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Question Number 120049 by bramlexs22 last updated on 29/Oct/20
f(x+2)+f(x−2)=f(x) f(1)=1 ,f(2)=2,f(3)=3,f(4)=4 then f(100)=?
f(x+2)+f(x−2)=f(x)f(1)=1,f(2)=2,f(3)=3,f(4)=4thenf(100)=?
Answered by bemath last updated on 29/Oct/20
⇒f(x−2)−f(x)+f(x+2)=0 replace x by x+2 ⇒f(x)−f(x+2)+f(x+4)=0 (i) replace again x by x+2⇒f(x+2)−f(x+4)+f(x+6)=0 (ii) adding(i) and (ii) ⇒f(x)+f(x+6) = 0 (iii) replace x by x+6 ⇒f(x+6)+f(x+12)=0 (iv) substract (iii) by (iv) ⇒ f(x+12)=f(x) it does meant f(x) is periodic function with period 12. ⇒f(100)=f(8.12+4)=f(4)=4
⇒f(x−2)−f(x)+f(x+2)=0replacexbyx+2⇒f(x)−f(x+2)+f(x+4)=0(i)replaceagainxbyx+2⇒f(x+2)−f(x+4)+f(x+6)=0(ii)adding(i)and(ii)⇒f(x)+f(x+6)=0(iii)replacexbyx+6⇒f(x+6)+f(x+12)=0(iv)substract(iii)by(iv)⇒f(x+12)=f(x)itdoesmeantf(x)isperiodicfunctionwithperiod12.⇒f(100)=f(8.12+4)=f(4)=4
Answered by Ar Brandon last updated on 29/Oct/20
f(x+2)+f(x−2)=f(x)...eqn(1) Replacing x with x−2 we have f(x)+f(x−4)=f(x−2)...eqn(2) eqn(1)+eqn(2) ⇒f(x+2)=−f(x−4) ⇒f(x)=−f(x−6){replacing x=x−2} ⇒f(x+6)=−f(x)=f(x−6) ⇒f(x)=f(x−12) Period is 12⇒f(100)=f[4+12(8)] ⇒f(100)=f(4)=4
f(x+2)+f(x−2)=f(x)…eqn(1)Replacingxwithx−2wehavef(x)+f(x−4)=f(x−2)…eqn(2)eqn(1)+eqn(2)⇒f(x+2)=−f(x−4)⇒f(x)=−f(x−6){replacingx=x−2}⇒f(x+6)=−f(x)=f(x−6)⇒f(x)=f(x−12)Periodis12⇒f(100)=f[4+12(8)]⇒f(100)=f(4)=4

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