Find the points on the graph of the function that are closest to the given point.

Question

$$f(x)=x^2\ ,\ (0,4)$$

I have attempted to plug this in and found the derivative but can not find the answer.

Any help would be appreciated!

Answer 1

Hint: This is just a reframed optimization problem. What you're looking to optimize–specifically minimize, in this case, is distance. Remember that the distance formula is $\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$ where one pair of $x$s and $y$s refers to the given point $(0, 4)$, and the other pair, some $x$ and $x^2$, refering to the point $(x, x^2)$ on the function:

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

Solve for $f'(x)=0$ and find the minimum(s). These will give you the $x$-coordinates of the closest ordered-pairs.

Answer 2

Distance formula!

Let's take a point on $f(x)$, that's $(x,x^2)$.

The distance formula tells us that the closest distance is given by $\sqrt{(x-0)^2+(x^2-4)^2}$.

That simplifies to $\sqrt{x^2+x^4-8x^2+16}=\sqrt{x^4-7x^2+16}=g(x)$.

So for any point $(x,f(x))$, our distance is $g(x)$.

You mentioned a derivative. Let's do that. The derivative should be zero when we get an extrema (minimum/maximum). We have to have a minimum because the function grows infinitely far from the point.

$g'(x) = \displaystyle \frac{4x^3-14x}{2\sqrt{x^4-7x^2+16}}$.

As we can see, $g'(x)=0$ when $4x^3-14x=0$, that's when $x^2=\frac{7}{2}$.

The closest points are thus $\displaystyle\left(\pm \frac{\sqrt{14}}{2},\frac{7}{2}\right)$

Answer 3

Anytime that you want to minimize the distance, just minimize the distance squared. The result will be the same.

In this case, the distance squared from $(0, 4)$ to $(x, x^2)$ is

$\begin{array}\\ D &=x^2+(x^2-4)^2\\ &=x^2+x^4-8x^2+16\\ &=x^4-7x^2+16\\ D' &=4x^3-14x\\ &=0 \qquad\text{when}\\ 0 &=4x^3-14x\\ \text{so}\\ x &= 0\\ \text{or}\\ 4x^2 &= 14\\ \text{so}\\ x &= \pm\sqrt{\frac{7}{2}}\\ \end{array} $

Now we have to check.

If $x = 0$ then $D = 16 $.

If $x =\pm\sqrt{\frac{7}{2}} $ then $x^2-4 =\frac{7}{2}-4 =-\frac14 $.

$\begin{array}\\ x^2 &= \frac{7}{2}\\ x^4 &= \frac{49}{4}\\ D &=x^4-7x^2+16\\ &=\frac{49}{4}-7\frac{7}{2}+16\\ &=-\frac{49}{4}+16\\ &=\frac{64-49}{4}\\ &=\frac{15}{4}\\ &\lt 16\\ \end{array} $

so the closest points are $(\pm\sqrt{\frac{7}{2}}, -\frac14)$.

Answer 4

The basic idea is the line joining the point $(0,4)$ will be normal to the curve. The reason is perpendicular is the shortest line joining two points.

Slope of any point $(h,h^2)$ lying on the curve is $2h$. Hence the slope of the normal is $\frac{-1}{2h}$. So the perpendicular has the given slope and passes through $(0,4)$. The equation of the normal is $2h(y-4)+x=0$. This line also passes through $(h,h^2)$.

Hence, $2h^3-8h+h=0$. Solve for $h$ and you have the answer.