Diketahui fungsi bilangan real f(x)=x/(1−x), untuk x≠1 Tentukan nilai dari f(2.021)+f(2.020)+…+f(3)+f(2)+f(1/2)+f(1/3)+…+f(1/2.020)+f(1/2.021)

Halo Fania K :) Jawaban: -2.020 perhatikan konsep berikut: f(x) = ax + b x = c maka: f(c) = ac + b Diketahui fungsi bilangan real f(x)=x/(1−x), untuk x≠1 perhatikan pola berikut: f(2) = 2/(1-2) = 2/(-1) = -2 f(1/2) = (1/2)/(1-(1/2)) =(1/2)/(1/2) = 1 f(2) + f(1/2) = -2 + 1 = -1 f(3) = 3/(1-3) = 3/(-2) = -3/2 f(1/3) = (1/3)/(1-(1/3)) =(1/3)/(2/3) = 1/2 f(3) + f(1/3) = -3/2 + 1/2 = -2/2 = -1 f(4) = 4/(1-4) = 4/(-3) = -4/3 f(1/4) = (1/4)/(1-(1/4)) =(1/4)/(3/4) = 1/3 f(4) + f(1/4) = -4/3 + 1/3 = -3/3 = -1 Karena: f(2) + f(1/2) = -1 f(3) + f(1/3) = -1 f(4) + f(1/4) = -1 Maka membentuk pola yakni sebagai berikut: f(n) + f(1/n) = -1 sehingga nilai dari f(2.021)+f(2.020)+…+f(3)+f(2)+f(1/2)+f(1/3)+…+f(1/2.020)+f(1/2.021) adalah.... banyak n = 2021 - 1 = 2020 karena n=1 atau x =1 tidak memenuhi n = 2.020 f(2.021)+f(2.020)+…+f(3)+f(2)+f(1/2)+f(1/3)+…+f(1/2.020)+f(1/2.021) = [-1(n)] = [-1(2020)] = -2.020 Jadi, nilai dari f(2.021)+f(2.020)+…+f(3)+f(2)+f(1/2)+f(1/3)+…+f(1/2.020)+f(1/2.021) adalah -2.020