The discussion focuses on finding the point on the graph of the function f(x) = √x that is closest to the point (4,0). The square of the distance D(x) is defined as D(x) = ((x-4)^2 + x). To find the minimum distance, the derivative D'(x) is set to zero, leading to the solution x = 7/2. The corresponding point on the graph is (7/2, √(7/2)), confirming that minimizing the square of the distance effectively minimizes the actual distance.
PREREQUISITESStudents studying calculus, particularly those focusing on optimization problems, as well as educators looking for examples of applying derivatives to real-world scenarios.
D(x) doesn't represent the distance - it's the square of the distance. As it turns out, though, mininimizing the distance is equivalent to minimizing the square of the distance.chapsticks said:Homework Statement
Find the point on the graph of f(x)=√x that is closest to the value (4,0).
(differentiate the square of the distance from a point (x,√x) on the graph of f to the point (4,0).)
Homework Equations
f(x)=√x
The Attempt at a Solution
Square of distance from (4,0)
D(x)= ((x-4)^2+x)
This is the correct value of x, but you haven't answered the question, which is to find the point on the graph of f(x)=√x that is closest to the [STRIKE]value[/STRIKE] point (4,0).chapsticks said:For minimum, D'(x)=0
2(x-4)+1=0
x=7/2