Question

aflati m€R stiind ca valoarea minima a functiei f:R->R , f(x)=x^2-mx+m-1 este egal cu -1/4

Answer 1

f min=-Δ/4a

Δ=m²-4m+4

f min=-(m²-4m+4)/4<=>-1/4=-(m²-4m+4)/4<=>1=m²-4m+4=>m²-4m+3=0

Δ'=16-12=4

m₁=(4+2)/2=3

m₂=(4-2)/2=1

m∈{1;3}