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.

1 Answer
Feb 20, 2018

(sqrt(7)/sqrt(2),7/2)

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(x) 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^4-8x^2+16)
sqrt(x^4-7x^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:
d/dxf(g(x)) = f'(g(x))*g'(x), where f(x)=sqrt(x),g(x)=(x^4-7x^2+16),f'(x)=1/(2sqrt(x)) and g'(x)=(4x^3-14x). (g'(x) can be found using the sum rule and then the power rule.) Plugging it all back in...

(4x^3-14x)/(2sqrt(x^4-7x^2+16))=(2x^3-7x)/(sqrt(x^4-7x^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...

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

2x^2-7=0 -> (sqrt(2)x-sqrt(7))(sqrt(2)x+sqrt(7))=0. The root will then be x=+-sqrt(7)/sqrt(2). Plugging x=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 (0,4) will be:

(sqrt(7)/sqrt(2),7/2)

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