Let  $f\left(x\right)=|x^2-3x-4|,-1\le x\le 4$  Then

$x^2-3x-4=(x-4)(x+1)$.

So in  $-1\le x\le 4,x^2-3x-4\le 0$

$\therefore f\left(x\right)=-(x^2-3x-4)(4-x)(x+1)$

$\therefore f\left(x\right)=-(x+1)+4-x=3-2x$

$\therefore \text{in}-1\le x<\frac{3}{2},f\left(x\right)>0 \text{and in }\frac{3}{2}<x\le 4\text{,}$

 $f\left(x\right)<0 \text{As }f\left(x\right)$  is continuous at $x=\frac{3}{2} \text{we find }f\left(\frac{3}{2}\right)$  is maximum value.

The minimum value = the least among  $\left\{f\right(-1),f(4\left)\right\}$