f.4 amathz...

f(x)=2x^2+6x+8
=2(x^2+3x+(3/2)^2-(3/2)^2)+8
=2(x+(3/2))^2 -(9/2) +8
=2(x+(3/2))^2 +(7/2)
the minimum value of 2(x+(3/2))^2 can only be 0
so the minimum value of 2x^2+6x+8 is 7/2
so f(x) can't be negative.
and the maximum value of 1/(2x^2+6x+8) is 2/7