Answer:
[tex]\large\boxed{g\left(\dfrac{2}{3}\right)=-\dfrac{2}{9}}[/tex]
Step-by-step explanation:
[tex]g(x)=x^2-x=x(x-1)\\\\g\left(\dfrac{2}{3}\right)\to\text{put }\ x=\dfrac{2}{3}\ \text{to the eqation of the function:}\\\\g\left(\dfrac{2}{3}\right)=\left(\dfrac{2}{3}\right)\left(\dfrac{2}{3}-1\right)=\left(\dfrac{2}{3}\right)\left(\dfrac{2}{3}-\dfrac{3}{3}\right)=\left(\dfrac{2}{3}\right)\left(-\dfrac{1}{3}\right)=-\dfrac{(2)(1)}{(3)(3)}=-\dfrac{2}{9}[/tex]
