Question 1083811
(a)  |z-2|+|z-3|
<pre>
We consider 4 potential cases

 Case 1:  both expressions within absolute value bars 
are non-negative:

 {{{matrix(2,3,

z-2>=0, and, z-3>=0,

z>=2,   and,    z>=3)}}} 

That's the same as {{{z>=3}}}

So |z-2|+|z-3| becomes  

(z-2)+(z-3) = z-2+z-3 = 2z-5

We build up 2z-5 from {{{z>=3}}}

{{{z>=3}}}
{{{2*z>=2*3}}}
{{{2z>=6}}}
{{{2z-5>=6-5}}}
{{{2z-5>=1}}}

So the minimum value is 1
and there is no maximum value.

Case 2:  the first expressions within absolute value bars 
is non-negative and the second expression within absolute
value bars is non-positive

{{{matrix(2,3,
z-2>=0, and, z-3<=0,
z>=2,   and,    z<=3)}}}

That's the same as {{{2<=z<=3}}}.

 So  |z-2|+|z-3| becomes 

(z-2)-(z-3) = z-2-z+3 = 1

So the minimum (and maximum) value for case 2 is 1.

Case 3:  the first expressions within absolute value bars 
is non-positive and the second expression within absolute
value bars is non-negative:

{{{matrix(2,3,
z-2<=0, and, z-3>=0,
z<=2,   and,    z>=3)}}}

That is impossible, so we rule out case 3

Case 4:  both expressions within absolute value bars 
are non-positive:

{{{matrix(2,3,

z-2<=0, and, z-3<=0,
z<=2,   and,    z<=3)}}} 

That's the same as {{{z<=2}}}

So  |z-2|+|z-3| becomes  

 -(z-2)-(z-3) = -z+2-z+3 = -2z+5

We build up -2z+5 from {{{z>=3}}}

{{{z<=2}}}
{{{-2z<=-2*2}}}  <--inequality reverses
{{{-2z<=-4}}}
{{{-2z+5<=-4+5}}}
{{{-2z+5>=1}}}

So for case 4, the minimum value is 1 and there is no maximum value.

Therefore for all the cases the minimum value is 1
and there is no maximum value. 

--------------------------------------
</pre>
(b) |2z-1|+|3z-2|
<pre> 
Case 1:  both expressions within absolute value bars 
are non-negative:

 

{{{matrix(3,3,

 

2z-1>=0, and, 3z-2>=0,
2z>=1,   and,    3z>=2,
z>=1/2,  and,     z>=2/3

)}}} 

That's the same as {{{z>=2/3}}}

So |2z-1|+|3z-2| becomes  

(2z-1)+(3z-2) = 2z-1+3z-2 = 5z-3

We build up 5z-3 from {{{z>=2/3}}}

{{{z>=2/3}}}
{{{5*z>=5*expr(2/3)}}}
{{{5z>=10/3}}}
{{{5z-3>=10/3-3}}}
{{{5z-3>=10/3-3/3}}}
{{{5z-3>=7/3}}}

That has a minimum value of 7/3
and no maximum value.
 
Case 2:  the first expressions within absolute value bars 
is non-negative and the second expression within absolute
value bars is non-positive

{{{matrix(2,3,

2z-1>=0, and, 3z-2<=0,
2z>=1,   and,   3z<=2)}}}
z>=1/2,  and     x<=2/3)}}}

 
That's the same as {{{1/2<=z<=2/3}}}.

So  |2z-1|+|3z-2| becomes 
    (2z-1)-(3z-2) = 2z-1-3z+2 = -z+1 
 
We build up -z+1 from {{{1/2<=z<=2/3}}}.

{{{1/2<=z<=2/3}}} 
{{{-1/2>=-z>=-2/3}}}   <---the inequality symbol reverses 
{{{-1/2+1>=-z+1>=-2/3+1}}}
{{{-1/2+2/2>=-z+2>=-2/3+3/3}}}
{{{1/2>=-z+2>=1/3}}}
So the minimum value for case 2 is 1/3
And the maximum value for case 2 is 1/2

Case 3:  the first expressions within absolute value bars 
is non-positive and the second expression within absolute
value bars is non-negative:

 {{{matrix(2,3,

2z-1<=0, and, 3z-2>=0,
2z<=1,   and,   3z>=2)}}}
z<=1/2,  and,    z>=2/3}}}
 
That is impossible, so we rule out case 3.

Case 4:  both expressions within absolute value bars 
are non-positive:

{{{matrix(3,3,

2z-1<=0, and, 3z-2<=0,
2z<=1,   and,    3z<=2,
z<=1/2,  and,     z<=2/3

)}}} 

That's the same as {{{z<=1/2}}}

So |2z-1|+|3z-2| becomes  

-(2z-1)-(3z-2) = -2z+1-3z+2 = -5z+3

We build up -5z+3 from {{{z<=1/2}}}

{{{z<=1/2}}}
{{{-5*z>=-5*expr(1/2)}}}   <--inequality symbol reverses
{{{-5z>=-5/2}}}
{{{-5z+3>=-5/2+3}}}
{{{-5z+3>=-5/2+6/2}}}
{{{-5z+3>=1/2}}}

That has a minimum value of 1/2
and no maximum value.

The minimum value from all the cases is 1/3 and
there is no maximum value.

Edwin</pre>