If Rolle's theorem holds for the function ${\text{⨍ (x)=x}}^3{\text{-ax}}^2\text{+bx-4, x∈}\left[1,2\right]$ with ${\text{⨍ }}^{\text{'}}\left(\frac{4}{3}\right)\text{=0}$ then the ordered pair (a, b) is equal to