {4, 4} Now, f ''( x) 12 x Thus, f ''(4) 48 0 Similarly , f ''(4) 48 0 Hence, the points for which x = -4 and x = 4 are the critical points and are the local maximum and minimum of the curve respectively 10. Page 4 of 10 Given, f ( x) 2 x3 ; x For , f '( x) 3x 2 0 x2 0 Thus, x1,2 0 x0 Now, f ''( x) 6 x Thus, f ''(0) 0 Hence, the function has ONLY one critical point for which x = 0 which coincidentally is also the global extreme. 12. Given, f ( x) x e x ; x For , f '( x) 1 e x 0 ex 1 x ln(1) 0 Thus, x0 Now, f ''( x) e x Thus, f ''(0) 1 0 Hence, the function has ONLY one critical point (maximum) for which x = 0 which coincidentally is also the global extreme. Solutions to questions on 4.pdf file 18. Page 5 of 10 Defining the area, A(x); x A( x) f (t )dt 0 But , f '(t ) (t 2.5)(t 5)(t 7.5) 0 Thus, 1 f (t ) (t 2.5)(t 5)(t 7.5) dt (2t 3 40t 2 275t 750) C 8 But , 750 f (0) C 0 8 375 C 4 Hence, 1 f (t ) (2t 3 40t 2 275t 750 750) 8 1 f (t ) (2t 3 40t 2 275t ) 8 x A( x) 0 x 1 1 40 275 2 f (t ) dt t 4 t 3 t 8 2 3 2 0 Hence, 1 1 40 3 275 2 374 A( x) ( x 4 x x ) 8 2 3 2 3 a) For minimum A(x): Page 6 of 10 1 A '( x) x(2 x 2 40 x 275) 0 8 2 x(2 x 40 x 275) 0 x1 0 x2 x3 10 Now, 1 A ''( x) (6 x 2 80 x 275) 8 Thus, 275 A ''(0) 0 8 75 A ''(10) 0 8 Therefore, Both values of x give a minimum A(x) b) For maximum A(x) 1 A '( x) x(2 x 2 40 x 275) 0 8 2 x(2 x 40 x 275) 0 x1 0 x2 x3 10 This yields similar results as above. Therefore, NO value of x within the domain gives maximum A(x) a) Critical points: Page 7 of 10 f '( x) 0 Thus, x {2, 7} b) Local extrema: Given, y ' f ''( x) x( x 5)( x 10) 0 Thus, x {0,5,10} y ''( x) 3 x 2 30 x 50 Now, y ''(0) 50 0 y ''(5) 25 0 y ''(10) 50 0 Hence, local maximum at x = 5 c) Local minima at x = 0 and x = 10 22. Given, f ( x ) 2 x3 15 x 2 6 For , f '( x) 6 x 2 30 x 0 x( x 5) 0 x {0,5} Now, f ''( x) 12 x 30 Testing ; f ''(0) 30 0 f ''(5) 30 0 Thus, point (0, 6) is a local maximum Similarly, the, point (5, -119) is a local minimum 24. Page 8 of 10 Given, h( x ) x 4 8 x 2 2 For , h '( x ) 4 x3 16 x 0 x {2, 0, 2} Now, h ''( x ) 12 x 2 16 Testing ; h ''(2) 32 0 h ''(0) 16 0 f ''(2) 32 0 Thus, point (0, -2) is a local maximum Similarly, the, points (-2, -18) and (2, -18) are local minima Solutions to questions on 5.pdf file 10. Maximum point at x = -1 and minimum point at x = 3 12. Page 9 of 10 Maximum point at x = 0 and minimum points at x = -2 and x = 2 Solutions to questions on 6.pdf file Solutions to questions on 7.pdf file Page 10 of 10