The value of $c$ in $(0,2)$ satisfying the Mean Value Theorem for the function $f(x)=x(x-1{)}^2,x\in [0,2]$ is equal to:

According to Lagrange's Mean Value Theorem, there exists $c\in (a,b)$ such that:
$$f^{\prime }(c)=\frac{f(b)-f(a)}{b-a}$$
Here $a=0,b=2$.
$$f(x)=x(x^2-2x+1)=x^3-2x^2+x$$
$$f(0)=0$$
$$f(2)=2(2-1{)}^2=2$$
$$f^{\prime }(x)=3x^2-4x+1$$
Substitute into the formula:
$$3c^2-4c+1=\frac{2-0}{2-0}=1$$
$$3c^2-4c=0$$
$$c(3c-4)=0$$
$$c=0\text{ or }c=4/3$$
Since $c$ must be in the open interval $(0,2)$, we choose $c=4/3$.