Question

If $0\le x\le 1$ and $f\left(x\right)=\left|\begin{array}{ccc}x& 1& 1\\ -1& x& 1\\ -1& -1& x\end{array}\right|$, then $f\left(x\right)$ has

Answer 1

$f\left(x\right)=\left|\begin{array}{ccc}x& 1& 1\\ -1& x& 1\\ -1& -1& x\end{array}\right|=x^3+3x$
And $f\left(x\right)$ is increasing function in $0\le x\le 1$
$f^{\prime }\left(x\right)=3x^2+3=3(x^2+1)$
Critical points are $\pm 1$
$f\left(1\right)=1^3+3.1=4$
$f(-1)={(-1)}^3+3.(-1)=-4$
A) Least values $f\left(x\right)$ is at $x=0$ that is $0$
B) Greatest value of $f\left(x\right)$ is at $x=1$ that is $4$
C) For $x=1$ $f\left(x\right)$ is maximum
D) For $x=-1f\left(x\right)$ is minimum