Differentiate the square of the distance

[chapsticks]

Homework Statement

Find the point on the graph of f(x)=√x that is closest to the value (4,0).

(differentiate the square of the distance from a point (x,√x) on the graph of f to the point (4,0).)

Homework Equations

f(x)=√x

The Attempt at a Solution

Square of distance from (4,0)
D(x)= ((x-4)^2+x)
For minimum, D'(x)=0
2(x-4)+1=0
x=7/2

 

[Mark44]

Homework Statement

Find the point on the graph of f(x)=√x that is closest to the value (4,0).

(differentiate the square of the distance from a point (x,√x) on the graph of f to the point (4,0).)

Homework Equations

f(x)=√x

The Attempt at a Solution

Square of distance from (4,0)
D(x)= ((x-4)^2+x)

D(x) doesn't represent the distance - it's the square of the distance. As it turns out, though, mininimizing the distance is equivalent to minimizing the square of the distance.

For minimum, D'(x)=0
2(x-4)+1=0
x=7/2

This is the correct value of x, but you haven't answered the question, which is to find the point on the graph of f(x)=√x that is closest to the [STRIKE]value[/STRIKE] point (4,0).

 

[chapsticks]

I don't get it?

 

[HallsofIvy]

Mark44 told you you have the right x value. Now what point is that on the graph?

 

[chapsticks]

Is it (7/2,(√7/2))?

 

[Mark44]

It depends on what you mean by (√7/2).

This is $\sqrt{7}/2$.

 

[chapsticks]

Oh I mean √7/√2

 

[Mark44]

Which is the same as √(7/2). So, yes, (7/2, √(7/2)) is the right point on the graph of f(x) = √x.

 

[chapsticks]

Yay thanks