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Question Number 12889 by sin (x) last updated on 05/May/17

f(x−1)=(2/3)+f(x) f(0)=36−f(21) ⇒f(0)=?

$${f}\left({x}−\mathrm{1}\right)=\frac{\mathrm{2}}{\mathrm{3}}+{f}\left({x}\right) \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{36}−{f}\left(\mathrm{21}\right) \\ $$$$\Rightarrow{f}\left(\mathrm{0}\right)=? \\ $$

Commented by prakash jain last updated on 06/May/17

f(x)−f(x−1)=−(2/3) f(0)=36−f(21) ....(1) f(1)−f(0)=−(2/3) .. f(21)−f(20)=−(2/3) ....(22) adding (1) to (22) f(21)=−(2/3)×21+36−f(21) 2f(21)=22 f(21)=11 substituting in (1) f(0)=36−11 f(0)=25

$${f}\left({x}\right)−{f}\left({x}−\mathrm{1}\right)=−\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{36}−{f}\left(\mathrm{21}\right)\:\:\:\:\:....\left(\mathrm{1}\right) \\ $$$${f}\left(\mathrm{1}\right)−{f}\left(\mathrm{0}\right)=−\frac{\mathrm{2}}{\mathrm{3}}\:\:\:\:\:\:\: \\ $$$$.. \\ $$$${f}\left(\mathrm{21}\right)−{f}\left(\mathrm{20}\right)=−\frac{\mathrm{2}}{\mathrm{3}}\:\:\:\:....\left(\mathrm{22}\right) \\ $$$${adding}\:\left(\mathrm{1}\right)\:{to}\:\left(\mathrm{22}\right) \\ $$$${f}\left(\mathrm{21}\right)=−\frac{\mathrm{2}}{\mathrm{3}}×\mathrm{21}+\mathrm{36}−{f}\left(\mathrm{21}\right) \\ $$$$\mathrm{2}{f}\left(\mathrm{21}\right)=\mathrm{22} \\ $$$${f}\left(\mathrm{21}\right)=\mathrm{11} \\ $$$${substituting}\:{in}\:\left(\mathrm{1}\right) \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{36}−\mathrm{11} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{25} \\ $$

Answered by prakash jain last updated on 06/May/17

f(0)=25

$${f}\left(\mathrm{0}\right)=\mathrm{25} \\ $$

Answered by Joel577 last updated on 07/May/17

f(x) − f(x − 1) = −(2/3) Let f(x) = ax + b f(x − 1) = a(x − 1) + b = ax − a + b (ax + b) − ax + a − b = −(2/3) a = −(2/3) f(21) + f(0) = 36 (−(2/3) . 21 + b) + (−(2/3) . 0 + b) = 36 −14 + 2b = 36 b = ((36 + 14)/2) = 25 f(0) = −(2/3) . 0 + 25 = 25

$${f}\left({x}\right)\:−\:{f}\left({x}\:−\:\mathrm{1}\right)\:=\:−\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\mathrm{Let}\:{f}\left({x}\right)\:=\:{ax}\:+\:{b} \\ $$$$\:\:\:\:\:\:\:\:{f}\left({x}\:−\:\mathrm{1}\right)\:=\:{a}\left({x}\:−\:\mathrm{1}\right)\:+\:{b}\:=\:{ax}\:−\:{a}\:+\:{b}\: \\ $$$$\left({ax}\:+\:{b}\right)\:−\:{ax}\:+\:{a}\:−\:{b}\:=\:−\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${a}\:=\:−\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${f}\left(\mathrm{21}\right)\:+\:{f}\left(\mathrm{0}\right)\:=\:\mathrm{36} \\ $$$$\left(−\frac{\mathrm{2}}{\mathrm{3}}\:.\:\mathrm{21}\:+\:{b}\right)\:+\:\left(−\frac{\mathrm{2}}{\mathrm{3}}\:.\:\mathrm{0}\:+\:{b}\right)\:=\:\mathrm{36} \\ $$$$−\mathrm{14}\:+\:\mathrm{2}{b}\:=\:\mathrm{36} \\ $$$${b}\:=\:\frac{\mathrm{36}\:+\:\mathrm{14}}{\mathrm{2}}\:=\:\mathrm{25} \\ $$$${f}\left(\mathrm{0}\right)\:=\:−\frac{\mathrm{2}}{\mathrm{3}}\:.\:\mathrm{0}\:+\:\mathrm{25}\:=\:\mathrm{25} \\ $$

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