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Question Number 11899 by ahmet last updated on 04/Apr/17

f′(x)={_(3     ;x>2) ^(2x  ; x≤2)   f(2)=1 ise f(1)+f(3)=?  czm∵ f(x)={_(3x+c_2  ; x>2) ^(x^2 +c_1  ;x≤2)   f(2)=x^2 +c_1  dir  1=4+c_1  => c_1 =−3  1=2.3+c_2  => c_2 =−5  f(x)={_(3x−5    x>2) ^(x^2 −3   ;x≤2)   f(1)=1^2 −3=−2, f(3)=3.3−5=4  f(1)+f(3)=−2+4=2

$${f}'\left({x}\right)=\left\{_{\mathrm{3}\:\:\:\:\:;{x}>\mathrm{2}} ^{\mathrm{2}{x}\:\:;\:{x}\leqslant\mathrm{2}} \right. \\ $$ $${f}\left(\mathrm{2}\right)=\mathrm{1}\:{ise}\:{f}\left(\mathrm{1}\right)+{f}\left(\mathrm{3}\right)=? \\ $$ $${czm}\because\:{f}\left({x}\right)=\left\{_{\mathrm{3}{x}+{c}_{\mathrm{2}} \:;\:{x}>\mathrm{2}} ^{{x}^{\mathrm{2}} +{c}_{\mathrm{1}} \:;{x}\leqslant\mathrm{2}} \right. \\ $$ $${f}\left(\mathrm{2}\right)={x}^{\mathrm{2}} +{c}_{\mathrm{1}} \:{dir} \\ $$ $$\mathrm{1}=\mathrm{4}+{c}_{\mathrm{1}} \:=>\:{c}_{\mathrm{1}} =−\mathrm{3} \\ $$ $$\mathrm{1}=\mathrm{2}.\mathrm{3}+{c}_{\mathrm{2}} \:=>\:{c}_{\mathrm{2}} =−\mathrm{5} \\ $$ $${f}\left({x}\right)=\left\{_{\mathrm{3}{x}−\mathrm{5}\:\:\:\:{x}>\mathrm{2}} ^{{x}^{\mathrm{2}} −\mathrm{3}\:\:\:;{x}\leqslant\mathrm{2}} \right. \\ $$ $${f}\left(\mathrm{1}\right)=\mathrm{1}^{\mathrm{2}} −\mathrm{3}=−\mathrm{2},\:{f}\left(\mathrm{3}\right)=\mathrm{3}.\mathrm{3}−\mathrm{5}=\mathrm{4} \\ $$ $${f}\left(\mathrm{1}\right)+{f}\left(\mathrm{3}\right)=−\mathrm{2}+\mathrm{4}=\mathrm{2} \\ $$ $$ \\ $$

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