3. Find the point P on the graph of closest to the point
$\displaystyle d^2(x)'=2(x-10)+1$

$\displaystyle \frac{-1}{2(x-10)+1} = (y-0)(x-10)$

$\displaystyle \Big\Downarrow$

$\displaystyle \frac{\frac{-1}{2(x-10)+1}}{y}=x-10$

$\displaystyle \Big\Downarrow$

$\displaystyle \frac{-1}{(2(x-10)+1)y}=x-10$

$\displaystyle \Big\Downarrow$

$\displaystyle \frac{-1}{(2(x-10)+1)\sqrt{x}}-x+10=0$

$\displaystyle \Big\Downarrow$

$\displaystyle -1-x+10=0$

$\displaystyle \Big\Downarrow$

$\displaystyle x=9$

$\displaystyle \Big\Downarrow$

$\displaystyle y=\sqrt{x}=\sqrt{9}=3$

so the point on the graph of $\displaystyle y=\sqrt{x}$ closest to the point (10,0) is (9,3)

Please let me know if this is right and if not where I have messed up at. Thank you