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

Staff Emeritus
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

chapsticks

Yay thanks

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