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