# Question dbd28

Jan 14, 2016

Define the distance between the graph and the point as a function and find the minimum.

The point is $\left(3.5 , 1.871\right)$

#### Explanation:

To know how close they are, you need to know the distance. The Euclidean distance is:

sqrt(Δx^2+Δy^2)#

where Δx and Δy are the differences between the 2 points. In order to be the nearest point, that point has to have the minimum distance. Therefore, we set:

$f \left(x\right) = \sqrt{{\left(x - 4\right)}^{2} + {\left({x}^{\frac{1}{2}} - 0\right)}^{2}}$

$f \left(x\right) = \sqrt{{x}^{2} - 8 x + 16 + {\left({x}^{\frac{1}{2}}\right)}^{2}}$

$f \left(x\right) = \sqrt{{x}^{2} - 8 x + 16 + {x}^{\frac{1}{2} \cdot 2}}$

$f \left(x\right) = \sqrt{{x}^{2} - 8 x + 16 + x}$

$f \left(x\right) = \sqrt{{x}^{2} - 7 x + 16}$

We now need to find the minimum of this function:

$f ' \left(x\right) = \frac{1}{2 \cdot \sqrt{{x}^{2} - 7 x + 16}} \cdot \left({x}^{2} - 7 x + 16\right) '$

$f ' \left(x\right) = \frac{2 x - 7}{2 \cdot \sqrt{{x}^{2} - 7 x + 16}}$

The denominator is always positive as a square root function. The numerator is positive when:

$2 x - 7 > 0$

$x > \frac{7}{2}$

$x > 3.5$

So the function is positive when $x > 3.5$. Similarly, it can be proved that it is negative when $x < 3.5$ Therefore, there function $f \left(x\right)$ has a minimum at $x = 3.5$, which means the distance is the least at $x = 3.5$ The y coordinate of $y = {x}^{\frac{1}{2}}$ is:

$y = {3.5}^{\frac{1}{2}} = \sqrt{3.5} = 1.871$

Finally, the point where the least distance from (4,0) is observed is:

$\left(3.5 , 1.871\right)$