Closest Pt of y = 1/x to Origin

**xwrathbringerx** · April 14th 2009, 08:54 PM

Hey guys

Find the coordinates of the point on the graph of y = 1/x (x>0) which is the closest to the origin.

Looking at a graph, the pt would b (1,1) i fink but i have not clue how i would prove it...i tried finding the turning pts like the previous problems but y = 1/x doesnt have any =p

Any hints please?

**Chris L T521** · April 14th 2009, 09:01 PM

Originally Posted by xwrathbringerx

Hey guys

Find the coordinates of the point on the graph of y = 1/x (x>0) which is the closest to the origin.

Looking at a graph, the pt would b (1,1) i fink but i have not clue how i would prove it...i tried finding the turning pts like the previous problems but y = 1/x doesnt have any =p

Any hints please?

Treat this as an optimization problem.

Let $D\left(x\right)=\sqrt{(x)^2+\left(\tfrac{1}{x}\rig ht)^2}$ (by distance formula from $(0,0)$ to $\left(x,\tfrac{1}{x}\right)$ )

To minimize the distance, we see that $D^{\prime}\!\left(x\right)=\tfrac{1}{2}\left(x^2+\ tfrac{1}{x^2}\right)^{-\frac{1}{2}}\cdot \left(2x-\frac{2}{x^3}\right)=\frac{x-\frac{1}{x^3}}{\sqrt{x^2+\frac{1}{x^2}}}$

Thus, $D^{\prime}\!\left(x\right)=0$ when $x-\frac{1}{x^3}=0\implies x^4=1\implies x=\pm1$ .

Since $x>0$ , we disregard the negative value. Thus, the x coordinate of the point that is closest to the origin is $x=1$ . As a result, the y coordinate is $y=\frac{1}{1}=1$ . Therefore, the point on $y=\frac{1}{x}$ that is closest to the origin is $\boxed{\left(1,1\right)}$ .

Does this make sense?

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