Question

let f(X)=5x^3-13x^2+11x+4 if the interval of difference in 1 then cosider following statement 1. $\Delta$ ^3f(x)=30 for all x 2. $\Delta$ ^2f(x)=34 for x=1 3 .$\Delta$f(x)= 4 foe x=1 which of the statement  c. if $\Delta$ ^nO^m = $\Delta$ ^nX^m then what is the value of O^6. what is the fun where first difference in 3x^2+7x+10 and assume h=1.

Answer 1

above statemer. let f@= 5x3-132²H1X+4. if the interval of difference in Lither compides the following statement, ® 13 f(1) = 30 for all D 02920 = 34 for 2=1 for x=) (3 D810 = y for 2-1 which of the f(x) = 5-13+11+4 20-13 = 7 azt correct aland 2only x=8 lantzanly fase) = 5(2)° -13(2)+11(2)74 = 14 @ 2 and 3 1, 2, 3 for) 55 = ope 2=3 Aspa) fost) 7 Too 2 14 प کد 02511) 1(21+2h) – 28(317h) + fear) 41-7 = 34.0 F(3) - 8H12) +501) ss-284 ] 3y foto-sfatt-3446177 to festg)= 5(n+3-13(74+39 +)(2+3) ty sn²73221768x+55 85(X) so statement @ is wrong

Caic f(at2) 5(472)3 - 13(2+2)? +1 (442) ty 5013 + 12 +1971 tiy no seati) = 5fantoj - 13/4+1) + 11(21+0+y sa²+ 20²7 and son 573 - 1321² tim74 2 3 21 5 اد 20 SS 32 -5) 68 -57 -15 -42 X-3 is 6 21 EX -5 13 -11 ay ex- 76-46 230

let the required fun. 121 € (2) given: AfC00 3262+75 +10 let 32827721710 = 32P+AS+B = 326421-1+A1217 B TB 2130 B 10 2-1 A + B = 377410 = 20 TA = 10 then sf120 = (0 tlose to by Interation fen2 = 323 + 10 x² +1071 tc a3 tsa²tion to

o - ND= E-1 given som = 3m loco then 36 30/000 (E-1} lase Ef(x) = futh)- £²5(1) = f(a+gh) 3E73E-1270 (x+3h) – 3(3+297 +3(0274). C36 (2+3) 3 (21+2) + 3(1+1)-7/9/2=0 6-3(204+ 300 729-19243 =E3 2 - - = Syo Honda = 1850