Question

Let, ( f(x)=(|x-1|-3)(|x+2|-5) )
( f(x)=left{begin{array}{ll}{[-(x-1)-3][-(x+2)-5]} & , x<-2 {[-(x-1)-3](x+2-5)} & ,-2 leq x<1 (x-1)-30[x+2-5] & x geq 1end{array}right. )
( f(x)=left{begin{array}{ll}(-x-2)(-x-1), & x<-2 (-x-2)(x-3) & -2 leq x<1 (x-4)(x-3), & x geq 1end{array}right. )
( F(x)=left{begin{array}{ll}x leqslant 4 & 0, x+F 4 (x+2)(x+7) & , x<-2 -(x+2)(x-3) & -2 leq x<1 (x-4)(x-3) & x geq 1end{array}right. )
A) ( f(x-2 leq x<1,-(x+2)(x-3)<0 )
( frac{partial}{2} sum_{3} x in(-infty,-2) cup(3,9) )
but for ( x )
( (x-3)<1 )
( x in(3,4) )
( x )
( x(-7,-2) cup(3,4) ) (ive your
terd back
q 3,0610 96 78

# (vii) a (1x – 11 – 3) (x + 2| - 5) < 0

Solution