Question

Se considera functia f:R->R, f(x)=mx²-(m+3)x+4. Sa se determine m ∈ R pentru care f(x)>0,∀ x ∈ R

Answer 1

$f(x)=mx^2-(m+3)x+4$

[tex]mx^2-(m+3)x+4=0 a=m b=-(m+3) c=4 delta=b^2-4ac= (-(m+3))^2-4*m*4 = -m^2-6m-9+16m= =-m^2+10m-9 [/tex]

Delta=0

[tex]-m^2+10m-9=0 /(-1) m^2-10m+9=0 a=1 b=-10 c=9 delta=b^2-4ac= (-10)^2-4*1*9=100-36=64 [/tex]

$m1= \frac{10+8}{2}= \frac{18}{2}=9$

$m2= \frac{10-8}{2}= \frac{2}{2}=1$