The function $f(x)=x^2-4x$, $x\in [0,4]$ attains minimum value at:

Step 1: Find the first derivative $f^{\prime }(x)$.
$$f(x)=x^2-4x⟹f^{\prime }(x)=2x-4$$
Step 2: Set $f^{\prime }(x)=0$ to find critical points.
$$2x-4=0⟹x=2$$
Step 3: Find the second derivative $f^{\prime \prime }(x)$.
$$f^{\prime \prime }(x)=2$$
Since $f^{\prime \prime }(2)=2>0$, the function has a local minimum at $x=2$.Step 4: Check boundaries $x=0$ and $x=4$.
$$f(0)=0^2-4(0)=0$$
$$f(4)=4^2-4(4)=0$$
$$f(2)=2^2-4(2)=4-8=-4$$
The absolute minimum is $-4$ at $x=2$.