Question 703463
<pre>
Let the desired point on the x axis be C(x,0)

{{{drawing(400,400,-2,8,-2,8,

red(line(-2,0,8,0),line(0,-2,0,8)),triangle(1,6,3,0,6,1),
locate(2.5,0,"C(x,0)"),
locate(0.1,8,y),locate(8,0,x),locate(1,6.5,"A(1,6)"),locate(6.2,1.5,"B(6,1)") )}}}

Using the distance formula:

D = AC+BC

D = {{{sqrt((x-1)^2+(6-0)^2)+sqrt((x-6)^2+(1-0)^2)}}}

D = {{{sqrt((x-1)^2+36)+sqrt((x-6)^2+1)}}}

We need to find {{{dD/dx}}}

Both those terms on the right are of the form 

u = {{{sqrt(v^2+k)}}}

Differentiate that and you get

{{{du/dx}}} = {{{expr(v/sqrt(v^2+k))*expr(dv/dx)}}}

So, using that on both terms: 

{{{dD/dx}}} = {{{(x-1)/sqrt((x-1)^2+36)}}} + {{{(x-6)/sqrt((x-6)^2+1)}}}

{{{dD/dx}}} = {{{(x-1)/sqrt(x^2-2x+1+36)}}} + {{{(x-6)/sqrt(x^2-12x+36+1)}}}

{{{dD/dx}}} = {{{(x-1)/sqrt(x^2-2x+37)}}} + {{{(x-6)/sqrt(x^2-12x+37)}}}

We set that = 0 to find minimum value:

{{{(x-1)/sqrt(x^2-2x+37)}}} + {{{(x-6)/sqrt(x^2-12x+37)}}} = 0

Clearing of fractions:

{{{(x-1)sqrt(x^2-12x+37)}}} + {{{(x-6)sqrt(x^2-2x+37)}}} = 0

{{{(x-1)sqrt(x^2-12x+37)}}} = {{{-(x-6)sqrt(x^2-2x+37)}}}

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:

x<sup>4</sup>-14x³+62x²-86x+37 =  x<sup>4</sup>-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={{{37/7}}} and x=7.

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

Thus x={{{37/7}}} is the only solution to {{{dD/dx}}} = 0 

x={{{37/7}}} is between 1 and 6. It's easy to show that 
{{{dD/dx}}} is negative when x=1 and positive when x=6, 
thus the desired point is C({{{37/7}}},0).

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

{{{drawing(400,400,-2,8,-2,8, graph(400,400,-2,8,-2,8),triangle(1,6,37/7,0,6,1),
locate(5,-.3,C(37/7,0)),locate(1,6.5,"A(1,6)"),locate(6.2,1.5,"B(6,1)") )}}}

Edwin</pre>