# How do I find the point on the graph f(x)=sqrt(x) closest to the point (4,0)? Please show the work

##### 1 Answer
Mar 13, 2016

$\left(\frac{7}{2} , 1.87\right)$

#### Explanation:

What we are being asked here is to simply minimize distance. Also, note that we can write $f \left(x\right) = \sqrt{x}$ as $y = \sqrt{x}$.

Now, what is this "distance?" How do we find it? Well, if you think back to Algebra I or Geometry, you'll remember that the distance between two points $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$ is given by: $\sqrt{{\left({y}_{2} - {y}_{1}\right)}^{2} + {\left({x}_{2} - {x}_{1}\right)}^{2}}$. For example, the distance between the points $\left(4 , 0\right)$ and $\left(0 , 3\right)$ would be:
$\sqrt{{\left(3 - 0\right)}^{2} + {\left(4 - 0\right)}^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5$

Ok, so what is $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$ in our example? $\left({x}_{1} , {y}_{1}\right)$ is simple - it's just the point given in the problem, $\left(4 , 0\right)$. Because we don't know what ${x}_{2}$ is, we'll just call it $x$ for now. As for ${y}_{2}$, we don't know that either; and since $y = \sqrt{x}$, we'll call it $\sqrt{x}$.

Our formula then becomes:
$\sqrt{{\left(\sqrt{x} - 0\right)}^{2} + {\left(x - 4\right)}^{2}} = \sqrt{\left({\sqrt{x}}^{2}\right) + {x}^{2} - 8 x + 16} = \sqrt{x + {x}^{2} - 8 x + 16} = \sqrt{{x}^{2} - 7 x + 16}$

We are being asked to minimize this distance, which we'll call $s$ to make the following calculations easier. To minimize something, we have to take its derivative, so let's start there:
$s = \sqrt{{x}^{2} - 7 x + 16} = {\left({x}^{2} - 7 x + 16\right)}^{\frac{1}{2}}$
$\frac{\mathrm{ds}}{\mathrm{dx}} = \left(2 x - 7\right) \cdot \frac{1}{2 {\left({x}^{2} - 7 x + 16\right)}^{\frac{1}{2}}} \to$Using power rule and chain rule
(ds)/dx=(2x-7)/(2sqrt(x^2-7x+16)

Now we set this equal to $0$ and solve for $x$:
0=(2x-7)/(2sqrt(x^2-7x+16)

$0 = 2 x - 7$

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

This is known as the critical value, and it represents the $x$-value for which the function is minimized. All we need to do now is find the corresponding $y$-value, using the definition of $y$: $y = \sqrt{x}$. Substituing $\frac{7}{2}$ for $x$:
$y = \sqrt{\frac{7}{2}}$
$y \approx 1.87$

And voila, the $y$-value. We can now say that the minimum distance between $f \left(x\right) = \sqrt{x}$ and the point $\left(4 , 0\right)$ (the place where these two are closest) occurs at $\left(\frac{7}{2} , 1.87\right)$. For a little extra fun, we can use the distance formula to see what the actual distance between the points is:
$s = \sqrt{{\left(1.87 - 0\right)}^{2} + {\left(\frac{7}{2} - 4\right)}^{2}} \approx 1.8$ units