# Homework Help: Differentiate the square of the distance

1. Dec 6, 2011

### chapsticks

1. The problem statement, all variables and given/known data

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).)

2. Relevant equations

f(x)=√x
3. 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

Last edited by a moderator: Dec 6, 2011
2. Dec 7, 2011

### Staff: Mentor

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.
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).

3. Dec 7, 2011

### chapsticks

I don't get it?

4. Dec 7, 2011

### HallsofIvy

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

5. Dec 7, 2011

### chapsticks

Is it (7/2,(√7/2))?

6. Dec 7, 2011

### Staff: Mentor

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

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

7. Dec 7, 2011

### chapsticks

Oh I mean √7/√2

8. Dec 7, 2011

### Staff: Mentor

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

9. Dec 7, 2011

Yay thanks