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ProMaster
a fost răspuns

AJUTOR VA ROG...

determinati legea de corespondenta a functiei : f:R->R stiind ca :
a) f(x-1)=x , oricare ar fi x E R;
b) f(2x-1)=4x , oricare ar fi x E R ;
c)f(2x-3)=1 , oricare ar fi x E R;
d) f(2x+3)=3x-2 , oricare ar fi x E R ;

Va rog , dau coroana !

Răspuns :

Matematician1234
a) -notez x-1=y⇒f(x-1)=f(y)=y+1 ⇔ f(x)=x+1
b) -notez 2x-1=y⇒f(2x-1)=f(y)=2(y+1) ⇔ f(x)=2x+2
c) -notez 2x-3=y⇒f(2x-3)=f(y)=1 ⇔ f(x)=1
d)-notez 2x+3=y⇒f(2x+3)=f(y)= (y-3)/2 - 2 = (y-3-4)/2= (y-13) / 2⇔ f(x)= (x-13)/2

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