The function $f(x)=\begin{vmatrix}x^2& x\\3 & 1\end{vmatrix}, x \in IR$ has a:

local minimum at $x=\frac{3}{2}$

The correct answer is Option (2) → local minimum at $x=\frac{3}{2}$

$f(x)=\begin{vmatrix}x^2& x\\3 & 1\end{vmatrix}=x^2-3x$

$f'(x)=2x-3⇒x=\frac{3}{2}$ at $f'(x)=0$

$f''(x)=2>0$ so $x=\frac{3}{2}$ is point of local maximum