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Thread: Closest Pt of y = 1/x to Origin

  1. #1
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    Exclamation Closest Pt of y = 1/x to Origin

    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?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by xwrathbringerx View Post
    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 $\displaystyle D\left(x\right)=\sqrt{(x)^2+\left(\tfrac{1}{x}\rig ht)^2}$ (by distance formula from $\displaystyle (0,0)$ to $\displaystyle \left(x,\tfrac{1}{x}\right)$)

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

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

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

    Does this make sense?
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