# Thread: Optimization

1. ## Optimization

Find the points on the graph f(x)=2x that is closest to the point (1,0) .

How would you go about this?

2. Originally Posted by av8or91
Find the points on the graph f(x)=2x that is closest to the point (1,0) .

How would you go about this?
The square distance of an arbitary point $(x,y)$ from $(1,0)$ is:

$d^2=(x-1)^2+y^2$

Now substitute from $y=2x$ for either $x$ or $y$ then differentiate and set the derivative to zero, and solve.

CB

3. Originally Posted by av8or91
Find the points on the graph f(x)=2x that is closest to the point (1,0) .

How would you go about this?

A general point on the graph of $f(x)$ is of the form $(x,f(x))$ , so the distance between such a point and (1,0) is;

$\sqrt{(x-1)^2+(f(x)-0)^2}=\sqrt{x^2-2x+1+4x^2}$ . Well, do some order here and check this is a function of x so to find is extreme points you have to...

Tonio