We are asked to find the point(s) closest to the point (0,-5) of the graph of `f(x)=x^2-7` .

We need to find a function that gives the distance from the point (0,-5) to a generic point on the graph of f(x). The distance between two points can be found using...

## Unlock

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

We are asked to find the point(s) closest to the point (0,-5) of the graph of `f(x)=x^2-7` .

We need to find a function that gives the distance from the point (0,-5) to a generic point on the graph of f(x). The distance between two points can be found using the distance formula (essentially the Pythagorean theorem) `d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)` .

One of the points is the given point (0,-5). The other point, (x,y), lies on the graph of f(x) so it can also be described as `(x,x^2-7)` .

Thus `d=sqrt((x-0)^2+((x^2-7)-(-5))^2)` or `d=sqrt(x^2+(x^2-2)^2)`

To find the minimum of this function, d(x), we take the first derivative with respect to x. The zeros of the derivative are critical points, and we can determine which of the critical points are minima.

`(dd)/(dx)=1/2(x^2+(x^2-2)^2)^(-1/2)(2x+2(x^2-2)(2x))`

`=(4x^3-6x)/sqrt(x^2+(x^2-2)^2)`

A fraction is zero if the numerator is zero, while the denominator is nonzero:

`(dd)/(dx)=0 => 4x^3-6x=0 => 2x(x^2-3)=0 => x=pm sqrt(3/2),0`

Using the first derivative test, we find that `x=-sqrt(3/2) " and " sqrt(3/2)` are minimums and x=0 is a local maximum.

(See the attachment for the graph of f(x) and d(x) (orange.))

Substituting the values for x at the minimums yields the points `(-sqrt(3/2),-11/2),(sqrt(3/2),-11/2)`

The point (0-7) is the furthest point on the graph in the interval `(-sqrt(3/2),sqrt(3/2))` .

Hope this helps.

**Further Reading**