Let the desired point on the x axis be C(x,0)



Using the distance formula:

D = AC+BC

D = 

D = 

We need to find 

Both those terms on the right are of the form 

u = 

Differentiate that and you get

 = 

So, using that on both terms: 

 =  + 

 =  + 

 =  + 

We set that = 0 to find minimum value:

 +  = 0

Clearing of fractions:

 +  = 0

 = 

Square both sides

(x-1)²(x²-12x+37) = (x-6)²(x²-2x+37)

(x²-2x+1)(x²-12x+37) = (x²-12x+36)(x²-2x+37)

Multiply that out and combine terms and get:

x4-14x³+62x²-86x+37 =  x4-14x³+97x²-516x+1332

That simplifies to:

35x²-430x+1295 = 0

Divide thru by 5: 

7x²-86x+259 = 0

That factors as

(7x-37)(x-7)=0

That has solutions x= and x=7.

We can show that x=7 is extraneous, since both terms of
 are positive wnen x=7, so it isn't 0 there.

Thus x= is the only solution to  = 0 

x= is between 1 and 6. It's easy to show that 
 is negative when x=1 and positive when x=6, 
thus the desired point is C(,0).

The graph showing the minimum of AC+BC drawn to scale is:



Edwin