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a) Consider a function f(x)= 1 X-4 +2. i) Simplify f(x)in the form of 0431-9(x)=0. (1 mark) ii) Find the vertical asymptote and horizontal asymptote and its analysis. (3 marks) iii) Find the interval(s) where f(x) is decreasing. (3 marks) iv) Find the interval(s) where f(x) is concave up or concave down. (3 marks)

Answer 1

Solution :- @is fG) +2 a fal= (21-4) (*-- 1 + 2 (7-4) 142a08 2007 (2-4) a-4 (i) The vertical asymptote will occur at those where denominator of f(n) ㅕ hou=0 N= =4 So there is one vertical asymptote 1. Now, lim fon) = lim 20-7 lim values of n Na 2-0 (2017) - Nylo horo no lim fenl=2 2 is asymptote on (-) (-2) 오 da The Horizontal wi) f'(n) f'(n) = -1 co EXER-{4} (n=4² a f is decreasing (-0,4) U (40) riv f" (n) 2 1-42 (1-4)? (1-4)3 In nou, (n-4) >0 too f''(n) > 0 in this case afis concane upward on (4,0) Similarly, for nau (x-4) 20 a (x-4) CO = (auco f" (nico f is concave down sign of f(n) he the + 4 on 60,4)