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 $\displaystyle (x,y)$ from $\displaystyle (1,0)$ is:
$\displaystyle d^2=(x-1)^2+y^2$
Now substitute from $\displaystyle y=2x$ for either $\displaystyle x$ or $\displaystyle y$ then differentiate and set the derivative to zero, and solve.
CB
A general point on the graph of $\displaystyle f(x)$ is of the form $\displaystyle (x,f(x))$ , so the distance between such a point and (1,0) is;
$\displaystyle \sqrt{(x-1)^2+(f(x)-0)^2}=\sqrt{x^2-2x+1+4x^2}$ . Well, do some order here and check this is a function of x so to find is extreme points you have to...
Tonio