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f:R->,f(x)=mx-3m-5 f(m)=-1

Răspuns :

f:R->,f(x)=mx-3m-5
f(m)=-1
m×m-3m-5=-1
m^2-3m=-1+5
m^2 -3m=4
m^2-3m-4=0
m^2+m-4m-4=0
m(m+1)-4(m+1)=0
(m+1)(m-4)=0
m+1=0
m=0-1
m=1

m-4=0
m=0+4
m=4
M(-1;-1)
M(4;-1)
f(m)=m²-3m-5=-1
m²-3m-4=0
rezolvand ecuatia (cu Δ sau descompunad in m²-4m+m-4) obtinem
m1=4
m2=-1
deci f(x) poate fi
4x-12-5=4x-17 si intr-adevar f(4)=-1
sau
-x+3-5=-x-2 si intr-adevar f(-1)=-1