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
ErmWhat
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.

Related questions
Impact of this question
47783 views around the world
You can reuse this answer
Creative Commons License