# Find the points on the graph of the function that are closest to the given point. f(x) = x^2, (0, 4)?

## Both smaller x and larger x. I have attempted to plug this in to the and found the derivative but can not find the answer.

Feb 20, 2018

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

#### Explanation:

The easiest way to solve it is to use the distance formula. Simply put, it is the pythagorean theorem but with any two points.

sqrt((x_2-x_1)^2+(y_2-y_1)^2

Given that $y$ (or in this case, $f \left(x\right)$ is equal to ${x}^{2}$, we can substitute it in for ${y}_{2}$. ${x}_{2}$ is just $x$, ${x}_{1}$ is the x-value of the point and ${y}_{1}$ is the y-value of the point.

sqrt((x-0)^2+(x^2-4)^2

To make the derivation simpler, expand it.
$\sqrt{{x}^{2} + {x}^{4} - 8 {x}^{2} + 16}$
$\sqrt{{x}^{4} - 7 {x}^{2} + 16}$

In this case, we solve for when this number is at the minimal point. The easiest way to do it is to integrate. Using chain rule, which is:
$\frac{d}{\mathrm{dx}} f \left(g \left(x\right)\right) = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$, where $f \left(x\right) = \sqrt{x}$,$g \left(x\right) = \left({x}^{4} - 7 {x}^{2} + 16\right)$,$f ' \left(x\right) = \frac{1}{2 \sqrt{x}}$ and $g ' \left(x\right) = \left(4 {x}^{3} - 14 x\right)$. ($g ' \left(x\right)$ can be found using the sum rule and then the power rule.) Plugging it all back in...

$\frac{4 {x}^{3} - 14 x}{2 \sqrt{{x}^{4} - 7 {x}^{2} + 16}} = \frac{2 {x}^{3} - 7 x}{\sqrt{{x}^{4} - 7 {x}^{2} + 16}}$. Now, we have to find when this equation is 0. It does not really matter what the denominator is, all that really matters is the numerator. So...

$\left(2 {x}^{3} - 7 x\right) = 0$, or $x \left(2 {x}^{2} - 7\right) = 0$. Looking at the graph itself, x≠0, so that can be ruled out as extraneous. So, all that is left is:

$2 {x}^{2} - 7 = 0 \to \left(\sqrt{2} x - \sqrt{7}\right) \left(\sqrt{2} x + \sqrt{7}\right) = 0$. The root will then be $x = \pm \frac{\sqrt{7}}{\sqrt{2}}$. Plugging $x = \frac{\sqrt{7}}{\sqrt{2}}$ this back into the equation $y = {x}^{2}$, outputs a value of $y = 3.5$ (since x is just $\sqrt{3.5}$). The points closest to $\left(0 , 4\right)$ will be:

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

Note this could be wrong since I have not done calculus in a long time.