Question

I need these questions all solved please.

(11) Let f(x) x3 - 1 and g(x) = 25 sin 3r (a) What is the range of g(f(x))? Justify your answer (b) If h(r) f(g(x)) compute h'(x) (c) If t(x) ef(r) compute t'(x). (d) Obtain f'() from first principles 6 marks 6 marks 6 marks 6 marks 3. (e) Obtain the equation of the line tangent to the curve y [6 marks f(x) at (12) In this question, f(x)= r+2)2 2r (a) Write down the domain, D, off and identify all horizontal and vertical asymp totes of the graph of y f(x) (b) Calculate the derivative f'(x) for in D, and write down the sets where f(x) is increasing and decreasing. 6 marks 6 marks (4 marks (c) Determine all the critical points of f(x) (d) Determine whether the points in part (c) above are local maxima, local minima points of inflection of the graph y f(x) (e) Sketch the graph of y f(x) [4 marks 10 marks or

Answer 1

f(x) = 431 and gen) = 2+5 sin zu (a) gifcm) = g(121) = 2+5 sin 3(421) an sin (3 (131) € 61,) je -5 <5sin (3421) 45 => 2-5 < 2+ 5 sin (31021)) < 5+2 => -3 { 2+5 sin (35(n1) 47. That is the range at gl fews) is [3,4] h() = f( gcm) = f (2+5 sin 3a) = (2+5sin 3x)21 = 8+125 sin? 31 +6031N3X+150 in 3x - 1 45 sim 4k +150 sinh 314 6 osin) 32% +. cih (n) = 3*125 sin32, (3 003 34) + 150.2 sin 34.1360532) +180105 32 a 1125 Sin 34 2033k + 900 sin 3420534 +1 so cos zu ☺ tew)= efen) is t'(W) = a (ess) - es sehr (n2-1) = 3nVeri f'(v) = lim feveth) --2cm) hoyon n70 = lim (n+)1 - 131 ๆ ใm) หวิห%, aาน 43kห้พ9 и h?ь = lim (67+3hx +32V) . hyo = 347

© =26 y = f(x) = 431. Then dy = f'()\n=3 =3"\u=3=27. Then the equation at the tangent at 1=3 amd Y=(3) 1 is 4-26) = f'(m) (K-3) => Y-26=27 (-3) => | 274Y = 55 12 © fen) = CH2)2 is defined for all her except at 15-2. The function f is defined. Then the domain ab tis R12-2}=(-4,-2) 01-2.0). vertical Asymptoteo Horizontal asymptotes y=fron) => y = (2+2) => gry + 400+ 40 = 27. => wy +147-2)x+ 4y=0. height of powers ab re five 4?) is Since the wefficient ab a horizontial asymptote, y. Then the line y=0 is therefore is (nr44x+4). heighest powers at y And the coefficient ab is a vertieul asymptote. ie (4+2)". Then the line 2=-2 F Note: The vertical asymptotes are obtained by equating to o heighest power ab a and the horizontal The coefficent af asymptote are obtained by equating to o the wefficient ab haighest powers" at ry

6 Here D= PRY2} and fen) = film = f'(m) = (+2) tri (2n) – au (2+2)V Tut2)4 5 2(+2) Y-24-24x+2) (1+2)" < 2x+4-WM her that is Tu +2)3 f' (n) is in oneoning at Thees points where from 70. and decreasing att Thoes points where then) 20. Now, f "() = -2 (2+ 2)2-(4-21)-3(x+2) (+2)6 - 20-4-1976 (4+2)4 Now, f"() 70 =1) arya hino => 1734 That is f' is increasing for all her with ny, 4 be on [4,0) and fren) so => GM 420= nga That is f' is dearening for all herst2} with x 24 i-e (-672)06241 obtained G Pritical points are advina poms Olesdoing by => 420 => 4-2020 => R=2 them y=& tejr is Bit) * The critical point at the function Rin) = at x=2, f^cm) = county a= 2 = Sun - - 32 <0. Hence at 422 The femetion of having local maxima. For finding. The point of inflection, we make f'(x)=0 lie autor sus = f(n) =0 => x=4 :. ya frollway - to - :: point at inflectton at the graph

© since the function y=f(x)= 21 22 23 hom vertical asymptote n=-2 and horizontal asymptote y=0. and me FO- 2 1 3 4 Horizontal asymptotes. KE-2 vertical asymptote Graph: y=fru) = (2012)