\(f\left(1\right)=3\cdot1+2=5\)
\(f\left(2\right)=3\cdot2+2=7\)
\(f\left(0\right)=3\cdot0+2=2\)
\(b,f\left(x_1\right)< f\left(x_2\right)\) do hs đồng biến
a). Thay f(1) vào f(x), ta có:
y=f(1)= 3 .1 + 2 = 5
y=f(2)= 3 .2 + 2 = 8
y=f(0)= 3 .0 + 2 = 2