Question

Fie functia f : R → R, f(x)= (2m-1)x+m-2. Scrieti legea de corespondenta in fiecare din cazurile :
a) (-5;2)∈ Gf;
b) (8; m-1)∈ Gf;
c) (m-1; m²=2)∈ Gf , m∈ Z.

Va rog ajutati ma si pe mine ! multumesc mult !

Answer 1

a. x= -5, f(x)=2
f(x)=(2m-1)* (-5)+m-2= 2
-10m+5+m-2=2
-9m+3=2
9m=3-2
9m=1
m=1/9

rezulta f(x)= ( 2*1/9 - 1)*x+m-2

f(x)= (2-9)/9*x+1/9-2
f(x)= -7/9*x+(1-18)/9
f(x)= -7/9*x-17/9

b. x=8 , f(x)=m-1
f(x)= (2m-1)*8+m-2=m-1
f(x)=16m-8+m-2=m-1
f(x)=17m-10=m-1
17m-m=10-1
16m=9
m=9/16
inlocuiesc in f(x)
f(x)=(2*9/16-1)*x+(9/16-2)
f(x)=(18-16)/16*x+(9-32)/16
f(x)=2/16*x-23/16
f(x)=1/8*x-23/16

c.  x=m-1, f(x)=m^2+2
f(x)=(2m-1)(m-1)+m-2=m^2+2
  f(x)=2m^2-2m-m+1+m-2=m^2+2
         2m^2-2m-1=m^2+2
         2m^2-m^2-2m-3=0
          m^2+m-3m-3=0
         m(m+1)-3(m+1)=0
         (m+1)(m-3)=0
rezulta ca sunt 2 solutii m= -1 si m=3
f(x)=[2*(-1)-1] *x-1-2
f(x)=-3x-3

si 
f(x)=(2*3-1)*x+3-2
f(x)=5x+1

Answer 2

a) f(-5)=2
(2m-1) * (-5)+m-2=2
-10m+5+m=4
-9m=-1
9m=1
m=1/9

f(x)=(2/9-1)x+1/9-2
f(x) =(-7/9)x-17/9

verificare
f(-5)=35/9-17/9=18/9=2 adevarat


b) f(8) =m-1
(2m-1) *8 +m-2=m-1
(2m-1)*8=1
2m-1=1/8
2m=1+1/8
2m=9/8
m=9/16


f(x) =(9/8-1)*x+9/16-2
f(x)=(1/8) *x-23/16
verificare
f(8)=1+m-2=m-1 adevarat, nu mai are sens sa inlocuiesc valoarea lui m



c) prea complicat calculul ..nu vad in fereastra, prefer sa il fac pe foaie